Differentiation, integration techniques, and series — the three pillars every differential equations course leans on from day one. Each module pairs a short explanation with a live diagram you can drag and slide, a worked example, and problems that hand you hints one at a time instead of the answer.
23 modules24 interactive labs207 practice problems3 checkpoint quizzesStaged hints · fully worked solutions
Part I · The derivative
Differentiation
Part II · The integral
Integration
Part III · Infinite sums
Series
↦How to use this
Play with the diagram before you read the algebra. Every module has an orange Interactive panel. Drag the point, slide the parameter, watch what breaks. The formula makes far more sense once you've seen the picture move.
Attempt every problem on paper first. Hints appear one at a time — take Hint 1, try again, and only take Hint 2 if you're still stuck. Every solution is then worked step by step, with the reasoning behind each move, not just the answer.
Tick "Mark complete" at the bottom of each module. Progress is saved in this browser.
Take the quiz at the end of each part. Under 8/10 means go back to the modules you missed — those exact gaps will bite you in the ODE course.
Why these three topics, specifically
Solving an ODE is almost always differentiate to check, integrate to solve. Separable equations are pure u-substitution; integrating factors are the product rule read backwards; partial fractions are how every Laplace transform gets inverted; and when no elementary solution exists at all, you fall back on a power series solution — which is Part III doing all the work.
Once this feels solid, the next step is the ODE Launchpad — a three-week course on the differential equations themselves.
Part I · Module 1 Differentiation
The derivative as a rate of change
A derivative is not a formula you memorize — it is the answer to one question: if I nudge the input, how fast does the output respond? Everything else in this toolkit is bookkeeping on top of that idea.
①Average rate → instantaneous rate
Pick two points on a curve and connect them. The slope of that secant line is the average rate of change between them:
Two readings of the same object, and you need both:
Geometric: \(f'(a)\) is the slope of the tangent line at \(x=a\).
Physical: \(f'(a)\) is the instantaneous rate at which \(f\) changes per unit of \(x\) — with units of output units per input unit.
InteractiveWatch the secant collapse into the tangentshrink \(h\) toward 0
secant slope —f′(a) —error —
The cyan line is the true tangent; the amber line is the secant through \((a,f(a))\) and \((a+h,f(a+h))\). Notice the error shrinks roughly proportionally to \(h\) — halve \(h\), halve the error. That is exactly what "limit" means here.
②Reading a derivative's sign
If…
then the graph…
and the quantity…
f′ > 0
rises
is increasing
f′ < 0
falls
is decreasing
f′ = 0
is momentarily flat
is at a max, min, or plateau
f″ > 0
curves upward (concave up)
is increasing faster and faster
f″ < 0
curves downward
is leveling off
Worked example · derivative straight from the definition
Find \(f'(a)\) for \(f(x) = x^2\) using the limit definition.
STEP 1
Write the difference quotient: \(\dfrac{(a+h)^2 - a^2}{h}\).
Factor and cancel the \(h\) — this cancellation is the whole trick: \(\dfrac{h(2a+h)}{h} = 2a + h\).
STEP 4
Now the limit is safe to take: \(\displaystyle\lim_{h\to0}(2a+h) = 2a\). So \(f'(x) = 2x\).
You cannot plug in h = 0 first
Before you cancel, the quotient is \(0/0\) — meaningless. The algebra step that removes the \(h\) from the denominator is not optional bookkeeping, it is the derivative computation.
Where this shows up in ODEs
A differential equation is a sentence about rates: \(\frac{dy}{dt} = 0.3y\) literally says "this population grows at 30% of its current size per year." Before you can solve one, you have to be fluent at reading \(\frac{dy}{dt}\) as a rate, not as a symbol.
③Practice
1
Use the limit definition to find \(f'(x)\) for \(f(x) = 3x^2 - 5x + 1\).
Hint 1Write out \(f(x+h)\) completely before subtracting: \(3(x+h)^2 - 5(x+h) + 1\).
Hint 2Every term without an \(h\) in it must cancel against \(-f(x)\). If something constant survives, you made an expansion error.
Hint 3You should reach \(\dfrac{6xh + 3h^2 - 5h}{h}\). Factor \(h\) out of the top.
Solution
STEP 1 · set up the definition
The derivative is the limit of the difference quotient, so write it down before doing any algebra:
\[ f'(x) = \lim_{h\to0}\frac{f(x+h)-f(x)}{h} \]
STEP 2 · build f(x+h)
Replace every \(x\) in the formula by \((x+h)\) — including inside the \(-5x\) term. Then expand:
Now \(h=0\) can be substituted safely, and the only surviving \(h\) term vanishes:
\[ f'(x) = \lim_{h\to0}(6x + 3h - 5) = 6x - 5 \]
CHECK
The power rule gives the same thing in one line: \(3x^2 \to 6x\), \(-5x \to -5\), \(1 \to 0\). The definition is what justifies the power rule — but once justified, you never do this by hand again.
2
A tank drains so that the volume remaining after \(t\) minutes is \(V(t) = 200 - 8t\) liters. What is \(V'(t)\), and what are its units? What does the sign mean?
Hint 1The graph of \(V\) is a straight line. What is the slope of a straight line, and how does that relate to the derivative?
Hint 2Units of a derivative are always output units per input unit — here, liters per minute.
Solution
STEP 1 · differentiate term by term
\(V(t) = 200 - 8t\) is a sum of two pieces. The constant \(200\) has derivative \(0\) (a constant never changes), and \(-8t\) is \(-8\) times \(t^1\), whose derivative is \(1\):
\[ V'(t) = 0 - 8(1) = -8 \]
STEP 2 · sanity-check geometrically
The graph of \(V\) is a straight line with slope \(-8\). The tangent line to a straight line is that line, so the derivative must be the constant slope \(-8\) at every \(t\) — no calculation needed.
STEP 3 · attach the units
A derivative always carries output units per input unit. Volume is in liters, time in minutes, so
\[ V'(t) = -8 \ \text{liters per minute} \]
STEP 4 · interpret the sign
Negative means decreasing: the tank is losing 8 L every minute. And because \(V'\) is constant, the drain rate never changes — the tank empties steadily, not faster at the start.
STEP 5 · read it as an ODE
The statement "the tank drains at 8 L/min" is itself a differential equation, \(\frac{dV}{dt} = -8\). Integrating both sides gives \(V = -8t + C\), and the initial condition \(V(0) = 200\) forces \(C = 200\), recovering the original formula. That round trip — rate ⟶ integrate ⟶ initial condition ⟶ formula — is the whole shape of a first-order ODE problem.
3
Use the definition to show that \(\dfrac{d}{dx}\left[\dfrac{1}{x}\right] = -\dfrac{1}{x^2}\).
Hint 1The difference quotient is \(\dfrac{1}{h}\left(\dfrac{1}{x+h} - \dfrac{1}{x}\right)\). Combine the inner fractions over a common denominator first.
Hint 2\(\dfrac{1}{x+h} - \dfrac{1}{x} = \dfrac{x - (x+h)}{x(x+h)} = \dfrac{-h}{x(x+h)}\). Now the \(h\) you need to cancel is sitting in the numerator.
Solution
STEP 1 · write the difference quotient
With \(f(x) = 1/x\), the quotient is a fraction whose numerator is itself a difference of fractions:
Two ways to be confident. (1) The power rule on \(x^{-1}\) gives \((-1)x^{-2} = -1/x^2\) ✓. (2) The sign is right: \(1/x\) is falling on both of its branches, and \(-1/x^2\) is negative for every \(x \ne 0\) ✓.
4
In the lab above, set \(f(x)=\sin x\) and \(a = 0\). Before touching the sliders, predict the tangent slope. Then check it.
Hint 1Near the origin, \(\sin x\) hugs a very familiar straight line.
Solution
STEP 1 · differentiate
\(\frac{d}{dx}\sin x = \cos x\) (module D4 — worth having memorized now).
STEP 2 · evaluate at the point
Evaluate the derivative at \(a = 0\), not the function: \(f'(0) = \cos 0 = 1\). So the predicted tangent slope is exactly \(1\).
STEP 3 · write the tangent line
Point-slope form through \((0, \sin 0) = (0,0)\) with slope 1:
\[ y - 0 = 1\,(x - 0) \quad\Longrightarrow\quad y = x \]
STEP 4 · check it in the lab
Set \(a = 0\) and shrink \(h\). The secant slope \(\frac{\sin h - \sin 0}{h} = \frac{\sin h}{h}\) creeps toward \(1\) — at \(h = 0.1\) it is already \(0.9983\).
WHY IT MATTERS
That one tangent line is the source of two facts you will use constantly: the small-angle approximation \(\sin x \approx x\), and the limit \(\lim_{x\to0}\frac{\sin x}{x} = 1\) — which is literally the statement "the derivative of \(\sin\) at 0 equals 1."
5
Use the limit definition to find \(f'(x)\) for \(f(x) = \sqrt{x}\), \(x>0\). (The cancellation needs one extra trick.)
Hint 1You get \(\dfrac{\sqrt{x+h}-\sqrt{x}}{h}\), and there is no \(h\) to factor out yet. Roots block ordinary factoring.
Hint 2Multiply top and bottom by the conjugate \(\sqrt{x+h}+\sqrt{x}\). Remember \((A-B)(A+B) = A^2 - B^2\), and squaring kills the roots.
Hint 3After the conjugate trick the numerator becomes \((x+h) - x = h\).
Solution
STEP 1 · set up
\[ f'(x) = \lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h} \]
Substituting \(h=0\) gives \(0/0\), and nothing factors — so manipulate first.
STEP 2 · multiply by the conjugate
Multiplying by \(\dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\) is multiplying by 1, so it changes nothing but the form:
Power rule: \(x^{1/2} \to \tfrac12 x^{-1/2} = \frac{1}{2\sqrt x}\) ✓. Notice the derivative blows up as \(x\to0^+\) — matching the vertical tangent the graph of \(\sqrt x\) has at the origin.
6
A cup of coffee cools according to \(T(t)\) °C after \(t\) minutes. You are told \(T(10) = 65\) and \(T'(10) = -1.8\). (a) What does \(-1.8\) mean in plain words? (b) Estimate \(T(11)\) and \(T(10.5)\). (c) Why is your estimate slightly too low?
Hint 1Read \(T'\) as "degrees per minute" — a rate, at that instant only.
Hint 2For a short step, new value ≈ old value + rate × elapsed time.
Hint 3For part (c), think about whether a cooling curve is concave up or concave down — a hot cup cools fastest at the start.
Solution
STEP 1 · translate the derivative (a)
Units are °C per minute. At exactly \(t=10\) minutes the coffee is cooling at \(1.8\) °C each minute — the minus sign says cooling, not warming. It is an instantaneous rate: nothing guarantees it stays \(1.8\) later on.
STEP 2 · linear estimate for one minute (b)
Use the tangent line: \(T(10 + \Delta t) \approx T(10) + T'(10)\,\Delta t\). With \(\Delta t = 1\):
This estimate is more trustworthy than the one-minute one, because you extrapolated the tangent line for less time.
STEP 4 · why it under-predicts (c)
Newton's law of cooling says the rate is proportional to how far above room temperature the coffee is. As it cools, that gap shrinks, so the cooling slows down — meaning \(T\) is concave up (\(T'' > 0\)) and the tangent line sits below the true curve. Real \(T(11)\) is a bit warmer than \(63.2\).
ODE CONNECTION
That verbal law is exactly the ODE \(\frac{dT}{dt} = -k\,(T - T_{\text{room}})\), one of the first equations you will solve in the course. Here it would give \(-1.8 = -k(65 - T_{\text{room}})\), letting you solve for \(k\) from a single measurement.
7
For \(f(x) = x^3 - 3x\): find where the tangent is horizontal, classify each such point as a maximum or minimum, and state where \(f\) is increasing.
Hint 1Horizontal tangent means \(f'(x) = 0\). Compute \(f'\) first, then solve.
Hint 2\(f'(x) = 3x^2 - 3 = 3(x^2-1)\). Factor the difference of squares.
Hint 3To classify, use the sign of \(f''(x) = 6x\): concave up at a minimum, concave down at a maximum.
\(f''(x) = 6x\). At \(x=-1\), \(f''=-6<0\) — concave down, so \((-1,2)\) is a local maximum. At \(x=1\), \(f''=6>0\) — concave up, so \((1,-2)\) is a local minimum.
STEP 5 · answer
Horizontal tangents at \(x=\pm1\); increasing on \((-\infty,-1)\) and \((1,\infty)\); decreasing on \((-1,1)\). Load this function into the lab above and drag \(a\) through \(-1\) and \(1\) to watch the tangent flatten and flip.
8
For \(f(x)=x^2\) on \([1,3]\): compute the average rate of change, then find the point \(c\) where the instantaneous rate equals it. Sketch what you have just proved.
Hint 1Average rate over \([a,b]\) is \(\dfrac{f(b)-f(a)}{b-a}\) — the slope of the secant joining the endpoints.
Hint 2Instantaneous rate is \(f'(c) = 2c\). Set it equal to your average and solve.
Solution
STEP 1 · average rate of change
\[ \frac{f(3)-f(1)}{3-1} = \frac{9-1}{2} = 4 \]
Geometrically: the secant line from \((1,1)\) to \((3,9)\) has slope 4.
STEP 2 · instantaneous rate
\(f'(x) = 2x\), so at the unknown point \(c\) the tangent slope is \(2c\).
STEP 3 · set them equal and solve
\[ 2c = 4 \quad\Longrightarrow\quad c = 2 \]
And \(c=2\) does lie inside \((1,3)\), as it must.
STEP 4 · what this means
At \(x=2\) the tangent line is parallel to the secant through the endpoints. That is the Mean Value Theorem: over any interval, at some interior instant the object was travelling at exactly its average speed. If you drove 120 km in 2 hours, at some moment your speedometer read exactly 60 km/h.
WHY IT MATTERS
The MVT is what proves the fact you use constantly in ODEs: if \(f' = g'\) everywhere, then \(f\) and \(g\) differ by a constant — which is why an antiderivative is unique only up to \(+C\).
9
A bacterial population satisfies \(\dfrac{dP}{dt} = 0.03P\), with \(t\) in hours. (a) Put the equation into words. (b) If \(P = 2000\), how fast is the population growing? (c) At \(P = 6000\)? (d) Why can't you answer "what is \(P\) after 10 hours?" from this information alone?
Hint 1Read the left side as "the rate of change of \(P\) per hour" and the right side as a formula for that rate.
Hint 2For (b) and (c) you are just evaluating the right-hand side — no solving required.
Hint 3For (d), ask what else a first-order ODE needs before it names a single function.
Solution
STEP 1 · translate the equation (a)
"The population grows at 3% of its current size per hour." The rate is not constant — it is proportional to how much is already there, which is the defining feature of exponential growth.
STEP 2 · evaluate at P = 2000 (b)
\[ \frac{dP}{dt} = 0.03(2000) = 60 \ \text{bacteria per hour} \]
STEP 3 · evaluate at P = 6000 (c)
\[ \frac{dP}{dt} = 0.03(6000) = 180 \ \text{bacteria per hour} \]
Triple the population, triple the growth rate — the signature of this model.
STEP 4 · why the future is undetermined (d)
The ODE only tells you the slope at each state, not the starting state. Its general solution is \(P(t) = Ce^{0.03t}\), a whole family of curves — one through every possible starting population. Without an initial condition such as \(P(0) = 500\), \(C\) is unknown and \(P(10)\) could be anything.
STEP 5 · with an initial condition
Given \(P(0) = 500\), you get \(C = 500\) and \(P(10) = 500e^{0.3} \approx 675\). This is the exact pattern of every first-order initial value problem: general solution first, then one condition to pin the constant.
Part I · Module 2 Differentiation
Position, velocity, acceleration
Motion is the canonical derivative story, and it is also the canonical ODE story: Newton's second law is a second-order differential equation in disguise. Get comfortable moving up and down this ladder in both directions.
Velocity is signed — it carries direction. Speed is \(|v|\).
The object is at rest / turning around where \(v(t) = 0\).
The object is speeding up when \(v\) and \(a\) have the same sign, and slowing down when they have opposite signs. This is the single most-missed exam question of the topic.
InteractiveThree graphs, one motionpress play, or scrub \(t\)
position —velocity —acceleration —motion —
The moving dot on the top track is the actual object. Watch the moment \(v\) crosses zero: the dot stops and reverses, while \(s\) is at a local max or min. Where \(a\) crosses zero, \(v\) is at its own max or min and the \(s\)-curve has an inflection point.
Worked example · reading the motion
A particle moves with \(s(t) = t^3 - 6t^2 + 9t\) meters, \(t\ge0\) seconds. Describe its motion on \([0,4]\).
STEP 1
\(v(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3)\). So \(v=0\) at \(t=1\) and \(t=3\) — two turning points.
STEP 2
Sign of \(v\): positive on \([0,1)\), negative on \((1,3)\), positive on \((3,4]\). Right, then left, then right.
STEP 3
\(a(t) = 6t - 12 = 6(t-2)\), so \(a<0\) before \(t=2\) and \(a>0\) after.
STEP 4
Combine signs: on \((1,2)\) both \(v<0\) and \(a<0\) — same sign, so it is speeding up while traveling left. On \((2,3)\), \(v<0\) but \(a>0\) — opposite signs, so it is slowing down, arriving at rest at \(t=3\).
STEP 5
Total distance is not \(|s(4)-s(0)|\); you must split at the turning points: \(|s(1)-s(0)| + |s(3)-s(1)| + |s(4)-s(3)| = 4 + 4 + 4 = 12\) m, while the net displacement is only \(4\) m.
Where this shows up in ODEs
Newton's law \(F = ma\) becomes \(m\,s''(t) = F(t, s, s')\) — a second-order ODE. A mass on a spring with damping is \(m s'' + c s' + k s = 0\), the equation you will spend weeks on. The three curves in the lab above are exactly the objects that equation relates.
②Practice
1
A ball is thrown upward with \(s(t) = 40 + 8t - 4.9t^2\) meters. When does it reach maximum height, and what is its velocity and acceleration at that instant?
Hint 1Maximum height happens where the ball is momentarily neither rising nor falling.
Hint 2Set \(v(t) = 8 - 9.8t = 0\).
Hint 3Acceleration here is constant — gravity does not pause at the top.
Solution
STEP 1 · differentiate to get velocity
Term by term, using the power rule:
\[ v(t) = s'(t) = 0 + 8 - 9.8t \]
STEP 2 · maximum height ⟺ v = 0
At the top the ball has stopped rising but not yet started falling, so its velocity passes through zero:
"It stops at the top, so acceleration is zero" is the classic error. Zero velocity is a statement about position changing; acceleration is about velocity changing — and velocity is changing right through the turnaround (from \(+\) to \(-\)). Gravity never switches off.
2
A particle has \(v(t) = t^2 - 4t + 3\). On which intervals in \([0,4]\) is it speeding up?
Hint 1You need the sign of \(v\) and the sign of \(a = v'\). Build a sign chart with both.
Hint 2\(v = (t-1)(t-3)\) and \(a = 2t - 4\). The critical marks are \(t = 1, 2, 3\).
Hint 3Speeding up ⟺ \(v\) and \(a\) share a sign.
Solution
STEP 1 · factor the velocity
\[ v(t) = t^2-4t+3 = (t-1)(t-3) \quad\Longrightarrow\quad v=0 \text{ at } t=1,\ 3 \]
The three critical times \(1, 2, 3\) cut \([0,4]\) into four intervals. Test one point inside each:
Interval
test t
v
a
same sign?
Behavior
(0, 1)
0.5
+1.25
−3
no
slowing down
(1, 2)
1.5
−0.75
−1
yes
speeding up
(2, 3)
2.5
−0.75
+1
no
slowing down
(3, 4)
3.5
+1.25
+3
yes
speeding up
STEP 4 · answer
Speeding up on \((1,2)\) and \((3,4)\).
WHY THE RULE WORKS
Speed is \(|v|\). When \(a\) pushes in the same direction the particle is already moving, \(|v|\) grows; when \(a\) opposes the motion, \(|v|\) shrinks. That is why the test is about matching signs, not about \(a\) being positive.
3
For the damped oscillation \(s(t) = 4e^{-t/3}\cos(3t)\), find \(v(t)\). (You will need the product and chain rules — peek ahead to D3 and D5 if needed.)
Hint 1Treat it as a product \(u\cdot w\) with \(u = 4e^{-t/3}\) and \(w = \cos 3t\).
Hint 2\(u' = -\tfrac43 e^{-t/3}\) (chain rule on the exponent) and \(w' = -3\sin 3t\).
Solution
STEP 1 · name the two factors
\(u = 4e^{-t/3}\) (the decaying envelope) and \(w = \cos 3t\) (the oscillation).
STEP 2 · differentiate each with the chain rule
For \(u\): the outer function is \(e^{\square}\), the inner is \(-t/3\) whose derivative is \(-\frac13\):
STEP 4 · factor the common exponential
\[ v(t) = -e^{-t/3}\left(\tfrac43\cos 3t + 12\sin 3t\right) \]
SANITY CHECK
At \(t=0\): \(v(0) = -\frac43\), small and negative — the curve starts at its peak \(s(0)=4\) and immediately turns down ✓. Also every term keeps the factor \(e^{-t/3}\), so the velocity dies out alongside the motion, as a damped system must.
ODE CONNECTION
This \(s\) is the general solution shape of an underdamped spring \(m s'' + c s' + k s = 0\). In the course you will write down this answer directly from the roots \(-\frac13 \pm 3i\) of a quadratic, without differentiating anything.
4
A particle moves with \(s(t) = t^2 - 4t + 3\) meters on \(0 \le t \le 5\). Find its displacement and its total distance travelled, and explain why they differ.
Hint 1Displacement is just the net change \(s(5)-s(0)\).
Hint 2Total distance must be broken at every turning point — find where \(v = 0\).
Hint 3\(v(t) = 2t-4\), so the particle reverses at \(t=2\).
Net movement: it ends 5 m to the right of where it started.
STEP 2 · find the turning point
\(v(t) = 2t - 4 = 0\) at \(t = 2\). Before that \(v<0\) (moving left); after, \(v>0\) (moving right).
STEP 3 · positions at the key times
\[ s(0) = 3, \qquad s(2) = 4-8+3 = -1, \qquad s(5) = 8 \]
STEP 4 · add the legs in absolute value
\[ |s(2)-s(0)| + |s(5)-s(2)| = |-1-3| + |8-(-1)| = 4 + 9 = 13\ \text{m} \]
STEP 5 · why they differ
The particle first walked 4 m left, then 9 m right. Displacement cancels those directions (\(-4+9 = 5\)); distance does not (\(4+9 = 13\)). Splitting at every zero of \(v\) is mandatory — integrating \(|v|\) instead of \(v\) is the integral version of exactly this idea, and you will meet it again in module I1's signed-area discussion.
5
A car brakes with velocity \(v(t) = 30 - 6t\) m/s. (a) When does it stop? (b) What is its acceleration? (c) How far does it travel while braking?
Hint 1"Stops" means \(v = 0\) — solve for \(t\).
Hint 2Acceleration is the derivative of velocity; here it is constant.
Hint 3Distance is the area under the velocity graph — and that graph is a triangle, so you don't even need to integrate.
Solution
STEP 1 · when it stops (a)
\[ 30 - 6t = 0 \quad\Longrightarrow\quad t = 5\ \text{s} \]
Constant deceleration: the car sheds 6 m/s of speed every second. Note \(v<0\)? No — \(v>0\) on \([0,5)\) while \(a<0\), opposite signs, so the car is slowing down the whole time ✓.
STEP 3 · distance, geometrically (c)
Distance travelled is the area under \(v\) from 0 to 5. That region is a triangle with base 5 s and height 30 m/s:
Because "area under the velocity curve = change in position" is the Fundamental Theorem of Calculus, previewed here in the easiest possible case. Module I2 makes that connection precise.
6
An object has acceleration \(a(t) = 4 - 2t\) m/s², with \(v(0) = 0\) and \(s(0) = 5\). Find \(v(t)\) and \(s(t)\), and the time at which the object is farthest to the right.
Hint 1Going up the ladder means antidifferentiating: what function has derivative \(4-2t\)?
Hint 2Each antidifferentiation introduces a new constant. Use \(v(0)=0\) to fix the first, \(s(0)=5\) to fix the second.
Hint 3"Farthest right" is a maximum of \(s\), so look for \(v = 0\) with \(v\) switching \(+\to-\).
Two antidifferentiations needed exactly two conditions. A second-order differential equation always works this way: \(s'' = a(t)\) is one, and \(v(0), s(0)\) are the two initial conditions that select a single motion out of the family.
7
A rock is dropped (initial velocity 0) from a cliff 80 m high, so \(s(t) = 80 - 4.9t^2\) with \(s\) measured upward from the ground. Find when it lands, its impact speed, and its average speed over the fall.
Hint 1"Lands" means \(s(t)=0\). Solve for \(t\) (take the positive root).
Hint 2Impact speed is \(|v|\) at that time, so compute \(v(t) = -9.8t\) first.
Hint 3Average speed is total distance ÷ total time — a single division, not a derivative.
Solution
STEP 1 · time of impact
\[ 80 - 4.9t^2 = 0 \ \Longrightarrow\ t^2 = \frac{80}{4.9} = 16.33 \ \Longrightarrow\ t \approx 4.04\ \text{s} \]
(The negative root is discarded — it refers to time before the drop.)
The average speed is exactly half the final speed. That is no accident: with constant acceleration, velocity grows linearly from 0, and the average of a linear ramp is its midpoint. It is also the Mean Value Theorem in action — at \(t \approx 2.02\) s the rock was moving at precisely its average speed.
8
Verify that \(s(t) = A\cos(\omega t)\) satisfies the spring equation \(s'' = -\omega^2 s\) for any constant \(A\). Then state the physical meaning of \(A\) and \(\omega\).
Hint 1Differentiate twice, using the chain rule each time — each derivative pulls out a factor \(\omega\).
Hint 2After two derivatives you should see the original function reappear, multiplied by something.
The equation holds for every \(A\) — the amplitude is arbitrary, which is why \(A\) survives as a free constant in the general solution.
STEP 4 · physical reading
\(A\) is the amplitude (how far the mass is pulled from equilibrium) and \(\omega\) is the angular frequency, with period \(T = 2\pi/\omega\). For a mass–spring system \(\omega = \sqrt{k/m}\): stiffer spring ⟹ faster oscillation, heavier mass ⟹ slower.
WHAT THE EQUATION SAYS
\(s'' = -\omega^2 s\) means "acceleration always points back toward equilibrium, in proportion to the displacement" — Hooke's law plus Newton's second law. Any function that solves it must oscillate, which is why sines and cosines are unavoidable in this course.
9
An object has constant acceleration \(a\). Starting from \(s''(t)=a\) with \(v(0)=v_0\) and \(s(0)=s_0\), derive the two standard kinematic equations.
Hint 1Antidifferentiate once to get velocity, using \(v(0)=v_0\) to fix the constant.
Hint 2Antidifferentiate again for position, using \(s(0)=s_0\).
Hint 3For the third equation, eliminate \(t\) between the two results.
Solution
STEP 1 · integrate the acceleration
\(a\) is constant, so its antiderivative is \(at\):
From step 2, \(t = \frac{v-v_0}{a}\). Substituting and simplifying:
\[ v^2 = v_0^2 + 2a\left(s-s_0\right) \]
WHY THIS MATTERS
Every "SUVAT" formula from physics is just two antiderivatives plus two initial conditions — and they are valid only for constant \(a\). The moment acceleration varies (air resistance, a spring), you must go back to \(s'' = a(t,s,s')\), which is a differential equation.
Part I · Module 3 Differentiation
Power, product & quotient rules
Three rules cover most of the algebra you will differentiate. The power rule you already trust; the product and quotient rules are the two people most often "simplify" into something false.
Say the quotient rule out loud every time until it is automatic: "low d-high minus high d-low, over low squared." The subtraction means order matters — swapping the two terms flips the sign of your entire answer.
The two false rules
\((uv)' \ne u'v'\) and \(\left(\frac{u}{v}\right)' \ne \frac{u'}{v'}\). Test it once on \(u=v=x\): \((x\cdot x)' = 2x\), but \(u'v' = 1\). Never again.
InteractiveRule assemblerpick \(u\) and \(v\), see the rule built and verified
Amber is the combined function, cyan is the derivative the rule produces, and the small white circles are a purely numerical derivative \(\frac{f(x+\varepsilon)-f(x-\varepsilon)}{2\varepsilon}\) computed from the amber curve alone. If the rule is right, the circles land on the cyan curve — that is your independent check on any derivative you compute by hand.
Worked example · product then quotient
Differentiate \(f(x) = \dfrac{x^2\sin x}{x+1}\).
STEP 1
Identify the outermost structure: this is a quotient, with \(u = x^2\sin x\) and \(v = x+1\). Always work from the outside in.
STEP 2
\(u\) itself needs the product rule: \(u' = 2x\sin x + x^2\cos x\). And \(v' = 1\).
Stop. Expanding the numerator here buys you nothing and costs you sign errors. Leave it factored unless the problem asks you to solve \(f'=0\).
Simplify before you differentiate
\(\dfrac{x^3 + 2x}{x}\) does not need the quotient rule — it is \(x^2+2\), so the derivative is \(2x\). Always spend five seconds looking for a division you can cancel or a product you can expand first.
Two negatives collided in the second term: the \(-3\) coefficient and the \(-2\) exponent multiplied to \(+6\). Getting \(-6x^{-3}\) there is the most common slip in this problem.
STEP 4 · convert back to readable form
\[ y' = 20x^3 + \frac{6}{x^3} + \frac{1}{\sqrt{x}} \]
CHECK
Every term of \(y'\) is positive for \(x>0\), so \(y\) should be increasing there — plausible, since \(5x^4\) and \(2\sqrt x\) both climb and \(-3/x^2\) rises toward 0 from below.
2
Differentiate \(g(t) = t^2 e^t\).
Hint 1Product rule with \(u = t^2\), \(v = e^t\).
Hint 2\(e^t\) is its own derivative, so \(v' = e^t\). Factor it out at the end.
Solution
STEP 1 · identify the two factors
\(u = t^2\) and \(v = e^t\). This is a genuine product of two non-constant functions, so the product rule is required — you cannot differentiate each piece and multiply.
The factored form hands you the zeros for free: \(g' = 0\) at \(t=0\) and \(t=-2\) (since \(e^t\) is never zero). Leaving the answer expanded would force you to solve \(t^2+2t=0\) later anyway.
3
Differentiate \(h(x) = \dfrac{2x - 1}{x^2 + 3}\) and find where the tangent is horizontal.
Hint 1Quotient rule: low d-high minus high d-low, over low squared.
Hint 2A fraction is zero exactly when its numerator is zero (and the denominator is not). So set the numerator \(=0\).
Hint 3The numerator simplifies to \(-2x^2 + 2x + 6\). Divide by \(-2\) before using the quadratic formula.
Solution
STEP 1 · label the pieces
\(u = 2x-1\) (high), \(v = x^2+3\) (low), so \(u' = 2\) and \(v' = 2x\).
Horizontal tangents at \(x \approx 2.30\) and \(x \approx -1.30\). The denominator \(x^2+3\) is never zero, so \(h\) is defined and smooth everywhere — no vertical asymptotes to exclude.
4
Given \(f(2) = 3\), \(f'(2) = -1\), \(g(2) = 5\), \(g'(2) = 4\), compute \((fg)'(2)\) and \((f/g)'(2)\).
Hint 1You never need the formulas for \(f\) and \(g\) — the rules only ask for values at the point.
Hint 2\((fg)' = f'g + fg'\) evaluated at 2 is \((-1)(5) + (3)(4)\).
Solution
STEP 1 · write the rule symbolically first
\[ (fg)'(x) = f'(x)g(x) + f(x)g'(x) \]
Only then substitute — plugging numbers in too early is how terms get swapped.
Derivative rules are local: they need only the value and slope of each function at the single point in question. This is why tables of values (or measured data) are enough to compute derivatives of combinations, with no formulas anywhere in sight.
5
Differentiate \(y = (x^2+1)(x^3-4x)\) two ways — with the product rule, and by expanding first — and confirm the answers agree.
Hint 1Product rule: \(u = x^2+1\), \(v = x^3-4x\).
Hint 2To expand, multiply out to \(x^5 - 3x^3 - 4x\) and differentiate term by term.
Hint 3To compare, expand your product-rule answer fully and collect like powers.
Identical answers, as they must be. For small polynomials, expanding is usually faster and less error-prone. The product rule earns its keep when a factor can't be expanded away — \(x^2\sin x\), \(te^{-t}\), \(x\ln x\) — which is most of what you meet in an ODE course.
6
Find the maximum value of \(f(x) = \dfrac{x}{x^2+1}\) on \([0,\infty)\).
Hint 1Maxima live where \(f'(x) = 0\) (or at endpoints). Use the quotient rule.
Hint 2The numerator of \(f'\) works out to \(1 - x^2\).
Hint 3Only \(x = 1\) is in the domain \([0,\infty)\); check the sign of \(f'\) on either side of it.
The denominator is always positive, so the sign of \(f'\) is the sign of \(1-x^2\):
\[ 1-x^2 = 0 \ \Longrightarrow\ x = 1 \ \ (\text{rejecting } x=-1, \text{ outside the domain}) \]
STEP 4 · confirm it's a maximum
For \(0 \le x < 1\), \(1-x^2 > 0\) so \(f\) is increasing; for \(x>1\), \(1-x^2<0\) so \(f\) is decreasing. A rise followed by a fall means \(x=1\) is a maximum.
Same answer, four times the algebra and three extra chances to drop a sign. Always spend five seconds looking for a simplification before choosing a rule. The same instinct saves you constantly in integration, where an unsimplified integrand can make an easy problem look impossible.
FINE PRINT
The simplified function \(x^2+2x\) is defined at \(x=0\) but the original is not, so strictly the two agree only for \(x\ne0\). That distinction matters when a problem asks about the domain.
8
Find the equation of the tangent line to \(y = \dfrac{x+1}{x-1}\) at \(x = 2\).
Hint 1A tangent line needs two things: a point on the curve, and the slope there.
Hint 2Get the point by evaluating \(y(2)\); get the slope by evaluating \(y'(2)\).
STEP 3 · the slope at x = 2
\[ y'(2) = \frac{-2}{(2-1)^2} = -2 \]
STEP 4 · point-slope form
\[ y - 3 = -2(x-2) \quad\Longrightarrow\quad y = -2x + 7 \]
CHECK
\(y'\) is negative for every \(x \ne 1\), so this curve is decreasing on both branches — consistent with a tangent of slope \(-2\). Also note the derivative blows up as \(x\to1\), which is the vertical asymptote showing itself in the derivative.
9
A student writes \(\dfrac{d}{dx}\left[x^2\sin x\right] = 2x\cos x\). Identify the error, and construct a one-line counterexample proving the rule they used is false.
Hint 1What rule would produce \(2x\cos x\)? Compare it to \(u'v'\).
Hint 2To disprove "the derivative of a product is the product of the derivatives," pick the simplest product you know the answer to.
Hint 3Try \(u = v = x\), so the product is \(x^2\).
Solution
STEP 1 · name the error
They multiplied the derivatives: \(u'v' = (2x)(\cos x)\). That is the false rule \((uv)' = u'v'\).
STEP 2 · the correct computation
\[ \frac{d}{dx}\left[x^2\sin x\right] = 2x\sin x + x^2\cos x \]
Two terms, not one — each factor gets a turn at being differentiated while the other is left alone.
STEP 3 · build the counterexample
Take \(u = v = x\). Then \(uv = x^2\), whose derivative is certainly \(2x\). But
\[ u'v' = (1)(1) = 1 \ne 2x \]
One line, and the false rule is dead.
STEP 4 · why the true rule has two terms
Think of a rectangle with sides \(u\) and \(v\); its area is \(uv\). Nudge both sides slightly: the area grows by a strip of size \(v\,du\) along one edge, plus a strip \(u\,dv\) along the other, plus a tiny corner \(du\,dv\) that is negligible in the limit. Those two strips are \(u'v + uv'\).
HABIT WORTH FORMING
Whenever you suspect a rule, test it on the simplest possible case (\(x\cdot x\), or \(x/x\)). It takes five seconds and catches most invented rules — including \(\left(\frac uv\right)' = \frac{u'}{v'}\), which the same test destroys.
Part I · Module 4 Differentiation
Derivatives of trig functions
Sine and cosine are the functions that come back to themselves after four derivatives. That closure is precisely why they solve oscillation problems — and why they dominate the second half of an ODE course.
①The six you must know cold
f(x)
f′(x)
Note
sin x
cos x
—
cos x
−sin x
the minus lives here
tan x
sec² x
always positive — tan is always increasing
cot x
−csc² x
co- functions get a minus
sec x
sec x tan x
—
csc x
−csc x cot x
co- functions get a minus
The pattern that halves the memorizing
Every co-function (cos, cot, csc) has a negative derivative formula, and its derivative is the "co-" version of its partner's. Learn three, flip signs, get six.
InteractiveThe four-step derivative cycledifferentiate repeatedly and watch it loop
current —next derivative —
The wheel on the right is the cycle \(\sin \to \cos \to -\sin \to -\cos \to \sin\). Four differentiations return you to where you started — which is the same as saying \(y'''' = y\), and also why \(y'' = -y\) is solved by sines and cosines.
Radians only
\(\frac{d}{dx}\sin x = \cos x\) is true only in radians. In degrees you would get \(\frac{\pi}{180}\cos x\). Put your calculator in radian mode for this entire course and leave it there.
Worked example · trig with the product and quotient rules
Expand the numerator: \(\cos x + \cos^2 x + \sin^2 x = \cos x + 1\).
STEP 4
Cancel one factor: \(z' = \dfrac{1+\cos x}{(1+\cos x)^2} = \dfrac{1}{1+\cos x}\). The Pythagorean identity is what made this collapse — always look for \(\sin^2+\cos^2\) hiding in a trig numerator.
②Practice
1
Differentiate \(f(x) = 3\sin x - 2\cos x\), then verify that it satisfies \(f'' + f = 0\).
Hint 1Differentiate twice and compare \(f''\) to \(f\) term by term.
Hint 2Two derivatives of \(\sin\) is \(-\sin\); two derivatives of \(\cos\) is \(-\cos\).
Solution
STEP 1 · first derivative
Differentiate each term, keeping the constants along for the ride:
\[ f'(x) = 3\cos x - 2(-\sin x) = 3\cos x + 2\sin x \]
Note the sign flip: \(-2\cos x\) differentiates to \(+2\sin x\), because \(\frac{d}{dx}\cos = -\sin\) and the two minuses cancel.
STEP 2 · second derivative
\[ f''(x) = -3\sin x + 2\cos x \]
You verified a solution to a differential equation — the simple harmonic oscillator \(y''+y=0\). Its general solution is \(A\sin x + B\cos x\): two arbitrary constants for a second-order equation, and your \(A=3, B=-2\) is one member of that family.
2
Differentiate \(y = \sec x \tan x\).
Hint 1Product rule with \(u=\sec x\), \(v=\tan x\).
Hint 2\(u' = \sec x\tan x\) and \(v' = \sec^2 x\), so you will get \(\sec x\tan^2 x + \sec^3 x\).
Solution
STEP 1 · set up the product rule
\(u = \sec x\), \(v = \tan x\), with \(u' = \sec x\tan x\) and \(v' = \sec^2x\) straight from the table.
The identity \(\sin x + \cos x = \sqrt2\,\sin\!\left(x+\frac{\pi}{4}\right)\) says this is just a shifted sine of amplitude \(\sqrt2\) — so of course its extremes are \(\pm\sqrt2\), a quarter-period apart from the origin. That "combine two sinusoids into one" trick reappears when you interpret ODE solutions as a single oscillation with a phase shift.
4
Show that \(y = \tan x - x\) is increasing on every interval where it is defined.
Hint 1"Increasing" means \(y' \ge 0\). Compute \(y'\) first.
Hint 2\(y' = \sec^2x - 1\). Is there an identity that rewrites that?
Hint 3\(\sec^2x - 1 = \tan^2 x\), and squares are never negative.
Solution
STEP 1 · differentiate
\[ y' = \sec^2 x - 1 \]
STEP 2 · apply the Pythagorean identity
From \(1 + \tan^2 x = \sec^2 x\) we get \(\sec^2 x - 1 = \tan^2 x\), so
\[ y' = \tan^2 x \]
STEP 3 · read off the sign
A square is never negative, so \(y' \ge 0\) everywhere \(\tan x\) is defined — and \(y' = 0\) only at the isolated points \(x = 0, \pm\pi, \pm2\pi,\dots\), which do not stop the function from increasing overall.
STEP 4 · state the conclusion carefully
\(y\) is increasing on each interval \(\left(-\frac{\pi}{2}+k\pi,\ \frac{\pi}{2}+k\pi\right)\). You cannot say "increasing on \(\mathbb{R}\)" — the function is undefined at the odd multiples of \(\pi/2\), where it jumps from \(+\infty\) to \(-\infty\).
COROLLARY
Since \(y(0) = 0\) and \(y\) increases, \(\tan x > x\) for \(0 < x < \frac{\pi}{2}\). Proving inequalities by differentiating the difference is a standard and very reusable trick.
5
Find the 50th derivative of \(\sin x\), and the 101st derivative of \(\cos x\).
Hint 1Don't differentiate 50 times — find the pattern after 4.
Hint 2The cycle is \(\sin \to \cos \to -\sin \to -\cos \to \sin\), so only the remainder of \(n \div 4\) matters.
Twelve complete cycles change nothing, so only the leftover 2 matters.
STEP 3 · read the table
\[ \frac{d^{50}}{dx^{50}}\sin x = -\sin x \]
STEP 4 · repeat for cosine
\(101 = 4(25)+1\), remainder 1, so
\[ \frac{d^{101}}{dx^{101}}\cos x = -\sin x \]
CONNECTION
The four-step cycle is exactly the statement \(y'''' = y\) for sines and cosines, and it is why the characteristic equation \(r^4 = 1\) (roots \(\pm1, \pm i\)) produces solutions built from \(e^{x}, e^{-x}, \sin x, \cos x\) in a fourth-order ODE.
6
Differentiate \(y = \dfrac{1+\sin x}{\cos x}\). Simplify before differentiating if you can.
Hint 1Split the single fraction into two: \(\frac{1}{\cos x} + \frac{\sin x}{\cos x}\).
Hint 2Those are \(\sec x\) and \(\tan x\) — both are in the table of six.
Solution
STEP 1 · split the fraction
\[ y = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \sec x + \tan x \]
STEP 2 · differentiate the two standard forms
\[ y' = \sec x\tan x + \sec^2 x \]
Notice something elegant: \(y' = \sec x \cdot y\). That means \(y\) satisfies the differential equation \(y' = (\sec x)\,y\) — a separable ODE you could solve in Week 1 of the real course.
STEP 4 · what the quotient rule would have cost
Directly: \(\frac{(\cos x)(\cos x) - (1+\sin x)(-\sin x)}{\cos^2 x} = \frac{\cos^2x + \sin x + \sin^2 x}{\cos^2 x} = \frac{1+\sin x}{\cos^2 x}\) — correct, and equal to the answer above, but it takes an identity and two extra lines to see that.
HABIT
Splitting a fraction whose denominator is a single term is nearly always worth doing first — in differentiation and, even more so, in integration.
7
Find the equation of the tangent line to \(y = \tan x\) at \(x = \dfrac{\pi}{4}\).
Hint 1You need the point \(\left(\frac\pi4, \tan\frac\pi4\right)\) and the slope \(y'\left(\frac\pi4\right)\).
Hint 2\(y' = \sec^2 x\), and \(\sec\frac{\pi}{4} = \sqrt2\).
Solution
STEP 1 · the point
\[ \tan\frac{\pi}{4} = 1 \quad\Longrightarrow\quad \left(\frac{\pi}{4},\ 1\right) \]
STEP 2 · the derivative
\[ y' = \sec^2 x \]
STEP 3 · evaluate the slope
\(\cos\frac\pi4 = \frac{\sqrt2}{2}\), so \(\sec\frac\pi4 = \frac{2}{\sqrt2} = \sqrt2\), and
STEP 4 · point-slope form
\[ y - 1 = 2\left(x - \frac{\pi}{4}\right) \quad\Longrightarrow\quad y = 2x + 1 - \frac{\pi}{2} \approx 2x - 0.571 \]
CHECK
\(\sec^2 x \ge 1\) always, so every tangent line to \(\tan x\) has slope at least 1 — consistent with the fact that \(\tan\) is always increasing (problem 4).
8
Derive \(\dfrac{d}{dx}\tan x = \sec^2 x\) from the quotient rule, using only \(\frac{d}{dx}\sin x = \cos x\) and \(\frac{d}{dx}\cos x = -\sin x\).
Hint 1Start from the definition \(\tan x = \frac{\sin x}{\cos x}\).
Hint 2The numerator of the quotient rule will contain \(\cos^2 x + \sin^2 x\).
Solution
STEP 1 · write tan as a quotient
\[ \tan x = \frac{\sin x}{\cos x}, \qquad u = \sin x,\ v = \cos x \]
STEP 2 · apply the quotient rule
\[ \frac{d}{dx}\tan x = \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{\cos^2 x} \]
Watch the double negative: \(-u v' = -\sin x \cdot(-\sin x) = +\sin^2 x\).
STEP 3 · simplify with the Pythagorean identity
\[ = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} \]
Every one of the remaining four derivatives falls out of this same move: \(\cot x = \frac{\cos x}{\sin x}\), \(\sec x = \frac{1}{\cos x}\), \(\csc x = \frac{1}{\sin x}\). If you can derive them in about a minute each, you never have to trust your memory of the table under exam pressure.
9
Differentiate \(y=\sin x\cos x\) two ways — by the product rule, and by first using a double-angle identity — and reconcile the answers.
Hint 1Product rule: \(u=\sin x\), \(v=\cos x\).
Hint 2Identity route: \(\sin x\cos x = \tfrac12\sin2x\).
Hint 3To reconcile, recall \(\cos2x = \cos^2x-\sin^2x\).
Solution
METHOD A · product rule
\[ y' = \cos x\cos x + \sin x(-\sin x) = \cos^2x - \sin^2x \]
METHOD B · simplify first
\[ y = \tfrac12\sin2x \quad\Longrightarrow\quad y' = \tfrac12\cos2x\cdot2 = \cos2x \]
STEP 3 · reconcile
The double-angle identity \(\cos2x = \cos^2x-\sin^2x\) shows the two answers are the same function ✓
STEP 4 · which was easier
Method B took one line and produced the answer already simplified. Rewriting with an identity before differentiating usually beats grinding through the product rule.
STEP 5 · use it
Setting \(y'=0\) gives \(\cos2x=0\), so \(x = \frac\pi4 + \frac{k\pi}{2}\) — where \(\sin x\cos x\) reaches \(\pm\frac12\). The identity form makes those extremes obvious; the product-rule form hides them behind a trig equation.
Part I · Module 5 Differentiation
The chain rule
The single most-used rule in differential equations — and the one whose absence causes the most silent errors. Every time you write \(y'\) while differentiating something containing \(y\), you are using the chain rule.
The Leibniz form is the one worth internalizing: the \(du\)'s "cancel" formally, which is exactly why the notation was designed this way. Read it as rates multiply through a chain: if \(y\) changes 3× as fast as \(u\), and \(u\) changes 5× as fast as \(x\), then \(y\) changes 15× as fast as \(x\).
InteractiveThe composition machineslide \(x\) and watch the rates multiply
u = g(x) —g′(x) —f′(u) —product = dy/dx —
The top strip is the machine: \(x\) enters the inner box \(g\), comes out as \(u\), and enters the outer box \(f\). Below, the composite curve is drawn with its tangent. The tangent's slope always equals the product of the two rates shown in the readout — never the sum, and never just \(f'(u)\).
The forgotten inner derivative
\(\frac{d}{dx}\sin(3x) = 3\cos(3x)\), not \(\cos(3x)\). \(\frac{d}{dx}e^{-2t} = -2e^{-2t}\), not \(e^{-2t}\). If your ODE answer is off by a constant factor, an unpaid inner derivative is the first thing to check.
Worked example · peeling a three-layer onion
Differentiate \(y = \sqrt{\sin(3x^2)}\).
STEP 1
Name the layers from the outside in: outer \(=\sqrt{\ \cdot\ }\), middle \(=\sin(\cdot)\), inner \(=3x^2\).
STEP 2
Outer derivative, inner left alone: \(\dfrac{1}{2\sqrt{\sin(3x^2)}}\).
Two places, constantly. (1) Verifying a solution: to check that \(y = e^{-3t}\) solves \(y' + 3y = 0\) you differentiate through the exponent. (2) Separation of variables: the whole method is the chain rule read backwards — \(\frac{d}{dx}F(y) = F'(y)\,y'\) is what lets you write \(\int g(y)\,dy\) after multiplying through by \(dx\).
②Practice
1
Differentiate \(y = (x^3 - 4x)^{7}\).
Hint 1Outer function is "something to the 7th"; inner is \(x^3-4x\).
Hint 2Power rule on the outside keeps the inside untouched: \(7(x^3-4x)^6\), then multiply by the inner derivative.
Solution
STEP 1 · name the layers
Outer: \(f(u) = u^7\). Inner: \(g(x) = x^3-4x\). Always identify the outermost operation first — here, "raise to the 7th."
STEP 2 · differentiate the outer, leaving the inner alone
\[ f'(g(x)) = 7(x^3-4x)^6 \]
The inside is copied down unchanged. Expanding it would be a catastrophic waste of time.
\(y\) has degree \(3\times7 = 21\), so \(y'\) should have degree 20. Your answer: \(6\times3 + 2 = 20\) ✓ — a fast, surprisingly effective sanity check on chain-rule problems.
2
Differentiate \(y = e^{-x^2/2}\) (the shape of the normal distribution).
Hint 1Derivative of \(e^{\text{stuff}}\) is \(e^{\text{stuff}}\) times the derivative of stuff.
Since \(y = e^{-x^2/2}\), the result says \(y' = -xy\) — the function satisfies a first-order separable ODE. Solving \(\frac{dy}{y} = -x\,dx\) is exactly how the Gaussian bell curve is derived in probability. Also note \(y'=0\) only at \(x=0\): the single peak, as expected.
3
Differentiate \(y = \ln\!\left(\dfrac{x}{x+1}\right)\) two ways: directly with the chain rule, and after using log properties. Confirm they agree.
Hint 1Log properties first: \(\ln\frac{x}{x+1} = \ln x - \ln(x+1)\). That version needs no quotient rule at all.
Hint 2Direct route: \(\frac{d}{dx}\ln u = \frac{u'}{u}\) with \(u = \frac{x}{x+1}\), and \(u' = \frac{1}{(x+1)^2}\) by the quotient rule.
Solution
METHOD A · STEP 1
Use \(\ln\frac{A}{B} = \ln A - \ln B\):
\[ y = \ln x - \ln(x+1) \]
METHOD A · STEP 2
Differentiate each log (the second needs the chain rule, but its inner derivative is just 1):
Same answer, three times the work. Expand logs before differentiating, always. And note the answer \(\frac{1}{x(x+1)}\) — Part II's partial fractions is the machine that runs this identity in reverse.
4
If \(y = f(u)\), \(u = g(t)\), \(g(2) = 5\), \(g'(2) = -3\), and \(f'(5) = 4\), find \(\frac{dy}{dt}\) at \(t=2\).
Hint 1Write the chain rule with the arguments spelled out: \(\frac{dy}{dt}\big|_{t=2} = f'(g(2))\cdot g'(2)\).
Hint 2The trap is evaluating \(f'\) at 2 instead of at \(g(2)=5\).
Solution
STEP 1 · write the rule with arguments
\[ \frac{dy}{dt} = f'\big(g(t)\big)\cdot g'(t) \]
The arguments are the whole difficulty here — write them before substituting anything.
STEP 2 · find where to evaluate f′
At \(t=2\), the inner function outputs \(g(2) = 5\). So \(f'\) must be evaluated at 5, not at 2. That is why the problem gave you \(f'(5)\).
\(u\) is falling 3 units per unit of \(t\), and \(y\) rises 4 units per unit of \(u\). Net: \(y\) falls 12 units per unit of \(t\). Rates multiply along the chain, and the sign follows.
5
Differentiate \(y = \sin^3(2x)\). (Three layers.)
Hint 1Rewrite it as \(\left[\sin(2x)\right]^3\) so the outermost operation is obvious.
Hint 2Layers from the outside in: cube → sine → \(2x\). You will multiply three factors.
Hint 3The innermost derivative is just 2 — don't lose it.
Solution
STEP 1 · rewrite to expose the outer layer
\[ y = \left[\sin(2x)\right]^3 \]
The notation \(\sin^3\) hides the cube; rewriting it prevents the classic error of differentiating the sine first.
STEP 2 · outer layer (the cube)
\[ 3\left[\sin(2x)\right]^2 \]
\(e^{ax}\cos(bx)\) and \(e^{ax}\sin(bx)\) are precisely the solutions of second-order linear ODEs with complex roots \(a \pm bi\). Every time you check such a solution you will run this exact product-plus-chain computation, so it is worth getting fluent.
7
Find the constant \(k\) so that \(y = e^{kt}\) satisfies \(y' + 5y = 0\). Then explain why this trick turns a differential equation into an algebra problem.
Hint 1Differentiate \(e^{kt}\) with the chain rule — the inner derivative is \(k\).
Hint 2Substitute into the equation and factor out \(e^{kt}\).
Hint 3\(e^{kt}\) is never zero, so it can be divided away safely.
Solution
STEP 1 · differentiate the trial solution
\[ y = e^{kt} \quad\Longrightarrow\quad y' = k e^{kt} \]
STEP 2 · substitute into the equation
\[ y' + 5y = ke^{kt} + 5e^{kt} = 0 \]
\(e^{kt} > 0\) for every \(t\), so it can never be the zero factor. Therefore
\[ k + 5 = 0 \quad\Longrightarrow\quad k = -5, \qquad y = e^{-5t} \]
STEP 5 · why this is the whole method
The chain rule turns differentiation of \(e^{kt}\) into multiplication by \(k\). So substituting \(e^{kt}\) converts the differential equation into a polynomial equation in \(k\) — the characteristic equation. For \(y'' + 3y' + 2y = 0\) you would get \(k^2+3k+2=0\), roots \(-1\) and \(-2\), and the solution \(y = C_1e^{-t} + C_2e^{-2t}\) without integrating anything.
8
A spherical balloon is inflated so its radius grows at \(\frac{dr}{dt} = 2\) cm/s. How fast is the volume growing when \(r = 5\) cm? (\(V = \frac43\pi r^3\))
Hint 1You want \(\frac{dV}{dt}\), but the formula gives \(V\) in terms of \(r\). Chain them: \(\frac{dV}{dt} = \frac{dV}{dr}\cdot\frac{dr}{dt}\).
Hint 2\(\frac{dV}{dr} = 4\pi r^2\) — which happens to be the surface area.
Because \(\frac{dV}{dr}\) is the surface area, the volume rate equals (surface area) × (radial speed) — new volume is a thin shell of thickness \(dr\) spread over the surface. Note it grows with \(r^2\): the balloon swallows air far faster when it is already big, even though the radius climbs at a steady 2 cm/s.
9
Differentiate \(y = \dfrac{1}{(x^2+1)^3}\).
Hint 1Rewrite as a negative power instead of using the quotient rule.
Hint 2\(y = (x^2+1)^{-3}\); outer is "to the \(-3\)", inner is \(x^2+1\).
Solution
STEP 1 · rewrite
\[ y = (x^2+1)^{-3} \]
A reciprocal of a power is never worth the quotient rule.
For \(x>0\) the derivative is negative, and indeed \(y\) falls away from its peak at \(x=0\); for \(x<0\) it is positive, climbing toward that peak. Symmetric bump, symmetric derivative ✓.
Part I · Module 6 Differentiation
Implicit differentiation
Some curves refuse to be written as \(y = f(x)\). Implicit differentiation lets you find their slopes anyway — and it is the exact technique behind exact equations and integrating factors later on.
①The one new idea
Treat \(y\) as an unknown function of \(x\). Then every time you differentiate a term containing \(y\), the chain rule contributes a factor of \(\frac{dy}{dx}\):
Differentiate both sides with respect to \(x\), tagging every \(y\)-term with \(\frac{dy}{dx}\).
Collect every term containing \(\frac{dy}{dx}\) on one side, everything else on the other.
Factor out \(\frac{dy}{dx}\) and divide.
The answer usually contains both \(x\) and \(y\) — that is expected, not a mistake. A point on the curve needs both coordinates to have a slope.
InteractiveDrag a point around an implicit curvethe tangent updates from \(dy/dx = -F_x/F_y\)
point —dy/dx —drag the amber dot
Drag anywhere near the curve — the point snaps onto it. Look for the places where the tangent goes vertical: those are exactly the points where \(F_y = 0\), i.e. where \(dy/dx\) blows up and the curve fails the vertical line test. No explicit \(y=f(x)\) could ever describe those points.
Worked example · the folium of Descartes
Find \(\frac{dy}{dx}\) for \(x^3 + y^3 = 6xy\), then the tangent line at \((3,3)\).
STEP 1
Differentiate both sides. Left: \(3x^2 + 3y^2 y'\). Right: the product rule gives \(6y + 6xy'\).
At \((3,3)\): \(y' = \frac{6-9}{9-6} = -1\), so the tangent is \(y - 3 = -(x-3)\), i.e. \(y = 6 - x\). Verify with the lab above.
Where this shows up in ODEs
An exact equation \(M\,dx + N\,dy = 0\) is nothing but a curve \(F(x,y) = C\) that has already been implicitly differentiated. Solving it means running this module backwards: given the differentiated form, reconstruct \(F\). The test \(\partial M/\partial y = \partial N/\partial x\) is checking that such an \(F\) exists.
②Practice
1
Find \(\frac{dy}{dx}\) for \(x^2 + y^2 = 25\), and the slope at \((3,4)\).
Hint 1Differentiate term by term: \(2x + 2y\,y' = 0\).
Hint 2Geometrically, the radius to \((3,4)\) has slope \(4/3\); the tangent must be perpendicular to it.
Solution
STEP 1 · differentiate both sides with respect to x
The \(x^2\) is routine. The \(y^2\) needs the chain rule, because \(y\) is a function of \(x\):
The radius from the origin to \((3,4)\) has slope \(\frac43\). The tangent to a circle is perpendicular to the radius, and \(-\frac34\) is exactly the negative reciprocal of \(\frac43\) ✓. Note also that at \((5,0)\) the formula gives division by zero — precisely where the tangent is vertical.
2
Find \(\frac{dy}{dx}\) for \(e^{xy} = x + y\).
Hint 1The left side needs the chain rule and the product rule: the exponent \(xy\) differentiates to \(y + xy'\).
Hint 2You get \(e^{xy}(y + xy') = 1 + y'\). Now expand and collect the \(y'\) terms.
Solution
STEP 1 · differentiate the left side
Outer function \(e^{\square}\), inner function \(xy\) — and that inner piece is itself a product:
Steps 4–5 are the entire method: collect, factor, divide. It never changes, no matter how ugly the equation — which is why implicit differentiation is more mechanical than it first looks.
3
Find where the ellipse \(x^2 + xy + y^2 = 12\) has a horizontal tangent.
Hint 1Implicitly differentiate to get \(2x + y + xy' + 2yy' = 0\).
Hint 2Horizontal tangent ⟹ \(y' = 0\) ⟹ the numerator of your expression vanishes: \(2x + y = 0\).
Hint 3Substitute \(y = -2x\) back into the original equation and solve for \(x\).
Solution
STEP 1 · differentiate implicitly
The middle term \(xy\) needs the product rule:
\[ 2x + \big(y + xy'\big) + 2yy' = 0 \]
STEP 2 · collect and solve for y′
\[ y'(x + 2y) = -(2x+y) \quad\Longrightarrow\quad y' = -\frac{2x+y}{x+2y} \]
STEP 3 · impose the horizontal condition
A fraction is zero when its numerator is:
\[ 2x + y = 0 \quad\Longrightarrow\quad y = -2x \]
The denominator \(x^2+3y^2\) is positive except at the origin, so this curve has no vertical tangents away from there — and the negative slope at \((1,2)\) says \(y\) must fall as \(x\) grows, which makes sense: to keep \(x^2y + y^3\) fixed at 10, a larger \(x\) demands a smaller \(y\).
5
For the circle \(x^2 + y^2 = 25\), find \(\dfrac{d^2y}{dx^2}\) and simplify it using the original equation.
Hint 1Start from \(y' = -x/y\) (problem 1) and differentiate again with the quotient rule.
Hint 2When you differentiate \(y\) in the denominator you get another \(y'\) — substitute \(-x/y\) for it.
Hint 3At the end, \(x^2+y^2\) will appear; replace it by 25.
Solution
STEP 1 · start from the first derivative
\[ y' = -\frac{x}{y} \]
STEP 2 · differentiate again (quotient rule)
Treat \(-x/y\) as a quotient with \(u=-x\), \(v=y\), remembering \(\frac{d}{dx}y = y'\):
On the upper semicircle \(y>0\), so \(y''<0\) — concave down, as a dome should be. On the lower semicircle \(y<0\) and \(y''>0\) — concave up ✓. Substituting the original relation at the end is standard practice and usually produces the tidiest form.
6
Find where \(x^2 + xy + y^2 = 12\) has a vertical tangent.
Hint 1Vertical tangent means the slope is undefined — so the denominator of \(y'\) is zero (while the numerator isn't).
Hint 2From problem 3, \(y' = -\frac{2x+y}{x+2y}\). Set \(x + 2y = 0\).
Hint 3Substitute \(x = -2y\) into the original equation.
STEP 2 · set the denominator to zero
\[ x + 2y = 0 \quad\Longrightarrow\quad x = -2y \]
STEP 3 · substitute into the curve
\[ (-2y)^2 + (-2y)(y) + y^2 = 4y^2 - 2y^2 + y^2 = 3y^2 = 12 \ \Longrightarrow\ y = \pm2 \]
STEP 4 · list the points and confirm
\(y=2 \Rightarrow x=-4\); \(y=-2 \Rightarrow x=4\). Check the numerator is not also zero at \((-4,2)\): \(2(-4)+2 = -6 \ne 0\) ✓, so the slope genuinely blows up rather than being \(0/0\).
\[ (-4,\,2) \quad\text{and}\quad (4,\,-2) \]
WHY THIS MATTERS
These are exactly the points where no function \(y = f(x)\) could describe the curve — it fails the vertical line test there. Implicit differentiation is what lets you talk about tangents at such points at all, and \(F_y = 0\) is the general signal that you have found one.
7
Use logarithmic differentiation to find \(\dfrac{dy}{dx}\) for \(y = x^x\), \(x>0\).
Hint 1Neither the power rule nor the exponential rule applies — the base and the exponent both vary.
Hint 2Take \(\ln\) of both sides first: \(\ln y = x\ln x\). Now differentiate implicitly.
Hint 3The left side gives \(\frac{y'}{y}\) by the chain rule; the right side needs the product rule.
Solution
STEP 1 · why the usual rules fail
\(\frac{d}{dx}x^n = nx^{n-1}\) needs a constant exponent; \(\frac{d}{dx}a^x = a^x\ln a\) needs a constant base. Here both move, so neither applies.
STEP 2 · take logs of both sides
The log turns the exponent into a coefficient — that is the whole point:
\[ \ln y = \ln\left(x^x\right) = x\ln x \]
STEP 3 · differentiate implicitly
Left side: chain rule, \(\frac{d}{dx}\ln y = \frac{y'}{y}\). Right side: product rule on \(x\cdot\ln x\):
\[ \frac{y'}{y} = (1)\ln x + x\cdot\frac1x = \ln x + 1 \]
STEP 4 · solve for y′
\[ y' = y\left(\ln x + 1\right) \]
STEP 5 · substitute y back
\[ \frac{dy}{dx} = x^x\left(\ln x + 1\right) \]
BONUS
Setting \(y'=0\) gives \(\ln x = -1\), i.e. \(x = 1/e \approx 0.368\) — the famous minimum of \(x^x\), with value \((1/e)^{1/e} \approx 0.6922\). The same log trick also tames products of many factors, e.g. \(y = \frac{(x+1)^3\sqrt{x-2}}{(x^2+1)^5}\).
8
Show that the circles \(x^2+y^2 = C\) and the lines \(y = kx\) always meet at right angles (they are orthogonal trajectories).
Hint 1Find the slope of each family at a general point \((x,y)\) — one implicitly, one directly.
Hint 2For the line \(y = kx\), note \(k = y/x\), so its slope at the point is \(y/x\).
Hint 3Perpendicular means the two slopes multiply to \(-1\).
Solution
STEP 1 · slope of the circles
Differentiate \(x^2+y^2 = C\) implicitly (the constant \(C\) dies, which is why the answer works for the whole family at once):
A product of \(-1\) means the tangents are perpendicular at every intersection point (away from the axes, where one slope is undefined). The families are orthogonal.
ODE CONNECTION
This is a standard first-order ODE application. To find the orthogonal trajectories of a family, you (1) write its ODE by eliminating the constant, (2) replace \(y'\) with \(-1/y'\), and (3) solve the new ODE. Here: circles satisfy \(y' = -x/y\); flipping gives \(y' = y/x\), which separates to \(\ln|y| = \ln|x| + c\), i.e. \(y = kx\) ✓. Field lines and equipotential curves in physics are exactly this construction.
9
A 5 m ladder leans against a wall, so \(x^2+y^2=25\). The base slides away at \(\frac{dx}{dt}=0.6\) m/s. How fast is the top falling when \(x=3\)?
Hint 1Differentiate the constraint with respect to time, treating both \(x\) and \(y\) as functions of \(t\).
Hint 2\(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\).
Hint 3Find \(y\) when \(x=3\) from the constraint itself.
Solution
STEP 1 · differentiate implicitly in t
Every term gets the chain rule, since \(x\) and \(y\) both depend on \(t\):
\[ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \]
STEP 2 · solve for the unknown rate
\[ \frac{dy}{dt} = -\frac{x}{y}\cdot\frac{dx}{dt} \]
STEP 3 · find y at the instant in question
\[ 3^2+y^2 = 25 \ \Longrightarrow\ y = 4 \]
As \(y\to0\) the factor \(\frac xy\to\infty\), so the top appears to fall infinitely fast — the model's way of saying the ladder has left the wall. Notice this is module D6's \(\frac{dy}{dx}=-\frac xy\) with an extra chain-rule factor of \(\frac{dx}{dt}\): related rates is implicit differentiation, with time as the hidden variable.
Part I · Module 7 Differentiation
Partial derivatives
When a function has two inputs, "the" rate of change is ambiguous — you must say with respect to which variable. A partial derivative freezes all the other variables and differentiates in one direction only.
Notice the \(y\) sits unchanged in the first limit. Practically: to compute \(f_x\), treat \(y\) as a constant — like a 7. Everything you already know about differentiation applies unchanged.
Geometrically, \(f_x\) is the slope of the surface along a slice with \(y\) held fixed — you are walking due east on the hillside and asking how steeply you are climbing.
InteractiveTwo slices through a surfacedrag the point on the contour map
point —f(x,y) —∂f/∂x —∂f/∂y —
Left: a contour map (warm = high, cool = low) with your point and the two slice directions. Right: the two one-variable functions you actually differentiate — top is \(f(x, y_0)\) with its tangent slope \(f_x\), bottom is \(f(x_0, y)\) with slope \(f_y\). On the saddle, find the point where both slopes are zero and notice it is neither a peak nor a pit.
Second partials and the mixed-partial test
You can differentiate twice: \(f_{xx}\), \(f_{yy}\), and the mixed partials \(f_{xy} = \partial_y(\partial_x f)\) and \(f_{yx}\). For every function you will meet in this course, \(f_{xy} = f_{yx}\) (Clairaut's theorem). That symmetry is the entire basis of the exactness test for ODEs.
Worked example · the exactness test
Is \((2xy + 3)\,dx + (x^2 - 1)\,dy = 0\) exact?
STEP 1
Name the pieces: \(M = 2xy + 3\), \(N = x^2 - 1\).
STEP 2
\(\dfrac{\partial M}{\partial y} = 2x\) (treat \(x\) as constant; the \(3\) dies).
STEP 3
\(\dfrac{\partial N}{\partial x} = 2x\) (treat \(y\) as constant; the \(-1\) dies).
STEP 4
They match, so the equation is exact: there is a potential \(F\) with \(F_x = M\) and \(F_y = N\). Integrating \(M\) in \(x\) gives \(F = x^2y + 3x + g(y)\); matching \(F_y = x^2 + g'(y)\) to \(N = x^2-1\) gives \(g(y) = -y\). The solution is \(x^2y + 3x - y = C\).
②Practice
1
For \(f(x,y) = x^3y - 4xy^2 + y\), find \(f_x\), \(f_y\), and confirm \(f_{xy} = f_{yx}\).
Hint 1For \(f_x\), the \(y\)'s are just coefficients — \(x^3y\) differentiates like \(3x^2\) times the constant \(y\). The lone \(+y\) has no \(x\), so it contributes 0.
Hint 2Then differentiate \(f_x\) with respect to \(y\), and separately differentiate \(f_y\) with respect to \(x\).
Solution
STEP 1 · f_x — freeze y, treat it like the number 7
Term by term: \(x^3y \to 3x^2y\); \(-4xy^2 \to -4y^2\) (the \(x\) has exponent 1); \(+y \to 0\) (no \(x\) present at all):
STEP 4 · differentiate that F in y and match N
\[ F_y = x^3 + g'(y) \quad\text{must equal}\quad N = x^3 + 4y \]
\[ \Longrightarrow\ g'(y) = 4y \ \Longrightarrow\ g(y) = 2y^2 \]
STEP 5 · write the solution
The solution curves are the level sets \(F = C\):
\[ x^3y + 2x + 2y^2 = C \]
WHY IT WORKS
The equation is really \(dF = F_x\,dx + F_y\,dy = 0\), i.e. "\(F\) doesn't change along the curve." Recognizing that shape is exactly module D8's total differential, and the test \(M_y=N_x\) is just Clairaut's theorem read backwards.
3
Verify that \(u(x,t) = e^{-t}\sin x\) satisfies the heat equation \(u_t = u_{xx}\).
Hint 1\(u_t\) treats \(x\) as constant, so \(\sin x\) is just a number multiplying \(e^{-t}\).
Hint 2\(u_{xx}\) means differentiate twice in \(x\), holding \(t\) fixed — two derivatives of \(\sin x\) give \(-\sin x\).
Solution
STEP 1 · differentiate in t
Freeze \(x\), so \(\sin x\) is a constant multiplier:
\[ u_t = \sin x \cdot \frac{\partial}{\partial t}e^{-t} = -e^{-t}\sin x \]
STEP 2 · differentiate once in x
Now freeze \(t\), so \(e^{-t}\) is the constant:
\[ u_x = e^{-t}\cos x \]
STEP 3 · differentiate again in x
\[ u_{xx} = -e^{-t}\sin x \]
A temperature profile shaped like one arch of a sine, decaying in time. Physically: the hotter middle bleeds heat into the cooler ends, and the whole profile flattens exponentially without changing shape. Because two independent variables appear, this is a partial differential equation — the contrast that gives "ordinary differential equations" its name.
4
For \(f(x,y) = e^{xy}\), find \(f_x\), \(f_y\), \(f_{xx}\), and \(f_{xy}\).
Hint 1For \(f_x\), the exponent is \(xy\) with \(y\) frozen, so the inner derivative is just \(y\).
Hint 2\(f_x = ye^{xy}\). For \(f_{xy}\), differentiate that in \(y\) — and notice it is now a product of \(y\) and \(e^{xy}\).
Solution
STEP 1 · f_x
Chain rule with \(y\) held constant — the inner derivative of \(xy\) with respect to \(x\) is \(y\):
\[ f_x = y\,e^{xy} \]
STEP 2 · f_y
By the symmetry of the formula, freezing \(x\) gives inner derivative \(x\):
\[ f_y = x\,e^{xy} \]
STEP 3 · f_xx
Differentiate \(f_x = ye^{xy}\) again in \(x\). Here \(y\) is still a constant coefficient:
\[ f_{xx} = y\cdot y e^{xy} = y^2 e^{xy} \]
STEP 4 · f_xy
Now differentiate \(f_x = ye^{xy}\) with respect to \(y\). This time \(y\) appears twice — as the coefficient and inside the exponent — so the product rule is required:
Differentiating \(f_y = xe^{xy}\) with respect to \(x\) gives \(e^{xy} + x(ye^{xy}) = e^{xy}(1+xy)\) — the same ✓.
WATCH FOR
The step where a variable is both a coefficient and inside the function is exactly where people forget the product rule. Scan for repeated variables before differentiating.
5
Test \((3xy + y^2)\,dx + (x^2 + xy)\,dy = 0\) for exactness. What does your answer mean for how you'd solve it?
Hint 1Same test as before: compute \(M_y\) and \(N_x\).
Hint 2\(M_y = 3x + 2y\) and \(N_x = 2x + y\). Are they equal for all \((x,y)\)?
Hint 3"Equal at one lucky point" is not enough — the test must hold identically.
Solution
STEP 1 · label and differentiate
\[ M = 3xy + y^2 \ \Longrightarrow\ M_y = 3x + 2y \]
\[ N = x^2 + xy \ \Longrightarrow\ N_x = 2x + y \]
STEP 2 · compare
\[ 3x + 2y \ne 2x + y \quad\text{(they agree only along the line } x + y = 0\text{)} \]
Since the test fails identically, the equation is not exact.
STEP 3 · what that rules out
There is no function \(F\) with \(F_x = M\) and \(F_y = N\), so you cannot solve it by reconstructing a potential. Stopping here and writing "not exact" is a complete answer to the question as asked.
STEP 4 · what you would do instead
Look for an integrating factor \(\mu\) making \(\mu M\,dx + \mu N\,dy = 0\) exact. The standard test:
Because that depends on \(x\) alone, \(\mu = e^{\int \frac1x dx} = x\) works. Multiplying through by \(x\) gives \((3x^2y + xy^2)dx + (x^3+x^2y)dy = 0\), and now \(M_y = 3x^2 + 2xy = N_x\) ✓ exact.
TAKEAWAY
"Not exact" is rarely the end of the story — it is the trigger for the integrating-factor method, which is one of the workhorses of a first-order ODE chapter.
6
For \(z = x^2y - y^3\), compute \(z_x\) and \(z_y\) at the point \((2,1)\) and say in words what each number means.
Hint 1Differentiate symbolically first, and only substitute \((2,1)\) at the end.
Hint 2Each partial is a slope of a slice: \(z_x\) walking in the \(x\)-direction, \(z_y\) in the \(y\)-direction.
Solution
STEP 1 · z_x
Freeze \(y\): the term \(-y^3\) has no \(x\), so it dies.
\[ z_x = 2xy \]
STEP 2 · z_y
Freeze \(x\): now \(x^2\) is a constant coefficient.
Standing on the surface above \((2,1)\): walking one unit in the \(+x\) direction, the height climbs at about 4 units per unit of \(x\). Walking in the \(+y\) direction, it climbs at only 1 unit per unit of \(y\). The surface is four times steeper east–west than north–south at that spot.
STEP 5 · combine them
The total differential packages both: \(dz = 4\,dx + 1\,dy\). So a small step \((dx,dy) = (0.1, 0.2)\) changes \(z\) by roughly \(0.4 + 0.2 = 0.6\). That linear estimate is exactly module D8's idea in two variables.
7
Verify that \(u(x,t) = \sin(x - 3t)\) satisfies the wave equation \(u_{tt} = 9\,u_{xx}\). What is the physical meaning of the 3?
Hint 1Differentiate twice in \(t\) — each \(t\)-derivative pulls out a factor \(-3\) by the chain rule.
Hint 2Differentiate twice in \(x\) — each \(x\)-derivative pulls out a factor 1.
Hint 3Compare the two results; the constant that appears is \((-3)^2\).
Solution
STEP 1 · derivatives in t
Inner function \(x-3t\) has \(t\)-derivative \(-3\):
The combination \(x - 3t\) means the whole waveform slides to the right: the point where the argument equals a fixed value satisfies \(x = 3t + \text{const}\). So the shape travels at speed 3, and the equation's coefficient is the speed squared.
GENERAL FACT
Any \(u = f(x - ct)\) with \(f\) twice differentiable solves \(u_{tt} = c^2u_{xx}\) — travelling waves of any shape, which is d'Alembert's solution. Verifying it takes only the chain rule you already have.
8
For \(f(x,y) = x^3y^2\), compute all four second partials \(f_{xx}, f_{yy}, f_{xy}, f_{yx}\).
Hint 1Get \(f_x\) and \(f_y\) first, then differentiate each of those twice.
True value: \(1.1^2+1.9^2 = 1.21+3.61 = 4.82\). The estimate is low by \(0.02\) — second-order error, exactly as in the one-variable case (module D8).
THE CONNECTION
The tangent plane is the total differential written out: \(dz = f_x\,dx + f_y\,dy\) with \(dx = 0.1\), \(dy = -0.1\) gives \(dz = 0.2-0.4 = -0.2\), and \(5 - 0.2 = 4.8\) ✓ Same computation, two pieces of notation.
Part I · Module 8 Differentiation
Differentials: dx, dy, dz
The symbol \(dy\) is not just decoration inside \(\frac{dy}{dx}\) — it is a quantity in its own right, and treating it as one is what makes separation of variables and exact equations legal moves rather than hand-waving.
①dy is the tangent's rise; Δy is the curve's rise
The differential of a one-variable function
\[ dy = f'(x)\,dx \]
Here \(dx\) is any input change you choose, and \(dy\) is the corresponding change predicted by the tangent line. The true change in the function is
True change vs. linear estimate
\[ \Delta y = f(x + \Delta x) - f(x) \qquad\text{vs.}\qquad dy = f'(x)\,\Delta x \]
They are not equal — but their difference shrinks faster than \(\Delta x\) itself, which is what "differentiable" really means. That is the whole content of the linear approximation \(f(x+\Delta x) \approx f(x) + f'(x)\Delta x\).
Interactivedy versus Δyshrink \(\Delta x\) and watch the gap collapse
dy (tangent) —Δy (true) —error —error / Δx —
The right-hand column shows the error as a fraction of \(\Delta x\). Halve \(\Delta x\) and this ratio also roughly halves — the error is second order. That is why engineers can replace curves by tangent lines over small intervals and why Euler's method for ODEs works at all.
Read it as: the total change in \(z\) is the change caused by moving in \(x\), plus the change caused by moving in \(y\). This is the formula that connects Module 7 to differential equations directly — because if \(F(x,y) = C\) is constant along a curve, then \(dF = 0\), which says
Compare to \(V = \pi(25)(12) \approx 942\) cm³: about a 2.8% uncertainty. Note the radius term contributes more than twice as much as the height term — because \(r\) enters squared, its error is doubled in the propagation.
Where this shows up in ODEs
Separation of variables is a differentials argument: from \(\frac{dy}{dx} = g(x)h(y)\) you write \(\frac{dy}{h(y)} = g(x)\,dx\) and integrate both sides. That step is only legitimate because \(dy\) and \(dx\) are genuine differentials, not just notation.
③Practice
1
Use a differential to estimate \(\sqrt{16.4}\) without a calculator, then find the error.
Hint 1Anchor at a point where you know the answer exactly: \(a = 16\), \(\Delta x = 0.4\).
Hint 2\(f(x) = \sqrt x\), so \(f'(x) = \frac{1}{2\sqrt x}\) and \(f'(16) = \frac18\).
Solution
STEP 1 · choose the anchor point
Pick the nearest input whose value you know exactly. \(\sqrt{16} = 4\), so take \(a = 16\) and \(\Delta x = 0.4\).
STEP 3 · compute the differential
\[ dy = f'(a)\,\Delta x = \frac18(0.4) = 0.05 \]
STEP 4 · add it to the known value
\[ \sqrt{16.4} \approx f(16) + dy = 4 + 0.05 = 4.05 \]
STEP 5 · measure the error
The true value is \(4.0496913\ldots\), so the estimate is high by \(0.00031\) — about \(0.008\%\).
WHY IT OVERSHOOTS
\(\sqrt x\) is concave down (\(f'' = -\frac{1}{4}x^{-3/2} < 0\)), so its tangent line lies above the curve. Any linear estimate of a concave-down function is an overestimate — knowing the concavity tells you the direction of your error before you check.
2
Find the total differential of \(z = x^2\ln y\).
Hint 1Compute \(z_x\) treating \(\ln y\) as a constant, then \(z_y\) treating \(x^2\) as a constant.
Solution
STEP 1 · partial with respect to x
Freeze \(y\), so \(\ln y\) is just a number multiplying \(x^2\):
\[ z_x = 2x\ln y \]
STEP 2 · partial with respect to y
Freeze \(x\), so \(x^2\) is the constant and \(\frac{d}{dy}\ln y = \frac1y\):
Each term answers "how much does \(z\) change if I nudge only this variable?" At \((x,y) = (3, e)\): \(dz = 6\,dx + \frac{9}{e}\,dy \approx 6\,dx + 3.31\,dy\), so \(z\) is about twice as sensitive to a wobble in \(x\) as in \(y\) there.
3
Rewrite \(\dfrac{dy}{dx} = \dfrac{x^2}{y}\) in differential form \(M\,dx + N\,dy = 0\), then separate and solve it.
Hint 1Cross-multiply: \(y\,dy = x^2\,dx\), then move everything to one side.
Hint 2Integrate each side separately — one constant \(C\) on the right is enough.
Solution
STEP 1 · treat dy and dx as genuine quantities
Multiply both sides by \(y\,dx\). This is legitimate precisely because \(dy\) and \(dx\) are differentials, not decoration:
Each differential is paired with a function of its own variable only — \(x\) with \(dx\), \(y\) with \(dy\). That is exactly the condition for separation.
STEP 4 · integrate both sides
\[ \int y\,dy = \int x^2\,dx \quad\Longrightarrow\quad \frac{y^2}{2} = \frac{x^3}{3} + C \]
One constant suffices: a \(C_1\) on the left would just merge into \(C_2 - C_1\).
STEP 5 · tidy up
Multiply by 6 and rename the constant:
\[ 3y^2 - 2x^3 = C' \]
VERIFY
Differentiate implicitly: \(6yy' - 6x^2 = 0 \Rightarrow y' = x^2/y\) ✓ — back where you started. You have just solved a separable ODE using nothing beyond Part I.
4
Estimate \((2.02)^5\) using a differential, and state whether your estimate is high or low before checking.
Hint 1Anchor at \(a = 2\), where \(2^5 = 32\) is easy, with \(\Delta x = 0.02\).
Hint 2\(f'(x) = 5x^4\), so \(f'(2) = 80\).
Hint 3For high-or-low, check the sign of \(f''(x) = 20x^3\) at \(x=2\).
Solution
STEP 1 · anchor and step
\[ f(x) = x^5, \qquad a = 2, \qquad \Delta x = 0.02, \qquad f(2) = 32 \]
STEP 2 · derivative at the anchor
\[ f'(x) = 5x^4 \quad\Longrightarrow\quad f'(2) = 5(16) = 80 \]
\(f'' = 20x^3 = 160 > 0\) at \(x=2\), so the curve is concave up and its tangent line lies below it. The estimate should be a slight underestimate.
CHECK
True value: \(33.6242\ldots\) — the estimate is low by \(0.024\), as predicted, an error of \(0.07\%\).
5
A cube's edge is measured as \(10\) cm with a possible error of \(\pm 0.05\) cm. Estimate the resulting uncertainty in the volume, in cm³ and as a percentage.
Hint 1\(V = s^3\), so \(dV = 3s^2\,ds\) — the differential converts an input error into an output error.
Hint 2Percentage error is \(dV/V\); try simplifying \(\frac{3s^2\,ds}{s^3}\) before plugging in numbers.
Solution
STEP 1 · the model
\[ V = s^3 \quad\Longrightarrow\quad \frac{dV}{ds} = 3s^2 \quad\Longrightarrow\quad dV = 3s^2\,ds \]
STEP 2 · substitute the measurement and its error
\[ dV = 3(10)^2(0.05) = 15\ \text{cm}^3 \]
So \(V \approx 1000 \pm 15\) cm³.
STEP 3 · relative error, symbolically
Divide the differential by the quantity itself — the \(s\)'s largely cancel:
For \(V = s^n\), the relative error multiplies by \(n\): \(\frac{dV}{V} = n\frac{ds}{s}\). A 0.5% error in a length becomes 1.5% in a volume, and 1% in a radius becomes 4% in the area of a sphere. This is why precision matters most on quantities that enter with high powers.
6
Use a differential to estimate \(\sin(31^\circ)\), given \(\sin 30^\circ = 0.5\) and \(\cos 30^\circ = 0.8660\).
Hint 1The derivative formula \(\frac{d}{dx}\sin x = \cos x\) is only valid in radians, so convert \(\Delta x\) first.
Hint 2\(1^\circ = \frac{\pi}{180} \approx 0.01745\) radians.
Hint 3Then \(dy = \cos(30^\circ)\cdot\Delta x\) with \(\Delta x\) in radians.
Solution
STEP 1 · convert the step to radians
This is the step everyone skips, and it silently ruins the answer:
STEP 2 · the differential
\[ dy = \cos(30^\circ)\,\Delta x = (0.8660)(0.017453) \approx 0.015115 \]
STEP 3 · add to the known value
\[ \sin(31^\circ) \approx 0.5 + 0.015115 = 0.515115 \]
STEP 4 · check
True value: \(0.5150381\). The estimate is off by \(0.00008\) — five-figure accuracy from one multiplication.
WHAT GOES WRONG WITHOUT THE CONVERSION
Using \(\Delta x = 1\) (degrees) would give \(\sin 31^\circ \approx 1.366\) — impossible, since sine never exceeds 1. Absurd answers like that are usually a units error, not an algebra error.
7
For \(z = \dfrac{x}{y}\), find \(dz\), then estimate the change in \(z\) when \(x\) goes from \(10\) to \(10.2\) and \(y\) goes from \(5\) to \(4.9\).
Hint 1Write \(z = xy^{-1}\) to make \(z_y\) easier.
Hint 2\(z_x = \frac1y\) and \(z_y = -\frac{x}{y^2}\) — note the minus: raising \(y\) lowers \(z\).
Hint 3Here \(dx = +0.2\) and \(dy = -0.1\). Keep the signs.
Both effects push the same way: the numerator grew and the denominator shrank, so \(z\) rises by about \(0.08\) from its starting value \(z = 2\).
CHECK AGAINST THE TRUE CHANGE
\[ \frac{10.2}{4.9} - \frac{10}{5} = 2.08163 - 2 = 0.08163 \]
The linear estimate captured it to within \(0.0016\). Notice how the two signs conspired — dropping the minus in \(z_y\) would have given \(0\) and hidden the effect entirely.
8
Use differentials to take two steps of Euler's method on \(\dfrac{dy}{dx} = x + y\), \(y(0) = 1\), with step size \(h = 0.1\). Compare with the exact solution \(y = 2e^{x} - x - 1\).
Hint 1Euler's method is the differential: \(y_{\text{new}} \approx y_{\text{old}} + y'\cdot h\), where \(y'\) comes from the ODE itself.
Hint 2At \((0,1)\), the ODE gives \(y' = 0 + 1 = 1\). Step to \(x=0.1\).
Hint 3Recompute \(y'\) at the new point before taking the second step — that is the whole idea.
Solution
STEP 1 · the method in one line
The differential says \(dy = y'\,dx\). Replace \(dx\) by the finite step \(h\) and \(y'\) by whatever the ODE says the slope is at your current point:
Euler gave \(1.22\), low by about \(0.023\) (1.8%).
STEP 5 · why it under-shoots
The true solution is concave up here, so each straight tangent step cuts the corner and lands below the curve. Halving \(h\) roughly halves the error — Euler's method is first order, exactly the "error proportional to \(\Delta x\)" behavior the lab in this module displays.
WHY THIS MATTERS
Most differential equations have no closed-form solution. When that happens, this is what you do instead — and every numerical ODE solver, from Euler to Runge–Kutta, is a refinement of the single idea \(dy = f(x,y)\,dx\).
9
Resistance is computed from \(R = V/I\). If \(V\) is measured to within 2% and \(I\) to within 1%, estimate the worst-case percentage error in \(R\).
Hint 1Use the total differential, then divide by \(R\) to get relative error.
Hint 2\(R_V = \frac1I\) and \(R_I = -\frac{V}{I^2}\).
Hint 3Worst case: the two errors push the same way, so add their magnitudes.
Solution
STEP 1 · total differential
\[ dR = \frac{\partial R}{\partial V}dV + \frac{\partial R}{\partial I}dI = \frac{dV}{I} - \frac{V}{I^2}dI \]
STEP 2 · convert to relative error
Divide by \(R = V/I\):
\[ \frac{dR}{R} = \frac{dV}{V} - \frac{dI}{I} \]
STEP 3 · substitute the bounds
\(\left|\frac{dV}{V}\right| \le 0.02\) and \(\left|\frac{dI}{I}\right| \le 0.01\).
STEP 4 · worst case
The two terms have opposite signs, so the worst case is when the errors run in opposite directions — high \(V\) with low \(I\):
so exponents multiply the percentage errors and you add magnitudes for the worst case. (Statisticians add them in quadrature instead, \(\sqrt{2^2+1^2}\approx2.2\%\), when the errors are independent and random.)
Part I · Module 9 Differentiation
Notation: five ways to say the same thing
Your ODE textbook will switch notations mid-chapter without warning, and your physics course uses a sixth. None of them mean anything different — but you have to read all of them at a glance.
Flip through all five and read the same four statements each time. The goal is not to pick a favorite — it is to stop noticing the difference.
①The field guide
Notation
First derivative
Second
Where you'll meet it
Leibniz
\(\dfrac{dy}{dx}\), \(\dfrac{d}{dx}f(x)\)
\(\dfrac{d^2y}{dx^2}\)
Integration, separation of variables, chain rule, related rates. Best when the independent variable must be explicit.
Prime (Lagrange)
\(f'(x)\), \(y'\)
\(f''(x)\), \(y''\)
The default for ODEs: \(y'' + 3y' + 2y = 0\). Compact, but hides what you're differentiating with respect to.
Dot (Newton)
\(\dot{x}\)
\(\ddot{x}\)
Physics and mechanics — and always means the derivative with respect to time.
Subscript
\(f_x\), \(u_t\)
\(f_{xx}\), \(u_{xy}\)
Partial derivatives and PDEs. \(u_{xy}\) means \(x\) first, then \(y\).
Operator
\(D f\), \(D_x f\)
\(D^2 f\)
Operator methods, annihilators, and Laplace-transform shortcuts. Lets you factor: \((D^2+3D+2)y = (D+1)(D+2)y\).
②Reading the notation carefully
\(\dfrac{d^2y}{dx^2}\) — the "2" is upstairs on the \(d\) and downstairs on the \(x\). It is not \(\left(\frac{dy}{dx}\right)^2\), which would be the derivative squared. Those are wildly different objects.
\(f'(2)\) means "differentiate first, then evaluate at 2." Evaluating first would give the derivative of a constant, which is 0. This is the single most common notational error in a calculus course.
\(y^{(4)}\) with parentheses is the fourth derivative; \(y^4\) without them is the fourth power.
\(\dfrac{dy}{dx}\bigg|_{x=a}\) is the standard way to say "\(f'(a)\)" in Leibniz notation.
In an ODE like \(y'' + p(x)y' + q(x)y = 0\), the primes are all with respect to \(x\) — the equation tells you the independent variable by which letters appear in the coefficients.
Why Leibniz notation earns its clunkiness
The chain rule \(\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}\), the substitution rule \(du = g'(x)dx\), and separation of variables all look like fraction algebra in Leibniz notation. That is not a coincidence — Leibniz designed the notation so the formal manipulation would come out right.
③Practice
1
Translate into prime notation: \(\dfrac{d^2s}{dt^2} + 4\dfrac{ds}{dt} + 3s = \cos t\). Then into dot and operator notation.
Hint 1The independent variable is \(t\), so primes and dots both mean \(d/dt\) here.
Hint 2In operator form, treat \(D\) as an object you can factor like a polynomial.
Solution
STEP 1 · identify the independent variable
Everything is differentiated with respect to \(t\), and \(s\) is the unknown function. So \(s' \equiv \frac{ds}{dt}\).
STEP 2 · prime notation
\[ s'' + 4s' + 3s = \cos t \]
STEP 3 · dot notation
Newton's dots always mean time derivatives, so this translation is safe here:
\[ \ddot{s} + 4\dot{s} + 3s = \cos t \]
STEP 4 · operator notation
Write \(D = \frac{d}{dt}\) and factor out \(s\):
\[ \left(D^2 + 4D + 3\right)s = \cos t \]
STEP 5 · factor the operator
The polynomial in \(D\) factors just like an ordinary quadratic:
\[ (D+1)(D+3)\,s = \cos t \]
WHY THE LAST FORM IS USEFUL
The roots \(-1\) and \(-3\) are the exponents in the homogeneous solution \(s_h = C_1e^{-t} + C_2e^{-3t}\). The operator notation makes the algebraic structure of a differential equation visible — you are literally factoring the equation.
Hint 1In (b) and (c), what kind of object is \(f(2)\)?
Hint 2\(f(2) = 8\) is a number, and numbers have derivative zero.
Solution
STEP 1 · part (a) — differentiate, then evaluate
\[ f'(x) = 3x^2 \quad\Longrightarrow\quad f'(2) = 3(4) = 12 \]
STEP 2 · part (b) — evaluate, then differentiate
\(f(2) = 8\) is a constant. The derivative of a constant is zero:
\[ \left[f(2)\right]' = \left[8\right]' = 0 \]
STEP 3 · part (c) — same thing in Leibniz clothing
\[ \frac{d}{dx}f(2) = \frac{d}{dx}(8) = 0 \]
STEP 4 · the rule to remember
Differentiate first, evaluate second. \(f'(2)\) means "the slope of \(f\) at the point where \(x=2\)" — you cannot ask for the slope of a single number.
WHERE THIS BITES
In initial value problems you will write things like \(y'(0) = 3\). That means "the derivative function, evaluated at 0" — not the derivative of the number \(y(0)\), which would always be zero and make the condition meaningless.
3
What does \(u_{xt}\) mean, and what other symbol equals it for well-behaved \(u\)?
Hint 1Read subscripts left to right in the order the derivatives were taken.
Solution
STEP 1 · decode the subscripts
Subscripts are applied left to right: first \(x\), then \(t\).
By Clairaut's theorem, if both mixed partials are continuous then the order does not matter:
\[ u_{xt} = u_{tx} \]
STEP 4 · in words
"How fast is the \(x\)-slope changing as time passes?" — equivalently, "how fast is the time-rate changing as you move in \(x\)?" Those two questions have the same answer.
CAUTION
The reversal between subscript and Leibniz ordering is a genuine trap when reading textbooks. When in doubt, translate to words rather than trusting the visual order.
4
Expand the operator equation \((D-2)(D+3)y = 0\) into ordinary prime notation, and write down its solution.
Hint 1Multiply the two factors as if \(D\) were a variable: \((D-2)(D+3) = D^2 + D - 6\).
Hint 2Then translate: \(D^2y = y''\), \(Dy = y'\).
Hint 3Each factor \((D-r)\) contributes a solution \(e^{rx}\).
Solution
STEP 1 · multiply the operator factors
\[ (D-2)(D+3) = D^2 + 3D - 2D - 6 = D^2 + D - 6 \]
Factoring \(D^2+D-6\) is ordinary algebra, but it hands you the complete solution of a differential equation. This is the notation's whole selling point, and it extends to the annihilator method for nonhomogeneous problems.
5
Explain, with a concrete example, why \(\dfrac{d^2y}{dx^2}\) and \(\left(\dfrac{dy}{dx}\right)^2\) are completely different objects.
Hint 1Pick a simple function like \(y = x^2\) and compute both quantities.
Hint 2One is "differentiate twice"; the other is "differentiate once, then square."
Solution
STEP 1 · pick a test function
Take \(y = x^2\), so \(\frac{dy}{dx} = 2x\).
STEP 2 · compute the second derivative
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(2x) = 2 \qquad \text{(a constant)} \]
\(2 \ne 4x^2\) for all but two values of \(x\). One is constant, the other unbounded — they are not even the same shape.
STEP 5 · read the notation carefully
In \(\frac{d^2y}{dx^2}\), the "2" upstairs sits on the operator \(d\), and downstairs on the variable \(x\) — it is shorthand for \(\frac{d}{dx}\left(\frac{dy}{dx}\right)\), not an exponent on anything. Squaring would be written \(\left(\frac{dy}{dx}\right)^2\), with explicit parentheses.
WHY IT MATTERS FOR ODEs
\(y'' + y = 0\) is linear and easy; \((y')^2 + y = 0\) is nonlinear and generally hard. Misreading the notation changes not only the answer but the entire method you should reach for.
6
Rewrite the ODE \(y'' - 5y' + 6y = 0\) in (a) Leibniz notation with independent variable \(x\), (b) operator notation, factored, and (c) dot notation, explaining when (c) is appropriate.
Hint 1Primes here must mean \(d/dx\) — the equation shows no other variable.
Hint 2Factor \(r^2 - 5r + 6\) the ordinary way.
Hint 3Dots carry an assumption about which variable is which.
But this is only appropriate if the independent variable is time. Newton's dots are a physics convention meaning \(d/dt\) specifically; writing \(\dot y\) for \(dy/dx\) will confuse every reader with a mechanics background.
STEP 5 · which to use when
Situation
Best notation
Stating an ODE compactly
primes
Separating variables / integrating
Leibniz (you need \(dx\), \(dy\))
Mechanics, time evolution
dots
Factoring, annihilators
operator \(D\)
Several independent variables
subscripts
7
A quantity satisfies \(\dfrac{dy}{dx}\bigg|_{x=3} = -2\). Write the same statement in prime, dot, and operator notation, and say what it means physically if \(y\) is a temperature and \(x\) is time in minutes.
Hint 1The vertical bar with a subscript means "evaluated at."
Hint 2Careful with dot notation: it presumes the variable is time — which the physical reading here does supply.
Solution
STEP 1 · decode the original
"Take the derivative of \(y\) with respect to \(x\), then evaluate it at \(x=3\); the result is \(-2\)."
STEP 2 · prime notation
\[ y'(3) = -2 \]
STEP 3 · dot notation
Legitimate here only because the independent variable is time:
Note the parentheses: you apply \(D\) to the function first, then evaluate. \(D\big(y(3)\big)\) would be the derivative of a constant, i.e. 0.
STEP 5 · physical reading
At the 3-minute mark the temperature is falling at 2 degrees per minute. Over the next 30 seconds it should drop about \(2 \times 0.5 = 1\) degree — a differential estimate (module D8).
IN CONTEXT
Statements like this are exactly the initial conditions of an IVP. A second-order equation needs two of them, typically \(y(0)\) and \(y'(0)\) — position and velocity.
8
Decide what each symbol means and whether it is a function or a number: (a) \(y^{(4)}\), (b) \(y^4\), (c) \(f''(x)\), (d) \(f''(1)\), (e) \(\frac{d}{dx}\left[f(x)g(x)\right]\).
Hint 1Parentheses in a superscript signal a derivative order, not a power.
Hint 2Anything evaluated at a specific input is a number; anything left in terms of \(x\) is a function.
Solution
(a) y⁽⁴⁾
The fourth derivative of \(y\) — parentheses mark the order. It is a function. (For \(y=\sin x\) it equals \(\sin x\) again, by the cycle in module D4.)
(b) y⁴
The function \(y\) raised to the fourth power. Also a function — but a completely different one. For \(y = \sin x\) this is \(\sin^4 x\), not \(\sin x\).
(c) f″(x)
The second derivative, left as a function of \(x\). It tells you the concavity at every point.
(d) f″(1)
That same function evaluated at \(x=1\): a single number. Its sign says whether the graph is concave up or down at that one point.
(e) d/dx[f(x)g(x)]
An instruction to differentiate the product, whose result is the function \(f'g + fg'\). The \(\frac{d}{dx}[\ \cdot\ ]\) form is an operator awaiting an argument, which is why it is the clearest notation when you are stating a rule.
THE HABIT
Before computing anything, ask: is this a function or a number? That single question prevents most notation errors — including the very common \(f'(2)\) versus \([f(2)]'\) confusion from problem 2.
9
Write the chain rule in Leibniz, prime, and operator notation. Which form makes the "rates multiply" idea obvious, and why is that not an actual cancellation?
Hint 1Leibniz: \(\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}\).
Hint 2Prime form must name the argument at which the outer derivative is evaluated.
Leibniz, by a distance: the \(du\)'s appear to cancel, which correctly suggests "if \(y\) changes 3× as fast as \(u\), and \(u\) changes 5× as fast as \(x\), then \(y\) changes 15× as fast as \(x\)."
STEP 5 · why it isn't really cancellation
\(\frac{dy}{du}\) is a single symbol for a limit, not a quotient of two numbers, so nothing literally cancels. The notation was designed so the formal manipulation gives the right answer — a design success, not a theorem. The proof still requires the limit argument.
WHERE IT PAYS OFF
The same fiction makes \(u\)-substitution (\(du = g'(x)dx\)) and separation of variables (\(\frac{dy}{h(y)} = g(x)dx\)) look like algebra. Both are legitimate — but because of theorems, not because fractions were cancelled.
Part I · Checkpoint Quiz
Quiz A — Differentiation
Ten questions across the whole derivative toolkit. Tap an answer to lock it in and see the reasoning. Aim for 8/10 before starting Part II.
Question 1
The derivative \(f'(a)\) equals which of the following?
Why
A and B are average rates over an interval. The derivative is the limiting instantaneous rate, which geometrically is the tangent slope.
Question 2
\(\dfrac{d}{dx}\left[\dfrac{4}{x^3}\right] = \) ?
Why
Rewrite as \(4x^{-3}\). Power rule: \(4(-3)x^{-4} = -12x^{-4} = -\frac{12}{x^4}\). The negative exponent makes the derivative negative — and \(4/x^3\) is indeed decreasing for \(x>0\).
Question 3
If \(y = x^2\cos x\), then \(y' = \) ?
Why
Product rule: \(u'v + uv' = 2x\cos x + x^2(-\sin x)\). Choice A is the "multiply the derivatives" error; B loses the minus from \(\frac{d}{dx}\cos x\).
Question 4
\(\dfrac{d}{dx}\sin(5x) = \) ?
Why
Chain rule: outer derivative \(\cos(5x)\) times inner derivative \(5\). Forgetting that factor of 5 is the most common error in the entire course.
Question 5
A particle has \(v(t) < 0\) and \(a(t) < 0\) at time \(t_0\). At that instant the particle is:
Why
Negative velocity means moving left. Velocity and acceleration share a sign, so the speed \(|v|\) is increasing — it is speeding up (in the negative direction).
\(2x + 2y y' = 0 \Rightarrow y' = -x/y\). Choice D forgets that \(y\) is a function of \(x\) and skips the chain-rule factor entirely.
Question 7
For \(f(x,y) = x^3y^2\), the partial \(f_y\) is:
Why
Hold \(x\) constant: \(x^3\) is just a coefficient, and \(\frac{d}{dy}y^2 = 2y\). Choice A is \(f_x\); choice C is the total differential's pieces added together, which is not a partial derivative.
Question 8
The total differential of \(z = f(x,y)\) is:
Why
Contributions from each independent direction add. Setting \(dz=0\) in choice A is exactly the form \(M\,dx + N\,dy = 0\) of an exact ODE.
Question 9
Which statement about \(\dfrac{d^2y}{dx^2}\) is correct?
Why
The second derivative is a different object from the derivative squared. In dot notation it is \(\ddot y\) (two dots), and its sign tells you concavity — it can be anything.
Question 10
Using a differential, the best linear estimate of \(f(3.1)\) given \(f(3) = 7\) and \(f'(3) = 2\) is:
Why
\(dy = f'(3)\,\Delta x = 2(0.1) = 0.2\), so \(f(3.1) \approx 7 + 0.2 = 7.2\). Choice C adds \(f'\) itself instead of \(f'\Delta x\).
Part II · Module 1 Integration
Area under a curve
The integral starts life as an area problem: chop the region into rectangles you can measure, add them up, then let the rectangles get infinitely thin. Everything else — the antiderivative connection, the techniques — is machinery built on this limit.
①Riemann sums
Split \([a,b]\) into \(n\) strips of width \(\Delta x = \frac{b-a}{n}\), pick a sample point \(x_i^*\) in each, and add up the rectangle areas:
Riemann sum → definite integral
\[ \int_a^b f(x)\,dx \;=\; \lim_{n\to\infty}\sum_{i=1}^{n} f(x_i^*)\,\Delta x \]
The \(\int\) sign is an elongated S for "sum," and \(dx\) is the surviving trace of \(\Delta x\). The notation is a picture of the construction.
InteractiveRectangles converging on the true arearaise \(n\), compare the methods
approximation —exact —error —Δx —
On an increasing function the left sum always under-estimates and the right sum always over-estimates — the true area is trapped between them. Watch how much faster midpoint and trapezoid converge: doubling \(n\) cuts their error by a factor of ~4, versus ~2 for left/right.
②Signed area
A definite integral is signed area: regions below the \(x\)-axis count as negative. That single fact resolves most "my answer is zero and I don't know why" moments:
If a problem asks for "the area between the curve and the axis," you must split at every root and add absolute values — the same way total distance differs from displacement in Module D2.
Properties worth knowing without thinking
\(\displaystyle\int_a^a f = 0\); \(\displaystyle\int_b^a f = -\int_a^b f\); \(\displaystyle\int_a^b f + \int_b^c f = \int_a^c f\); constants pull out; sums split. But products do not split: \(\int fg \ne \left(\int f\right)\left(\int g\right)\).
③Practice
1
Estimate \(\int_0^2 x^2\,dx\) with \(n=4\) left rectangles by hand, then compare to the exact value \(8/3\).
Hint 1\(\Delta x = 0.5\), and left endpoints are \(x = 0, 0.5, 1, 1.5\).
Hint 2Sum \(= 0.5\left[0^2 + 0.5^2 + 1^2 + 1.5^2\right]\).
Solution
STEP 1 · find the strip width
\[ \Delta x = \frac{b-a}{n} = \frac{2-0}{4} = 0.5 \]
STEP 2 · list the sample points
Left endpoints means the left edge of each strip: \(x = 0,\ 0.5,\ 1,\ 1.5\). (The right edge \(x=2\) is never used — that is what makes it a left sum.)
STEP 5 · compare with the exact value
\[ \int_0^2 x^2dx = \frac{8}{3} \approx 2.667, \qquad \text{error} = 0.917 \]
WHY IT UNDERSHOOTS
\(x^2\) is increasing on \([0,2]\), so the left edge is the lowest point of each strip and every rectangle sits under the curve. Left sums always underestimate an increasing function. Bump \(n\) to 40 in the lab and watch the error drop by a factor of ten.
2
Without computing anything, order these from smallest to largest for \(\int_1^3 \ln x\,dx\): left sum, right sum, exact value.
Hint 1Is \(\ln x\) increasing or decreasing on \([1,3]\)?
Hint 2For an increasing function, each left rectangle's height is the smallest value on its strip.
Solution
STEP 1 · determine the monotonicity
\(\frac{d}{dx}\ln x = \frac1x > 0\) on \([1,3]\), so \(\ln x\) is increasing there.
STEP 2 · left rectangles
On each strip \([x_i, x_{i+1}]\), the smallest value of an increasing function is at the left edge. So each left rectangle fits under the curve, and \(L_n < \) true area.
STEP 3 · right rectangles
Symmetrically, the right edge gives the largest value on each strip, so each rectangle pokes above the curve: \(R_n > \) true area.
The true value is trapped between two computable numbers, so \(R_n - L_n\) bounds your error. For a monotone \(f\), that gap is exactly \(\left|f(b)-f(a)\right|\Delta x\) — here \((\ln 3 - \ln 1)\Delta x\). If the function were decreasing, the inequality would simply flip: "left sums underestimate" is a statement about increasing functions, not a universal law.
3
Given \(\int_0^4 f = 10\) and \(\int_0^7 f = 4\), find \(\int_4^7 f\) and \(\int_7^4 f\).
Hint 1Use additivity: \(\int_0^4 + \int_4^7 = \int_0^7\).
Hint 2Reversing the limits flips the sign.
Solution
STEP 1 · write the additivity property
Areas over adjacent intervals add:
\[ \int_0^4 f + \int_4^7 f = \int_0^7 f \]
STEP 2 · substitute the known values
\[ 10 + \int_4^7 f = 4 \]
STEP 3 · solve
\[ \int_4^7 f = 4 - 10 = -6 \]
STEP 4 · reverse the limits
Swapping the endpoints flips the sign:
\[ \int_7^4 f = -\int_4^7 f = +6 \]
INTERPRET
A negative value over \([4,7]\) means \(f\) spends most of that interval below the axis — enough negative area to cancel 6 units of the 10 accumulated earlier. Nothing here required knowing \(f\) itself; these three properties (additivity, reversal, \(\int_a^a = 0\)) are pure bookkeeping and appear constantly on exams.
4
Estimate \(\int_0^{\pi} \sin x\,dx\) using \(n=3\) midpoint rectangles, and compare with the exact value 2.
Hint 1\(\Delta x = \pi/3\). The midpoints are at \(\pi/6,\ \pi/2,\ 5\pi/6\).
Hint 2\(\sin\frac\pi6 = 0.5\), \(\sin\frac\pi2 = 1\), \(\sin\frac{5\pi}{6} = 0.5\).
Hint 3The exact value comes from \(\left[-\cos x\right]_0^\pi\).
Error: \(0.094\), about 4.7% — from only three rectangles. A left sum with \(n=3\) would be off by roughly 0.6.
WHY MIDPOINT IS SO GOOD
On each strip the rectangle cuts the curve twice, so the sliver it misses on one side is nearly cancelled by the sliver it adds on the other. That cancellation makes the midpoint error scale like \(\Delta x^2\) rather than \(\Delta x\) — doubling \(n\) cuts the error by four.
5
A flow meter records these rates (liters per minute) at one-minute intervals:
t (min)
0
1
2
3
4
r(t)
12
15
19
18
14
Estimate the total volume delivered over the four minutes using the trapezoid rule.
Hint 1Total volume is the area under the rate curve — this is a Riemann-sum problem with data instead of a formula.
The meter gives a rate in L/min. Multiplying rate by time gives volume, so the total volume is the area under \(r(t)\) from 0 to 4 — i.e. \(\int_0^4 r(t)\,dt\). No formula for \(r\) exists, so estimate numerically.
STEP 2 · why endpoints get weight 1 and interiors weight 2
Each trapezoid averages its two edge heights: \(\frac{\Delta t}{2}(r_i + r_{i+1})\). Every interior reading belongs to two adjacent trapezoids, so it appears twice; the two ends appear once.
STEP 3 · apply the rule with Δt = 1
\[ T = \frac{1}{2}\left[12 + 2(15) + 2(19) + 2(18) + 14\right] \]
Unknown — with data alone you cannot compute an error. But if the flow is smooth, the trapezoid rule's error scales like \(\Delta t^2\), so halving the sampling interval would cut it by about four. This "integrate measured data" situation is the most common real-world use of Riemann sums.
6
For \(f(x) = x^2 - 4\) on \([0,3]\), compute the signed area \(\int_0^3 f\,dx\) and the total (unsigned) area between the curve and the \(x\)-axis.
Hint 1Find where the curve crosses the axis — that's where you must split.
Hint 2\(x^2-4 = 0\) at \(x=2\) (only \(x=2\) is in \([0,3]\)). The function is negative before it and positive after.
Hint 3Total area \(= \left|\int_0^2 f\right| + \int_2^3 f\).
Signed area is to total area exactly as displacement is to distance travelled (module D2). In both cases the fix is the same: split at every sign change and take absolute values.
7
Compute \(\int_0^3 (2x+1)\,dx\) two ways: by finding the area of the resulting trapezoid geometrically, and by antidifferentiating. Then explain why any Riemann sum with equal strips gives this answer exactly.
Hint 1The region under a line is a trapezoid: area \(= \frac{\text{height}_1 + \text{height}_2}{2}\times\text{width}\).
Hint 2Heights are \(f(0)=1\) and \(f(3)=7\); width is 3.
Hint 3For the last part, think about where the midpoint height of a straight line sits relative to the average of the two edge heights.
Solution
STEP 1 · geometric route
The region is a trapezoid with parallel sides \(f(0)=1\) and \(f(3)=7\), width 3:
The trapezoid rule replaces the curve on each strip by a straight line through its endpoints. When the function is a straight line, that replacement changes nothing — so the rule is exact for any \(n\).
STEP 4 · why the midpoint rule is also exact
For a line, the height at the midpoint equals the average of the two edge heights. So each midpoint rectangle has exactly the same area as the trapezoid it replaces: the overshoot on one side cancels the undershoot on the other, precisely.
STEP 5 · but left and right sums are not
\(L_n\) uses the low edge every time and \(R_n\) the high edge, so they miss by \(\pm\frac{(b-a)^2}{2n}\cdot\text{slope}\). Verify in the lab: set \(f(x)=x^3-3x^2+4\), choose Trapezoid, and watch the error vanish only where the curve is nearly straight.
8
Suppose \(f\) is positive and decreasing on \([a,b]\). Which of \(L_n\), \(R_n\) overestimates the integral, and which underestimates? Justify in one sentence, then state what happens if \(f\) has a maximum in the middle of the interval.
Hint 1On each strip, where does a decreasing function attain its largest value — the left edge or the right?
Hint 2The "left underestimates" slogan is only true for increasing functions; re-derive rather than recall.
Solution
STEP 1 · reason on a single strip
On \([x_i, x_{i+1}]\) a decreasing \(f\) is largest at the left edge and smallest at the right edge.
STEP 2 · conclude for left sums
Every left rectangle is built at the strip's maximum height, so it contains the region under the curve:
\[ L_n > \int_a^b f \qquad\text{(overestimate)} \]
STEP 3 · conclude for right sums
Every right rectangle uses the strip's minimum height and sits inside the region:
\[ R_n < \int_a^b f \qquad\text{(underestimate)} \]
STEP 4 · the general statement
It is monotonicity, not the word "left," that decides. Increasing: \(L_n < I < R_n\). Decreasing: \(R_n < I < L_n\).
STEP 5 · a hump in the middle
If \(f\) rises then falls, neither sum is guaranteed to bound the integral — the strips on the rising part err one way and those on the falling part the other, and the errors partly cancel. You get no inequality, only an estimate. (Midpoint and trapezoid remain well-behaved regardless, which is one more reason to prefer them.)
9
Estimate \(\int_0^1e^{-x^2}dx\) with the trapezoid rule and \(n=4\), then compare with the true value \(0.746824\).
Hint 1\(\Delta x = 0.25\), and the nodes are \(0, 0.25, 0.5, 0.75, 1\).
Hint 2Weights: endpoints once, interior points twice, all times \(\frac{\Delta x}{2}\).
Hint 3\(e^{-0.0625}=0.9394\), \(e^{-0.25}=0.7788\), \(e^{-0.5625}=0.5698\), \(e^{-1}=0.3679\).
Solution
STEP 1 · list the values
\[ f(0)=1,\quad f(0.25)=0.9394,\quad f(0.5)=0.7788,\quad f(0.75)=0.5698,\quad f(1)=0.3679 \]
An underestimate, because \(e^{-x^2}\) is concave down over most of \([0,1]\) and the chords cut beneath the curve.
STEP 5 · why numerics at all here
\(e^{-x^2}\) has no elementary antiderivative, so the Fundamental Theorem cannot be used — numerical integration or the series method of module S5 are the only routes. Doubling to \(n=8\) would cut the error by about four (trapezoid is second order).
Part II · Module 2 Integration
Integration as the opposite of differentiation
Areas and antiderivatives look like unrelated problems. The Fundamental Theorem says they are the same problem — and that is the single most useful fact in calculus, because it means you can compute areas by reversing derivatives.
①The accumulation function
Fix a starting point \(a\) and let the upper limit move:
In words: the rate at which accumulated area grows is exactly the height of the curve. Wide curve, area piles up fast; curve dips below the axis, accumulated area decreases.
InteractiveArea accumulating in real timeslide \(x\) and watch \(A(x)\) draw itself
f(x) —A(x) = ∫₀ˣ f —A is —
Green shading counts positive, red counts negative. Two things to verify with your own eyes: where \(f\) crosses zero going down, \(A\) has a maximum; and where \(f\) is negative, \(A\) decreases. The bottom curve is the antiderivative of the top curve — you are watching the Fundamental Theorem happen.
②The evaluation theorem
Fundamental Theorem, Part 2
\[ \int_a^b f(x)\,dx = F(b) - F(a) \quad\text{where } F' = f \]
This is the version you compute with. Finding areas becomes an exercise in recognizing derivatives backwards, which is why the rest of Part II is a catalog of reversal techniques.
\(\int \frac{dx}{x} = \ln|x| + C\). Dropping the bars silently restricts you to \(x>0\) — and in ODEs you will integrate \(\frac{dy}{y}\) constantly, where \(y\) may well be negative. The bars are what let \(e^{C}\) absorb into an arbitrary constant of either sign.
Note the double negative in the last step — the single most common arithmetic slip in definite integrals.
2
If \(A(x) = \int_0^x f(t)\,dt\) and \(f\) is positive and decreasing on \([0,5]\), describe the shape of \(A\).
Hint 1\(A' = f\). What does \(f>0\) say about \(A\)?
Hint 2\(A'' = f'\). What does \(f\) decreasing say about \(A''\)?
Solution
STEP 1 · apply FTC Part 1
\[ A'(x) = f(x) \]
STEP 2 · use the sign of f
\(f > 0\) means \(A' > 0\): the accumulated area only ever grows, so \(A\) is increasing on \([0,5]\).
STEP 3 · differentiate once more
\[ A''(x) = f'(x) \]
STEP 4 · use the fact that f decreases
\(f\) decreasing means \(f' < 0\), so \(A'' < 0\): \(A\) is concave down.
STEP 5 · describe the graph
\(A\) starts at \(A(0)=0\), rises steadily, but its climb gradually flattens — the classic "diminishing returns" curve. Intuitively: area keeps piling up, but each new sliver is shorter than the last because the curve is dropping.
CHECK IN THE LAB
Choose \(f(t) = e^{-(t-3)^2}\) in the accumulation lab and slide \(x\) past \(t=3\): \(A\) rises fastest exactly where \(f\) peaks, then levels off as \(f\) fades — the shape you just deduced without a formula.
Growing \(x\) shrinks the interval \([x,5]\), so the accumulated amount should decrease where \(\cos(t^2)>0\). The minus sign matches ✓
NOTE
You never needed an antiderivative of \(\cos(t^2)\) — and that is fortunate, because it has no elementary one. FTC Part 1 sidesteps the antiderivative entirely.
4
Evaluate \(\int_0^{\pi/2}\left(2\cos x - \sin x\right)dx\).
Hint 1Antiderivative of \(\cos\) is \(\sin\); antiderivative of \(\sin\) is \(-\cos\).
Hint 2So \(F(x) = 2\sin x + \cos x\). Mind the sign flip on the second term: \(-\sin \to +\cos\).
Solution
STEP 1 · antidifferentiate
\(\int\cos = \sin\) and \(\int\sin = -\cos\), so the \(-\sin x\) term becomes \(+\cos x\):
\[ F(x) = 2\sin x + \cos x \]
STEP 2 · verify by differentiating back
\[ F'(x) = 2\cos x - \sin x \ ✓ \]
STEP 3 · evaluate at the upper limit
\[ F\!\left(\frac{\pi}{2}\right) = 2(1) + 0 = 2 \]
STEP 4 · evaluate at the lower limit
\[ F(0) = 2(0) + 1 = 1 \]
Splitting: \(\int_0^{\pi/2}2\cos x\,dx = 2\) and \(\int_0^{\pi/2}\sin x\,dx = 1\), so \(2-1 = 1\) ✓. Both pieces are positive areas, and the second partly cancels the first.
5
Let \(A(x) = \displaystyle\int_1^x (t^2-4)\,dt\). Find where \(A\) is increasing and decreasing, locate its minimum, and compute \(A(4)\).
Hint 1Don't integrate yet — FTC gives \(A'(x)\) instantly.
Hint 2\(A'(x) = x^2-4\), which is negative between \(-2\) and 2.
Hint 3For \(A(4)\) you do need the antiderivative \(\frac{t^3}{3}-4t\).
Solution
STEP 1 · get A′ for free
FTC Part 1 says the derivative of an accumulation function is the integrand with \(t\) replaced by \(x\):
\(A(1)=0\) by definition, and \(A\) dips to a minimum at \(x=2\) before climbing to 9 at \(x=4\). Indeed \(A(2) = F(2)-F(1) = -\frac{16}{3}+\frac{11}{3} = -\frac53 < 0\) ✓ — the negative area from \(t\in(1,2)\), where \(t^2-4<0\).
THE GENERAL FORMULA
\[ \frac{d}{dx}\int_{a}^{g(x)}f(t)\,dt = f\big(g(x)\big)\,g'(x) \]
And if both limits move: \(\frac{d}{dx}\int_{h(x)}^{g(x)} f = f(g)g' - f(h)h'\). Note again that no antiderivative of \(e^{-t^2}\) was needed — none exists in elementary terms.
7
Evaluate \(\displaystyle\int_{-1}^{2}|x|\,dx\).
Hint 1\(|x|\) is defined piecewise, so no single antiderivative works across \(x=0\).
Hint 2Split at 0: \(|x| = -x\) for \(x<0\) and \(|x| = x\) for \(x\ge0\).
Hint 3Or just compute the areas of two triangles.
Solution
STEP 1 · split at the corner
The formula for \(|x|\) changes at \(x=0\), so the integral must be split there:
The graph is two triangles: one with base 1 and height 1 (area \(\frac12\)), one with base 2 and height 2 (area 2). Total \(\frac52\) ✓
THE TRAP
Writing \(\left[\frac{x^2}{2}\right]_{-1}^{2} = 2 - \frac12 = \frac32\) treats \(|x|\) as \(x\) throughout and silently subtracts the left-hand area instead of adding it. Any integrand with an absolute value, a piecewise definition, or a sign change needs splitting.
8
Water enters a tank at \(r(t) = 6 - 0.5t\) L/min for \(0 \le t \le 12\). (a) Find the total volume added over the 12 minutes. (b) When is the tank filling fastest? (c) When does the flow reverse, and what is the net change from \(t=0\) to \(t=12\)?
Hint 1Total change of a quantity = integral of its rate. This is the "net change theorem."
Hint 2\(r(t)\) is decreasing and hits zero at \(t=12\)... check that: \(6-0.5t = 0\) gives \(t=12\).
Hint 3Fastest filling means the largest rate, not the largest volume.
The rate graph is a straight line from \(r(0)=6\) down to \(r(12)=0\) — a triangle of area \(\frac12(12)(6) = 36\) ✓
STEP 4 · fastest filling (b)
\(r\) is decreasing throughout (\(r' = -0.5 < 0\)), so the maximum rate is at the left endpoint: \(t = 0\), with \(r = 6\) L/min.
STEP 5 · reversal and net change (c)
\(r(t) = 0\) exactly at \(t=12\) and would go negative afterwards — so within the given window the flow never reverses, and the net change equals the total added, 36 L. (If the window ran to \(t=16\), the last four minutes would drain \(\int_{12}^{16}r = -4\) L, and net and total would differ.)
THE BIG IDEA
"Integrate the rate to get the total change" is the most-used sentence in applied calculus, and it is nothing but FTC Part 2 in words.
9
Find the average value of \(f(x)=\sin x\) on \([0,\pi]\), and find the point where \(f\) actually attains it.
Hint 1Average value is \(\frac{1}{b-a}\int_a^bf\,dx\).
Hint 2\(\int_0^\pi\sin x\,dx = 2\).
Hint 3Then solve \(\sin c = \) that average.
Solution
STEP 1 · compute the integral
\[ \int_0^\pi\sin x\,dx = \left[-\cos x\right]_0^\pi = 1+1 = 2 \]
STEP 2 · divide by the width
\[ \bar f = \frac{2}{\pi} \approx 0.6366 \]
STEP 3 · find where it is attained
\[ \sin c = \frac2\pi \quad\Longrightarrow\quad c = \arcsin(0.6366) \approx 0.690 \ \text{ or } \ \pi - 0.690 \approx 2.452 \]
STEP 4 · the theorem behind it
The Mean Value Theorem for Integrals guarantees such a \(c\) exists for any continuous \(f\): the rectangle of height \(\bar f\) has exactly the same area as the region under the curve.
STEP 5 · why 2/π is worth knowing
It is the average of a rectified sine — the DC value an AC signal delivers through a full-wave rectifier, and the reason \(0.637\times\) peak appears throughout electronics. (The RMS value \(1/\sqrt2 \approx 0.707\) is a different average, of the square.)
Part II · Module 3 Integration
Definite vs. indefinite integrals
One produces a number, the other produces a whole family of functions. Confusing them is harmless in calculus and fatal in differential equations, where that arbitrary constant is the thing your initial condition pins down.
①Two different objects
Definite \(\displaystyle\int_a^b f(x)\,dx\)
A number. Signed area from \(a\) to \(b\). The variable \(x\) is a dummy — \(\int_a^b f(x)dx = \int_a^b f(u)du\). No \(+C\): the constant cancels in \(F(b)-F(a)\).
Indefinite \(\displaystyle\int f(x)\,dx\)
A family of functions — every antiderivative of \(f\). Always carries \(+C\), because any two antiderivatives of the same function differ by a constant.
InteractiveThe +C family, and how an initial condition picks oneslide \(C\), then drag the condition
antiderivative —C from the condition —
The faint curves are the whole family \(F(x) + C\) — vertical shifts of one another, all with identical slopes at each \(x\). The amber curve is the one you selected with the \(C\) slider; the cyan curve is the unique member passing through the marked initial condition. That is exactly what "particular solution" means in an ODE.
Where this shows up in ODEs
A first-order ODE has a one-parameter family of solutions; second-order has two parameters. Those parameters are the constants of integration. When your ODE textbook writes \(y = Ce^{2x}\), the \(C\) came from an indefinite integral, and an initial condition like \(y(0)=5\) is what turns the family into a single curve.
Worked example · same integrand, three different questions
A
\(\displaystyle\int 2x\,dx = x^2 + C\) — a family of parabolas.
B
\(\displaystyle\int_1^3 2x\,dx = \left[x^2\right]_1^3 = 9 - 1 = 8\) — a number. Notice the \(C\) would have cancelled: \((9+C)-(1+C)=8\).
C
Solve \(y' = 2x\) with \(y(1) = 5\): \(y = x^2 + C\), and \(5 = 1 + C\) gives \(C = 4\), so \(y = x^2 + 4\) — one specific function.
Two constants are not needed
When you integrate both sides of a separable ODE, write \(+C\) on one side only. \(C_1\) and \(C_2\) would immediately merge into \(C_2 - C_1\); carrying both is bookkeeping you will regret.
②Practice
1
Find \(\int(4x^3 - 6x + 2)\,dx\) and then \(\int_0^2(4x^3 - 6x + 2)\,dx\).
Hint 1Antidifferentiate term by term; only the indefinite answer gets \(+C\).
Had you kept \(C\): \((8+C) - (0+C) = 8\). This is exactly why definite integrals never carry \(+C\).
TWO DIFFERENT OBJECTS
The first answer is a family of functions; the second is the number 8. Writing \(+C\) on a definite integral, or dropping it on an indefinite one, are both marked wrong — and the second mistake will cost you the arbitrary constant your ODE initial condition is supposed to determine.
2
Solve the initial value problem \(\dfrac{dy}{dx} = 3\cos x\), \(y(\pi) = 1\).
Hint 1Integrate both sides with respect to \(x\) — the left side gives you \(y\) itself.
Hint 2\(y = 3\sin x + C\). Now substitute \(x=\pi\), remembering \(\sin\pi = 0\).
Solution
STEP 1 · integrate both sides
\[ \int\frac{dy}{dx}\,dx = \int 3\cos x\,dx \quad\Longrightarrow\quad y = 3\sin x + C \]
This is the general solution: a one-parameter family of curves, all with identical slopes at each \(x\).
STEP 2 · impose the initial condition
Substitute \(x = \pi\), \(y = 1\):
\[ 1 = 3\sin\pi + C = 3(0) + C \]
STEP 3 · solve for the constant
\[ C = 1 \]
STEP 4 · write the particular solution
\[ y = 3\sin x + 1 \]
STEP 5 · verify both requirements
Differentiate: \(y' = 3\cos x\) ✓. Evaluate: \(y(\pi) = 0 + 1 = 1\) ✓. Always check both — the ODE and the condition.
3
An object has acceleration \(a(t) = 6t\), with \(v(0)=4\) and \(s(0)=2\). Find \(s(t)\).
Hint 1Integrate twice — each integration produces its own constant, and each initial condition kills one.
Hint 2\(v(t) = 3t^2 + C_1\), and \(v(0)=4\) forces \(C_1 = 4\). Now integrate again.
A second-order equation needed exactly two conditions to become a single function. That correspondence — order of the equation = number of arbitrary constants = number of conditions required — holds for every linear ODE in the course.
4
Find \(f(x)\) given \(f''(x) = 12x\), \(f'(0) = 1\), and \(f(0) = 2\).
Hint 1Antidifferentiate twice, resolving one constant at each stage.
Hint 2\(f'(x) = 6x^2 + C_1\) and \(f'(0)=1\) gives \(C_1\) immediately.
Solution
STEP 1 · integrate f″ to get f′
\[ f'(x) = \int 12x\,dx = 6x^2 + C_1 \]
STEP 2 · use f′(0) = 1
Substituting \(x=0\) kills the \(6x^2\) term, isolating the constant:
Here they were harmless, but in an ODE you will write \(\int\frac{dy}{y} = \ln|y|\) where \(y\) may be negative. Keeping the bars is what allows the final solution \(y = \pm e^{C}e^{kx} = Ae^{kx}\) with \(A\) of either sign.
6
Solve \(\dfrac{dy}{dx} = e^{2x}\) with \(y(0) = 3\), and sketch how the solution differs from the general solution.
Hint 1\(\int e^{2x}dx = \frac12e^{2x} + C\) — the \(\frac12\) comes from undoing the chain rule.
Hint 2Substitute \(x=0\), where \(e^0 = 1\).
Solution
STEP 1 · integrate
Differentiating \(e^{2x}\) produces a factor 2, so antidifferentiating must divide by 2:
The general solution is an infinite stack of identical curves, each shifted vertically. The condition \(y(0)=3\) selects the single one passing through \((0,3)\). Open the lab in this module, pick \(f(x)=e^{x/2}\), and drag the condition slider to watch that selection happen.
7
A student computes \(\displaystyle\int_{-1}^{1}\frac{dx}{x^2} = \left[-\frac1x\right]_{-1}^{1} = -1 - 1 = -2\). Explain what is wrong, and what the correct answer is.
Hint 1The integrand is positive everywhere it is defined. Can a positive function enclose negative area?
Hint 2Check whether the integrand is defined on the whole interval \([-1,1]\).
Hint 3FTC Part 2 requires the antiderivative to be continuous across the interval.
Solution
STEP 1 · spot the absurdity
\(\frac{1}{x^2} > 0\) everywhere it exists, so the answer cannot possibly be negative. Something structural is wrong, not just arithmetic.
STEP 2 · find the illegal step
The integrand blows up at \(x = 0\), which lies inside \([-1,1]\). FTC Part 2 requires \(f\) to be continuous on the closed interval, and \(-\frac1x\) is not continuous there either — it jumps from \(+\infty\) to \(-\infty\).
STEP 3 · treat it properly, as an improper integral
Split at the singularity and take one-sided limits:
The right-hand piece already diverges, so the whole integral diverges. There is no finite answer — the region has infinite area.
HABIT TO BUILD
Before writing \(\left[F\right]_a^b\), scan the interval for points where the integrand is undefined. If one is inside, you have an improper integral (module I6), not a routine one.
8
The family \(y = \frac{x^3}{3} + C\) solves \(y' = x^2\). (a) Which member passes through \((3,5)\)? (b) Can two different members ever cross? (c) What does that mean for initial value problems?
Hint 1For (a), substitute the point and solve for \(C\).
Hint 2For (b), suppose two members shared a point and see what that forces \(C_1\) and \(C_2\) to be.
Solution
STEP 1 · substitute the point (a)
\[ 5 = \frac{27}{3} + C = 9 + C \quad\Longrightarrow\quad C = -4 \]
\[ y = \frac{x^3}{3} - 4 \]
So they are the same curve. Distinct members never intersect — they are rigid vertical translates of one another.
STEP 3 · interpret geometrically
The family fills the plane like contour lines: through every point of the plane passes exactly one curve.
STEP 4 · consequence for IVPs (c)
Every initial condition \(y(x_0) = y_0\) selects exactly one solution — existence and uniqueness. That is why "solve the ODE, then apply the condition" always produces a single answer rather than several.
LOOKING AHEAD
For general equations \(y' = f(x,y)\) this is the Existence–Uniqueness Theorem, and it can fail when \(f\) misbehaves — e.g. \(y' = \sqrt{y}\) with \(y(0)=0\) has more than one solution. Your ODE course will make the precise conditions explicit.
9
A firm's marginal cost is \(C'(x)=3x^2+2\) dollars per unit. Find the increase in total cost when production rises from 10 to 20 units, and explain why \(C(0)\) is irrelevant.
Hint 1Total change = definite integral of the rate — the Net Change Theorem.
Hint 2\(\int_{10}^{20}(3x^2+2)dx = \left[x^3+2x\right]_{10}^{20}\).
Solution
STEP 1 · set up the net change integral
\[ C(20)-C(10) = \int_{10}^{20}C'(x)\,dx = \int_{10}^{20}\left(3x^2+2\right)dx \]
The constant of integration — here the fixed cost \(C(0)\) — cancels in the subtraction \(C(20)-C(10)\). A definite integral gives a change, and changes never see the constant.
STEP 5 · when you would need it
To find the total cost \(C(20)\) rather than the increase, you need the indefinite integral \(C(x)=x^3+2x+K\) plus one data point (say \(C(0)=5000\) fixed costs) to pin \(K\). Exactly the general-solution-plus-initial-condition structure of an ODE.
Part II · Module 4 Integration
u-substitution
The chain rule played backwards. If you can spot an inner function whose derivative is also sitting in the integrand — up to a constant — the integral collapses. This is the technique you will use most often in an ODE course.
①The mechanism
Substitution rule
\[ \int f\big(g(x)\big)\,g'(x)\,dx \;=\; \int f(u)\,du \qquad\text{with } u = g(x),\ du = g'(x)\,dx \]
Procedure:
Choose \(u\): the inside of a composition — under a root, in an exponent, inside a trig function, or the denominator.
Compute \(du = g'(x)\,dx\) and solve for \(dx\) if needed.
Replace everything. If any \(x\) survives, your choice of \(u\) was wrong (or you need to solve for \(x\) in terms of \(u\)).
Integrate in \(u\), then either substitute back (indefinite) or change the limits (definite).
InteractiveTwo regions, one areastep through the substitution
Left is the original integrand over \([a,b]\) in \(x\); right is the transformed integrand over \([g(a),g(b)]\) in \(u\). The two shaded regions look nothing alike — different widths, different heights — but their areas are identical. Substitution is a change of coordinates that preserves area, and \(du = g'(x)dx\) is exactly the bookkeeping that makes it so.
Worked example · definite integral, limits changed
(1) Forgetting to change the limits on a definite integral, then plugging \(x\)-values into a \(u\)-expression. (2) Leaving a stray \(x\) in the integrand. (3) Treating \(du\) as if it were \(dx\) — the constant factor matters: \(du = 3\,dx\) means \(dx = \frac{du}{3}\).
Where this shows up in ODEs
Every separable equation ends in two integrals, and at least one of them is usually a substitution. Solving \(\frac{dy}{dx} = xy\) gives \(\int\frac{dy}{y} = \int x\,dx\); solving \(\frac{dP}{dt} = kP(1-P)\) (logistic growth) needs partial fractions and substitution. The integrating-factor method produces \(\int \mu(x)q(x)dx\), which is a substitution problem more often than not.
②Practice
1
Evaluate \(\displaystyle\int (3x+5)^{8}\,dx\).
Hint 1Let \(u = 3x+5\). What is \(du\)?
Hint 2\(du = 3\,dx\), so \(dx = du/3\) — that \(\frac13\) rides along outside the integral.
Solution
STEP 1 · choose u
Take the inside of the composition: \(u = 3x+5\).
STEP 2 · compute du and solve for dx
\[ du = 3\,dx \quad\Longrightarrow\quad dx = \frac{du}{3} \]
STEP 4 · integrate in u
\[ \frac13\cdot\frac{u^9}{9} = \frac{u^9}{27} \]
STEP 5 · substitute back
\[ \int(3x+5)^8dx = \frac{(3x+5)^9}{27} + C \]
CHECK
Differentiate: \(\frac{9(3x+5)^8\cdot 3}{27} = (3x+5)^8\) ✓. For any linear inside, the rule is simply "integrate as usual, then divide by the coefficient of \(x\)."
STEP 4 · substitute back
\[ = \frac12\ln(x^2+4) + C \]
The bars can be dropped because \(x^2+4 > 0\) always.
CONTRAST THIS
\[ \int\frac{dx}{x^2+4} = \frac12\arctan\frac{x}{2} + C \]
Same denominator, completely different technique — and the only difference is that single factor of \(x\) in the numerator. Before choosing a method, always ask: is the derivative of the inside present?
STEP 4 · integrate term by term
\[ \int\left(u^{3/2}+3u^{1/2}\right)du = \frac{2}{5}u^{5/2} + 3\cdot\frac{2}{3}u^{3/2} = \frac25u^{5/2} + 2u^{3/2} \]
STEP 5 · substitute back
\[ \int x\sqrt{x-3}\,dx = \frac25(x-3)^{5/2} + 2(x-3)^{3/2} + C \]
WHEN TO USE THIS
Whenever the substitution leaves a stray \(x\) and the inside is linear, solving for \(x\) in terms of \(u\) rescues the problem. It fails for nonlinear insides — \(\int x\sqrt{x^2-3}\,dx\) doesn't need it, and \(\int \sqrt{x^3-3}\,dx\) is beyond it.
5
Evaluate \(\displaystyle\int\cos(5x)\,dx\) and \(\displaystyle\int_0^{\pi/10}\cos(5x)\,dx\).
Hint 1\(u = 5x\), \(du = 5\,dx\).
Hint 2For the definite version, either change the limits to \(u\)-values or substitute back first — both work, but pick one and be consistent.
Solution
STEP 1 · substitute
\[ u = 5x, \qquad du = 5dx, \qquad dx = \frac{du}{5} \]
STEP 2 · indefinite integral
\[ \int\cos(5x)dx = \frac15\int\cos u\,du = \frac15\sin u + C = \frac{\sin 5x}{5} + C \]
STEP 3 · definite version — change the limits
\[ x=0 \Rightarrow u=0, \qquad x = \frac{\pi}{10} \Rightarrow u = \frac{\pi}{2} \]
Writing \(\left[\frac{\sin u}{5}\right]_0^{\pi/10}\) — mixing a \(u\)-expression with \(x\)-limits — gives \(\frac{\sin(0.314)}{5} = 0.062\), which is wrong. Either convert the limits or convert the function, never neither.
STEP 4 · substitute back
\[ \int x^2\sqrt{x^3+1}\,dx = \frac{2}{9}\left(x^3+1\right)^{3/2} + C \]
CHECK
Differentiate: \(\frac29\cdot\frac32(x^3+1)^{1/2}\cdot 3x^2 = x^2\sqrt{x^3+1}\) ✓ — the chain rule regenerating the \(3x^2\) is exactly why the substitution worked.
7
Evaluate \(\displaystyle\int e^{\sin x}\cos x\,dx\) and \(\displaystyle\int \frac{dx}{x\ln x}\).
Hint 1In the first, ask what sits in the exponent — and whether its derivative is elsewhere in the integrand.
Hint 2In the second, \(\ln x\) has derivative \(\frac1x\), and there is a \(\frac1x\) available.
Solution
FIRST · STEP 1
\[ u = \sin x \quad\Longrightarrow\quad du = \cos x\,dx \]
The \(\cos x\,dx\) in the integrand is precisely \(du\) — a perfect match.
FIRST · STEP 2
\[ \int e^{\sin x}\cos x\,dx = \int e^u\,du = e^u + C = e^{\sin x} + C \]
SECOND · STEP 1
Rewrite to see the structure: \(\frac{1}{x\ln x} = \frac{1}{\ln x}\cdot\frac1x\). Take
\[ u = \ln x \quad\Longrightarrow\quad du = \frac{dx}{x} \]
SECOND · STEP 2
\[ \int\frac{dx}{x\ln x} = \int\frac{du}{u} = \ln|u| + C = \ln\left|\ln x\right| + C \]
PATTERN
Both are instances of \(\int f(g)g' = F(g)\). Train yourself to scan an integrand for a function and its derivative sitting side by side — that recognition is 90% of substitution.
8
Evaluate \(\displaystyle\int_0^{2}\frac{x\,dx}{\sqrt{x^2+5}}\) using a substitution with changed limits.
Hint 1\(u = x^2+5\), so \(du = 2x\,dx\).
Hint 2Limits: \(x=0 \Rightarrow u=5\); \(x=2 \Rightarrow u=9\).
Hint 3\(\int u^{-1/2}du = 2u^{1/2}\).
Solution
STEP 1 · substitute
\[ u = x^2+5, \qquad du = 2x\,dx, \qquad x\,dx = \tfrac12 du \]
STEP 2 · convert the limits immediately
\[ x=0 \Rightarrow u = 5, \qquad x=2 \Rightarrow u = 9 \]
Doing this now means you never have to substitute back at the end.
The integrand is positive and at most \(\frac{2}{\sqrt5} \approx 0.89\) on \([0,2]\), so the area is under \(1.8\) — and \(0.764\) is comfortably in range ✓
9
Solve the separable ODE \(\dfrac{dy}{dx} = 2xy^2\) with \(y(0) = 1\).
Hint 1Separate: get all the \(y\)'s with \(dy\) and all the \(x\)'s with \(dx\).
Hint 2\(\frac{dy}{y^2} = 2x\,dx\), i.e. \(\int y^{-2}dy = \int 2x\,dx\).
Hint 3\(\int y^{-2}dy = -y^{-1}\). Don't forget the minus when solving for \(y\).
Solution
STEP 1 · separate the variables
Divide by \(y^2\) and multiply by \(dx\) — legal because \(dy\) and \(dx\) are differentials (module D8):
\[ \frac{dy}{y^2} = 2x\,dx \]
STEP 2 · integrate both sides
\[ \int y^{-2}\,dy = \int 2x\,dx \quad\Longrightarrow\quad -\frac1y = x^2 + C \]
One constant on the right is enough.
STEP 3 · apply the initial condition before solving for y
Substituting \(x=0\), \(y=1\) now is easier than after rearranging:
\[ -1 = 0 + C \quad\Longrightarrow\quad C = -1 \]
STEP 4 · solve for y
\[ -\frac1y = x^2 - 1 \quad\Longrightarrow\quad y = \frac{1}{1-x^2} \]
STEP 5 · check and note the domain
Verify: \(y' = \frac{2x}{(1-x^2)^2} = 2xy^2\) ✓, and \(y(0)=1\) ✓.
But the solution blows up at \(x = \pm1\), so it is only valid on \((-1,1)\) — the interval containing the initial point. A perfectly innocent-looking ODE can produce a solution that escapes to infinity in finite "time," which is why stating the interval of validity matters.
Part II · Module 5 Integration
Integration by parts
The product rule played backwards. Use it when the integrand is a product of two unlike things — a polynomial times an exponential, a log times anything, an inverse trig function on its own.
①The formula and where it comes from
Start with the product rule \((uv)' = u'v + uv'\), integrate both sides, and rearrange:
You are trading one integral for another. The trade only helps if \(\int v\,du\) is simpler than what you started with, and that depends entirely on how you split the integrand.
Whatever appears earlier in that list becomes \(u\) (it will be differentiated, and these get simpler when differentiated); the rest becomes \(dv\).
InteractiveChoose u, see whether the trade helpstry both splits
The picture is the geometric identity behind the formula: for a monotone curve, the rectangle \(u_2v_2\) minus the rectangle \(u_1v_1\) splits into the region \(\int v\,du\) (below the curve) plus \(\int u\,dv\) (left of the curve). Rearranged, that is \(\int u\,dv = uv - \int v\,du\).
Take \(u = \sin x\), \(dv = e^x dx\). Then \(du = \cos x\,dx\), \(v = e^x\), giving \(I = e^x\sin x - \int e^x\cos x\,dx\).
STEP 2
Apply parts again, keeping the same style of choice: \(u=\cos x\), \(dv = e^x dx\). Then \(\int e^x\cos x\,dx = e^x\cos x + \int e^x\sin x\,dx = e^x\cos x + I\).
STEP 3
Substitute back: \(I = e^x\sin x - e^x\cos x - I\). The original integral has reappeared — treat it as an unknown.
STEP 4
Solve algebraically: \(2I = e^x(\sin x - \cos x)\), so \(I = \tfrac12 e^x(\sin x - \cos x) + C\).
STEP 5
If you had switched your choice on the second pass, you would have unwound back to \(I = I\) — true but useless. Be consistent.
The "invisible dv = dx" trick
\(\int \ln x\,dx\) looks like it has only one factor. Take \(u = \ln x\) and \(dv = dx\); then \(v = x\) and \(\int\ln x\,dx = x\ln x - \int x\cdot\frac1x dx = x\ln x - x + C\). The same trick handles \(\int\arctan x\,dx\) and \(\int\arcsin x\,dx\).
Where this shows up in ODEs
Everywhere in Laplace transforms — \(\mathcal{L}\{f\} = \int_0^\infty e^{-st}f(t)\,dt\) is an integration-by-parts factory, and the transform rule \(\mathcal{L}\{f'\} = sF(s) - f(0)\) is derived by one application of parts. Variation of parameters also routinely produces \(\int t e^{-2t}dt\)-shaped integrals.
②Practice
1
Evaluate \(\displaystyle\int x\cos x\,dx\).
Hint 1LIATE: Algebraic beats Trig, so \(u = x\).
Hint 2\(du = dx\) and \(v = \sin x\). Assemble \(uv - \int v\,du\).
Solution
STEP 1 · split the integrand
LIATE ranks Algebraic above Trig, so:
\[ u = x, \qquad dv = \cos x\,dx \]
STEP 2 · differentiate u, integrate dv
\[ du = dx, \qquad v = \sin x \]
No \(+C\) is needed for \(v\) — any antiderivative works, and the constant cancels.
STEP 3 · apply the formula
\[ \int u\,dv = uv - \int v\,du = x\sin x - \int \sin x\,dx \]
STEP 4 · finish the easier integral
\[ = x\sin x + \cos x + C \]
STEP 5 · verify by differentiating
\[ \frac{d}{dx}\left(x\sin x + \cos x\right) = \sin x + x\cos x - \sin x = x\cos x \ ✓ \]
The trade worked because differentiating \(x\) made it disappear, leaving a pure trig integral.
2
Evaluate \(\displaystyle\int_0^1 x e^{2x}\,dx\).
Hint 1\(u = x\), \(dv = e^{2x}dx\). Careful with \(v\): it is \(\frac12 e^{2x}\), not \(e^{2x}\).
Hint 2You get \(\left[\frac{x}{2}e^{2x}\right]_0^1 - \frac12\int_0^1 e^{2x}dx\).
Solution
STEP 1 · choose the split
\[ u = x, \qquad dv = e^{2x}dx \]
STEP 2 · find du and v
\[ du = dx, \qquad v = \int e^{2x}dx = \frac12e^{2x} \]
The \(\frac12\) comes from undoing the chain rule; dropping it doubles your final answer.
STEP 3 · apply the definite-integral form
\[ \int_0^1 xe^{2x}dx = \left[\frac{x}{2}e^{2x}\right]_0^1 - \int_0^1 \frac12 e^{2x}\,dx \]
STEP 4 · evaluate the boundary term
\[ \left[\frac{x}{2}e^{2x}\right]_0^1 = \frac{e^2}{2} - 0 = \frac{e^2}{2} \]
STEP 5 · evaluate the remaining integral
\[ \frac12\int_0^1 e^{2x}dx = \frac12\left[\frac{e^{2x}}{2}\right]_0^1 = \frac{e^2-1}{4} \]
Hint 1Log beats Algebraic in LIATE, so \(u = \ln x\) even though \(x^2\) looks like the "obvious" choice.
Hint 2\(du = \frac{dx}{x}\), \(v = \frac{x^3}{3}\). The remaining integral becomes \(\int\frac{x^3}{3}\cdot\frac1x dx = \int\frac{x^2}{3}dx\).
Solution
STEP 1 · choose u by LIATE
L (log) outranks A (algebraic):
\[ u = \ln x, \qquad dv = x^2dx \]
STEP 2 · differentiate and integrate
\[ du = \frac{dx}{x}, \qquad v = \frac{x^3}{3} \]
STEP 3 · apply the formula
\[ \int x^2\ln x\,dx = \frac{x^3}{3}\ln x - \int \frac{x^3}{3}\cdot\frac{1}{x}\,dx \]
STEP 4 · simplify the new integrand
This is where the choice pays off — the log is gone and the powers cancel:
\[ \int\frac{x^2}{3}dx = \frac{x^3}{9} \]
STEP 5 · assemble
\[ \int x^2\ln x\,dx = \frac{x^3}{3}\ln x - \frac{x^3}{9} + C \]
WHY THE OTHER CHOICE FAILS
Taking \(u = x^2\), \(dv = \ln x\,dx\) requires \(v = \int\ln x\,dx = x\ln x - x\) — you'd have to solve a harder problem just to start. LIATE exists to steer you away from exactly that.
4
Evaluate \(\displaystyle\int x^2 e^{-x}\,dx\).
Hint 1You will need parts twice — each pass knocks the power of \(x\) down by one.
Hint 2First pass: \(u = x^2\), \(dv = e^{-x}dx\), \(v = -e^{-x}\). Mind the minus signs.
Hint 3After the first pass you should have \(-x^2e^{-x} + 2\int xe^{-x}dx\).
Solution
STEP 1 · first pass
\[ u = x^2,\ dv = e^{-x}dx \quad\Longrightarrow\quad du = 2x\,dx,\ v = -e^{-x} \]
\[ \int x^2e^{-x}dx = -x^2e^{-x} + 2\int xe^{-x}dx \]
The minus in \(v\) meets the minus in \(-\int v\,du\), producing the \(+2\).
STEP 2 · second pass on the leftover
\[ u = x,\ dv = e^{-x}dx \quad\Longrightarrow\quad du = dx,\ v = -e^{-x} \]
\[ \int xe^{-x}dx = -xe^{-x} + \int e^{-x}dx = -xe^{-x} - e^{-x} \]
STEP 3 · substitute back into step 1
\[ \int x^2e^{-x}dx = -x^2e^{-x} + 2\left(-xe^{-x} - e^{-x}\right) \]
\(\int x^ne^{-x}dx\) always gives \(-e^{-x}\) times a degree-\(n\) polynomial, and requires \(n\) passes. Each pass reduces the power by one — a useful thing to recognize before you start, so you know how much work you signed up for.
5
Evaluate \(\displaystyle\int \arctan x\,dx\).
Hint 1There seems to be only one factor — use the invisible \(dv = dx\).
Hint 2\(u = \arctan x\), \(dv = dx\), so \(du = \frac{dx}{1+x^2}\) and \(v = x\).
Hint 3The leftover \(\int\frac{x}{1+x^2}dx\) is a substitution problem.
Solution
STEP 1 · manufacture a second factor
Write the integrand as \(\arctan x \cdot 1\):
\[ u = \arctan x, \qquad dv = dx \]
STEP 2 · differentiate and integrate
\[ du = \frac{dx}{1+x^2}, \qquad v = x \]
STEP 3 · apply the formula
\[ \int\arctan x\,dx = x\arctan x - \int\frac{x}{1+x^2}\,dx \]
STEP 4 · handle the leftover by substitution
\[ w = 1+x^2,\ dw = 2x\,dx \quad\Longrightarrow\quad \int\frac{x\,dx}{1+x^2} = \frac12\ln(1+x^2) \]
STEP 5 · assemble
\[ \int\arctan x\,dx = x\arctan x - \frac12\ln\left(1+x^2\right) + C \]
SAME TRICK, THREE PLACES
\(\int\ln x\,dx\), \(\int\arcsin x\,dx\), \(\int\arctan x\,dx\) all yield to \(dv = dx\). Whenever an integrand is a single "hard to integrate but easy to differentiate" function, this is the move.
6
Evaluate \(\displaystyle\int x\sec^2 x\,dx\).
Hint 1LIATE: \(u = x\), \(dv = \sec^2x\,dx\).
Hint 2What function has derivative \(\sec^2 x\)?
Hint 3The leftover \(\int\tan x\,dx\) you computed back in module I4.
Solution
STEP 1 · choose the split
\[ u = x, \qquad dv = \sec^2x\,dx \]
STEP 2 · du and v
\[ du = dx, \qquad v = \tan x \]
(Because \(\frac{d}{dx}\tan x = \sec^2x\), from module D4.)
STEP 3 · apply the formula
\[ \int x\sec^2x\,dx = x\tan x - \int\tan x\,dx \]
STEP 4 · use the known tangent integral
\[ \int\tan x\,dx = -\ln|\cos x| \]
STEP 5 · assemble
\[ \int x\sec^2x\,dx = x\tan x + \ln|\cos x| + C \]
NOTE
Two techniques stacked: parts to strip the \(x\), then substitution inside \(\int\tan x\,dx\). Multi-technique problems are the norm on exams, so practise naming the outer method first.
7
Evaluate \(\displaystyle\int t^2\sin t\,dt\).
Hint 1Algebraic beats Trig, so \(u = t^2\) — and expect to repeat.
Hint 2First pass gives \(-t^2\cos t + 2\int t\cos t\,dt\), and you already solved \(\int t\cos t\,dt\) in problem 1.
Solution
STEP 1 · first pass
\[ u = t^2,\ dv = \sin t\,dt \quad\Longrightarrow\quad du = 2t\,dt,\ v = -\cos t \]
\[ \int t^2\sin t\,dt = -t^2\cos t + 2\int t\cos t\,dt \]
STEP 2 · reuse problem 1
\[ \int t\cos t\,dt = t\sin t + \cos t \]
STEP 3 · substitute back
\[ \int t^2\sin t\,dt = -t^2\cos t + 2\left(t\sin t + \cos t\right) + C \]
STEP 4 · tidy
\[ = -t^2\cos t + 2t\sin t + 2\cos t + C = \left(2-t^2\right)\cos t + 2t\sin t + C \]
STEP 5 · spot-check at t = 0
The derivative should be \(t^2\sin t\), which is 0 at \(t=0\). Differentiating: \(-2t\cos t - (2-t^2)\sin t + 2\sin t + 2t\cos t = t^2\sin t\) ✓
8
Evaluate \(\displaystyle\int_1^{e}\ln x\,dx\), and interpret the answer as an area.
Hint 1Use the invisible \(dv = dx\) trick with \(u = \ln x\).
Hint 2The antiderivative is \(x\ln x - x\).
Hint 3Recall \(\ln e = 1\) and \(\ln 1 = 0\).
Solution
STEP 1 · integrate by parts
\[ u = \ln x,\ dv = dx \quad\Longrightarrow\quad du = \frac{dx}{x},\ v = x \]
\[ \int\ln x\,dx = x\ln x - \int x\cdot\frac1x\,dx = x\ln x - x \]
STEP 2 · evaluate at the upper limit
\[ \left(e\ln e - e\right) = e - e = 0 \]
STEP 3 · evaluate at the lower limit
\[ \left(1\ln 1 - 1\right) = 0 - 1 = -1 \]
The area under \(\ln x\) from 1 to \(e\) is exactly 1 — a pleasingly clean result. Geometric check: the region sits inside the rectangle \([1,e]\times[0,1]\), whose area is \(e-1 \approx 1.718\), and the curve is concave down, so an answer near 60% of the rectangle is entirely plausible ✓
9
Show that \(\displaystyle\int_0^{\infty} t\,e^{-st}\,dt = \frac{1}{s^2}\) for \(s>0\) — the Laplace transform of \(f(t)=t\).
Hint 1Write the improper integral as a limit: \(\lim_{b\to\infty}\int_0^b\).
Hint 2Integrate by parts with \(u = t\), \(dv = e^{-st}dt\), so \(v = -\frac{1}{s}e^{-st}\).
Hint 3At the upper limit you need \(\lim_{b\to\infty}be^{-sb}\) — exponentials beat polynomials.
Solution
STEP 1 · replace ∞ with a limit
\[ \int_0^\infty te^{-st}dt = \lim_{b\to\infty}\int_0^b te^{-st}dt \]
STEP 2 · integrate by parts
\[ u = t,\ dv = e^{-st}dt \quad\Longrightarrow\quad du = dt,\ v = -\frac{1}{s}e^{-st} \]
\[ \int_0^b te^{-st}dt = \left[-\frac{t}{s}e^{-st}\right]_0^b + \frac1s\int_0^b e^{-st}dt \]
STEP 3 · the boundary term
\[ \left[-\frac{t}{s}e^{-st}\right]_0^b = -\frac{b}{s}e^{-sb} - 0 \xrightarrow[b\to\infty]{} 0 \]
because \(e^{-sb}\) decays faster than \(b\) grows, for any \(s>0\).
STEP 4 · the remaining integral
\[ \frac1s\int_0^\infty e^{-st}dt = \frac1s\cdot\frac1s = \frac{1}{s^2} \]
(using \(\int_0^\infty e^{-st}dt = 1/s\), the worked example in module I6).
For \(s \le 0\) the integrand doesn't decay and the integral diverges — which is why every Laplace table entry comes with a domain restriction. Repeating this computation with \(t^n\) gives \(\mathcal{L}\{t^n\} = n!/s^{n+1}\), one integration by parts per power.
Part II · Module 6 Integration
Improper integrals
An integral is improper when the region is infinitely long or infinitely tall. Sometimes the area is still finite — and knowing which is which is what makes Laplace transforms legal.
\[ \int_a^{b} f(x)\,dx = \lim_{t\to a^+}\int_t^{b}f(x)\,dx \quad\text{if } f \text{ blows up at } a \]
If the limit is a finite number, the integral converges; otherwise it diverges. You must actually write the limit — evaluating "at \(\infty\)" directly is not a legal step.
InteractivePush the boundary to infinitywatch the area settle — or not
b —∫₁ᵇ f dx —as b→∞ —limit value —
The \(p\)-test made visible. For \(p>1\) the running area climbs and then flattens against its limit \(\frac{1}{p-1}\). For \(p\le1\) it keeps climbing forever, however slowly — at \(p=1\) it grows like \(\ln b\), which is the classic example of "goes to infinity so slowly you'd never notice numerically."
The two halves point opposite directions, and that is not a typo: out at infinity you need the tail to shrink fast; near zero you need the spike to grow slowly.
Worked example · the Laplace transform of 1
Evaluate \(\displaystyle\int_0^{\infty} e^{-st}\,dt\) for \(s>0\).
STEP 1
Write the limit: \(\displaystyle\lim_{b\to\infty}\int_0^b e^{-st}dt\).
Take the limit. Because \(s>0\), \(e^{-sb}\to0\), so the value is \(\frac1s\).
STEP 4
You have just computed \(\mathcal{L}\{1\} = \frac1s\). The condition \(s>0\) is not decoration — for \(s\le0\) the integral diverges, which is why every Laplace-transform table comes with a domain restriction.
Split at the singularity
\(\int_{-1}^{1}\frac{dx}{x^2}\) is not \(\left[-\frac1x\right]_{-1}^1 = -2\) — a positive integrand cannot have negative area. The function blows up at \(x=0\) inside the interval, so you must split at 0 and take two limits, both of which diverge. Always scan the interval for interior singularities before you antidifferentiate.
②Practice
1
Determine whether \(\displaystyle\int_1^{\infty}\frac{dx}{x^3}\) converges, and find its value if it does.
Hint 1\(p = 3 > 1\), so predict convergence — then confirm it with the limit.
Hint 2Antiderivative: \(-\frac{1}{2x^2}\).
Solution
STEP 1 · write the limit form
You may not evaluate "at infinity" — replace the upper limit by \(b\) and take a limit at the end:
STEP 3 · evaluate between the limits
\[ \left[-\frac{1}{2x^2}\right]_1^b = -\frac{1}{2b^2} + \frac12 \]
STEP 4 · take the limit
\[ \lim_{b\to\infty}\left(\frac12 - \frac{1}{2b^2}\right) = \frac12 \]
STEP 5 · conclude
The limit is finite, so the integral converges to \(\frac12\) — consistent with the \(p\)-test, since \(p = 3 > 1\). An infinitely long region can still have finite area, provided the tail thins out fast enough.
This is \(\mathcal{L}\{t\}\) evaluated at \(s=2\), matching the general formula \(1/s^2\) ✓
3
Does \(\displaystyle\int_0^{1}\frac{dx}{\sqrt x}\) converge?
Hint 1The trouble is at \(x=0\), not at infinity — this is a Type 2 improper integral with \(p = \frac12\).
Hint 2Near zero, convergence needs \(p<1\).
Solution
STEP 1 · identify the trouble spot
The integrand \(x^{-1/2}\) blows up as \(x\to0^+\). The interval is finite, so this is Type 2 — an infinite height, not an infinite width.
STEP 2 · replace the bad endpoint by a limit
\[ \int_0^1\frac{dx}{\sqrt x} = \lim_{t\to0^+}\int_t^1 x^{-1/2}dx \]
STEP 3 · antidifferentiate and evaluate
\[ \left[2\sqrt x\right]_t^1 = 2 - 2\sqrt t \]
STEP 4 · take the limit
\[ \lim_{t\to0^+}\left(2 - 2\sqrt t\right) = 2 \]
STEP 5 · conclude
Converges to 2. An infinitely tall spike can enclose finite area if it is thin enough fast enough — here \(p = \frac12 < 1\) ✓ by the near-zero half of the \(p\)-test.
CONTRAST
The same integrand on \([1,\infty)\) diverges. Same function, opposite verdicts, because the two halves of the \(p\)-test point in opposite directions.
4
Evaluate \(\displaystyle\int_1^{\infty}e^{-2x}\,dx\), and compare its rate of convergence with \(\int_1^\infty x^{-2}dx\).
Hint 1\(\int e^{-2x}dx = -\frac12e^{-2x}\).
Hint 2For the comparison, compute how much "tail" is left beyond \(x=10\) for each.
Solution
STEP 1 · limit form and antiderivative
\[ \lim_{b\to\infty}\left[-\frac12 e^{-2x}\right]_1^b = \lim_{b\to\infty}\left(-\frac12e^{-2b} + \frac12e^{-2}\right) \]
STEP 2 · take the limit
\[ = \frac{e^{-2}}{2} = \frac{1}{2e^2} \approx 0.0677 \]
STEP 3 · the tail of each integral beyond x = 10
\[ \int_{10}^\infty e^{-2x}dx = \frac{e^{-20}}{2} \approx 1.0\times10^{-9} \]
\[ \int_{10}^\infty x^{-2}dx = \frac{1}{10} = 0.1 \]
STEP 4 · compare
Both converge, but the exponential's remaining tail is a hundred million times smaller. Exponential decay is in a different league from any power decay.
WHY THIS MATTERS FOR LAPLACE TRANSFORMS
The factor \(e^{-st}\) in \(\mathcal{L}\{f\} = \int_0^\infty e^{-st}f(t)dt\) is exactly this kind of crusher: it makes the integral converge even when \(f\) grows, so long as \(f\) grows slower than the exponential decays. That is the "exponential order" condition in the existence theorem.
Hint 1The integrand decays faster than \(1/x\) — but is that enough?
Hint 2Substitute \(u = \ln x\), \(du = \frac{dx}{x}\).
Hint 3The integral becomes \(\int\frac{du}{u}\), whose antiderivative grows without bound.
Solution
STEP 1 · limit form
\[ \lim_{b\to\infty}\int_2^b\frac{dx}{x\ln x} \]
STEP 2 · substitute
\[ u = \ln x, \qquad du = \frac{dx}{x} \quad\Longrightarrow\quad \int\frac{du}{u} = \ln|u| \]
STEP 3 · back-substitute and evaluate
\[ \left[\ln(\ln x)\right]_2^b = \ln(\ln b) - \ln(\ln 2) \]
STEP 4 · take the limit
\[ \lim_{b\to\infty}\ln(\ln b) = \infty \quad\Longrightarrow\quad \textbf{diverges} \]
STEP 5 · how slowly?
Absurdly slowly. To make the integral reach 3 you need \(\ln(\ln b) \approx 3\), i.e. \(b \approx e^{e^3} \approx 5\times10^{8}\). Numerically it looks convergent for any range you could plot — which is exactly why divergence must be proved, not eyeballed.
MORAL
Decaying faster than \(1/x\) is not sufficient. The borderline between convergence and divergence is razor thin, and \(\frac{1}{x\ln x}\) sits just on the wrong side of it. (Note \(\int_2^\infty\frac{dx}{x(\ln x)^2}\) does converge.)
6
Evaluate \(\displaystyle\int_0^{1}\ln x\,dx\).
Hint 1\(\ln x \to -\infty\) as \(x\to0^+\), so this is improper at the lower endpoint.
Hint 2Antiderivative from module I5: \(x\ln x - x\).
Hint 3You will need \(\lim_{t\to0^+}t\ln t\), which is 0 — the linear factor wins over the logarithm.
Solution
STEP 1 · spot the impropriety
The integrand is unbounded near \(x = 0\), so write
STEP 2 · antiderivative
\[ \int\ln x\,dx = x\ln x - x \]
STEP 3 · evaluate between t and 1
\[ \left[x\ln x - x\right]_t^1 = (0 - 1) - \left(t\ln t - t\right) = -1 - t\ln t + t \]
STEP 4 · handle the tricky limit
\(t\ln t\) is of the form \(0\cdot(-\infty)\). Rewriting as \(\frac{\ln t}{1/t}\) and applying L'Hôpital gives \(\frac{1/t}{-1/t^2} = -t \to 0\):
\[ \lim_{t\to0^+} t\ln t = 0 \]
STEP 5 · conclude
\[ \int_0^1\ln x\,dx = -1 \]
Negative, as it must be: \(\ln x < 0\) on \((0,1)\), so all the area lies below the axis. And it is finite despite the infinite spike — the spike is logarithmic, which is far too gentle to produce infinite area.
Both halves converged here, so the total is unambiguous. If even one half diverges, the whole integral diverges — you may not rescue it by letting both limits run to infinity together. That symmetric shortcut (the "principal value") would wrongly assign the value 0 to \(\int_{-\infty}^\infty x\,dx\), which genuinely diverges.
BONUS
Dividing by \(\pi\) turns \(\frac{1}{1+x^2}\) into a probability density — the Cauchy distribution, famous for having no mean, because \(\int x\cdot\frac{dx}{1+x^2}\) diverges.
8
Show that \(\mathcal{L}\{e^{at}\} = \displaystyle\int_0^{\infty}e^{-st}e^{at}\,dt = \frac{1}{s-a}\), and state the condition on \(s\).
Hint 1Combine the exponentials first: \(e^{-st}e^{at} = e^{-(s-a)t}\).
Hint 2Then it is the same computation as \(\int_0^\infty e^{-kt}dt\) with \(k = s-a\).
Hint 3Convergence requires the exponent to be genuinely negative.
\[ \mathcal{L}\{e^{at}\} = \frac{1}{s-a}, \qquad s > a \]
If \(s \le a\) the exponent is non-negative, the integrand does not decay, and the integral diverges.
READ THE RESULT
The transform has a pole exactly at \(s = a\) — and \(a\) is the growth rate of the original function. In the ODE course you will run this backwards: seeing \(\frac{1}{s+3}\) in a transform tells you \(e^{-3t}\) is in the solution.
9
Show that \(\displaystyle\int_1^{\infty}e^{-x^2}dx\) converges, without evaluating it.
Hint 1You cannot antidifferentiate \(e^{-x^2}\) — use a comparison instead.
Hint 2For \(x\ge1\), how do \(x^2\) and \(x\) compare? What does that do to \(e^{-x^2}\) versus \(e^{-x}\)?
Hint 3\(\int_1^\infty e^{-x}dx\) is easy to evaluate.
Solution
STEP 1 · establish the inequality
For \(x\ge1\) we have \(x^2\ge x\), so \(-x^2 \le -x\) and therefore
\[ 0 < e^{-x^2} \le e^{-x} \]
STEP 2 · evaluate the comparison integral
\[ \int_1^\infty e^{-x}dx = \lim_{b\to\infty}\left[-e^{-x}\right]_1^b = e^{-1} \approx 0.368 \]
Finite, so it converges.
STEP 3 · apply the comparison test
A positive integrand bounded above by a convergent one must itself converge:
\[ \int_1^\infty e^{-x^2}dx \ \text{ converges, and is } \le 0.368 \]
STEP 4 · check the direction of the inequality
Smaller than convergent ⟹ converges ✓ (Larger than divergent ⟹ diverges. The other two combinations prove nothing — the same directional rule as for series.)
STEP 5 · the actual value
\(\int_1^\infty e^{-x^2}dx \approx 0.1394\), comfortably under the bound. The full integral \(\int_0^\infty e^{-x^2}dx = \frac{\sqrt\pi}{2}\) is famous, but needs a two-dimensional polar trick — comparison tells you convergence for free, long before you can find the value.
Part II · Module 7 Integration
Trigonometric substitution
When a square root of a quadratic blocks every other method, a trig substitution turns it into a bare trig expression via the Pythagorean identities. The right triangle is the bookkeeping device that gets you back.
①The three patterns
Pattern
Substitute
Root becomes
Identity used
\(\sqrt{a^2-x^2}\)
\(x = a\sin\theta\)
\(a\cos\theta\)
\(1-\sin^2 = \cos^2\)
\(\sqrt{a^2+x^2}\)
\(x = a\tan\theta\)
\(a\sec\theta\)
\(1+\tan^2 = \sec^2\)
\(\sqrt{x^2-a^2}\)
\(x = a\sec\theta\)
\(a\tan\theta\)
\(\sec^2-1 = \tan^2\)
Memorize the shape, not the list: whichever identity turns your expression into a perfect square is the substitution you want.
InteractiveThe reference trianglepick a case, drag \(\theta\)
x —the radical —dx —
Build the triangle from the substitution, and every trig function of \(\theta\) can be read straight off the sides — that is how you convert your answer back to \(x\) at the end. The radical is always one of the three sides, which is the whole point of the method.
Convert to sines and cosines: \(\frac{\sec\theta}{\tan^2\theta} = \frac{\cos\theta}{\sin^2\theta}\). Substitute \(w = \sin\theta\): \(\frac19\int w^{-2}dw = -\frac{1}{9\sin\theta}\).
STEP 5
Back to \(x\) using the triangle: \(\tan\theta = x/3\) means opposite \(=x\), adjacent \(=3\), hypotenuse \(=\sqrt{x^2+9}\). So \(\sin\theta = \frac{x}{\sqrt{x^2+9}}\), and
\[ \int\frac{dx}{x^2\sqrt{x^2+9}} = -\frac{\sqrt{x^2+9}}{9x} + C \]
Complete the square first
\(\sqrt{x^2 + 6x + 13} = \sqrt{(x+3)^2 + 4}\) — now it matches pattern 2 with \(a=2\) and \(u = x+3\). Most exam problems hide the pattern behind a shift like this, so completing the square is step zero.
Don't reach for it too early
\(\int\frac{x\,dx}{\sqrt{x^2+9}}\) needs no trig at all — the numerator's \(x\) makes it a plain \(u\)-substitution with \(u = x^2+9\). Trig substitution is for when there is no convenient \(x\,dx\) available. Always try \(u\)-sub first.
②Practice
1
Evaluate \(\displaystyle\int\frac{dx}{\sqrt{4-x^2}}\) with a trig substitution, and confirm it matches the arcsine formula.
Hint 1Pattern 1 with \(a=2\): \(x = 2\sin\theta\), \(dx = 2\cos\theta\,d\theta\).
Hint 2The radical becomes \(2\cos\theta\), so almost everything cancels.
Solution
STEP 1 · match the pattern
\(\sqrt{a^2-x^2}\) with \(a = 2\) calls for the sine substitution:
\[ x = 2\sin\theta \quad\Longrightarrow\quad dx = 2\cos\theta\,d\theta \]
From \(x = 2\sin\theta\) we get \(\sin\theta = \frac{x}{2}\), hence \(\theta = \arcsin\frac x2\):
\[ \int\frac{dx}{\sqrt{4-x^2}} = \arcsin\frac{x}{2} + C \]
CHECK
Differentiating \(\arcsin\frac x2\) gives \(\frac{1/2}{\sqrt{1-x^2/4}} = \frac{1}{\sqrt{4-x^2}}\) ✓ — matching the standard formula \(\int\frac{dx}{\sqrt{a^2-x^2}} = \arcsin\frac xa + C\), which you have now derived rather than memorized.
2
Evaluate \(\displaystyle\int\sqrt{9-x^2}\,dx\).
Hint 1\(x = 3\sin\theta\) makes the radical \(3\cos\theta\) and \(dx = 3\cos\theta\,d\theta\).
Hint 2You land on \(9\int\cos^2\theta\,d\theta\) — use the half-angle identity \(\cos^2\theta = \frac{1+\cos2\theta}{2}\).
Hint 3To convert \(\sin2\theta\) back, use \(\sin2\theta = 2\sin\theta\cos\theta\) and read both factors off the triangle.
Over \([-3,3]\) this gives \(\frac92\left(\frac\pi2 + \frac\pi2\right) = \frac{9\pi}{2}\) — exactly half the area of a circle of radius 3 ✓, which is what the integrand describes.
3
Which technique fits \(\displaystyle\int\frac{x\,dx}{\sqrt{4-x^2}}\)? Evaluate it.
Hint 1Look at the numerator before choosing a method — is the derivative of the inside already present?
Hint 2\(u = 4-x^2\) gives \(du = -2x\,dx\), and \(x\,dx\) is right there.
Solution
STEP 1 · diagnose
The radical is the same as problem 1, but now there is an \(x\) in the numerator — and \(\frac{d}{dx}(4-x^2) = -2x\). That match means plain substitution, no triangles.
STEP 2 · substitute
\[ u = 4-x^2, \qquad du = -2x\,dx \quad\Longrightarrow\quad x\,dx = -\tfrac12 du \]
STEP 3 · rewrite and integrate
\[ \int\frac{-\frac12du}{\sqrt u} = -\frac12\int u^{-1/2}du = -\frac12\left(2u^{1/2}\right) = -\sqrt u \]
Identical radical, completely different method — decided entirely by one factor of \(x\). Always try \(u\)-substitution first; trig substitution is the fallback for when no such factor exists.
As \(x\to\infty\) the answer approaches \(\frac19\), so the total area from 3 to \(\infty\) is finite — plausible, since the integrand decays like \(x^{-3}\) ✓
\[ \int\frac{dx}{(x^2+4)^{3/2}} = \frac{x}{4\sqrt{x^2+4}} + C \]
WHERE THIS APPEARS
This exact integral gives the electric field of a charged rod and the magnetic field on the axis of a current loop. The \((\text{something}^2 + a^2)^{3/2}\) denominator is the fingerprint of an inverse-square law integrated along a line.
At \(\theta = \frac\pi2\): \(\sin\pi = 0\), so the bracket is \(\frac\pi2\). At 0 it is 0.
\[ = 2\left(\frac{\pi}{2}\right) = \pi \]
STEP 5 · geometric verification
\(y = \sqrt{4-x^2}\) with \(0\le x\le2\) is the quarter circle of radius 2 in the first quadrant:
\[ \frac14\pi r^2 = \frac14\pi(4) = \pi \ ✓ \]
WHY BOTHER WITH CALCULUS THEN
Because the same method handles \(\int_0^1\sqrt{4-x^2}dx\), where the region is not a nice circular sector and geometry alone gets you nowhere.
8
For each integral, name the best technique without evaluating it: (a) \(\int\frac{x\,dx}{x^2+9}\), (b) \(\int\frac{dx}{x^2+9}\), (c) \(\int\frac{dx}{\sqrt{9-x^2}}\), (d) \(\int\frac{x\,dx}{\sqrt{9-x^2}}\), (e) \(\int\frac{dx}{x^2-9}\).
Hint 1First question every time: is the derivative of the inside sitting in the numerator?
Hint 2Second question: does the denominator factor over the reals?
Hint 3\(x^2+9\) is irreducible; \(x^2-9 = (x-3)(x+3)\) is not.
Solution
(a) ∫ x dx/(x²+9)
u-substitution with \(u = x^2+9\); the \(x\,dx\) is half of \(du\). Answer type: logarithm.
(b) ∫ dx/(x²+9)
Standard arctangent form (or a tangent substitution, which derives it). No \(x\) available, and the quadratic doesn't factor. Answer type: arctangent.
(c) ∫ dx/√(9−x²)
Sine substitution — or recognize the arcsine form directly. Answer type: arcsine.
(d) ∫ x dx/√(9−x²)
u-substitution with \(u = 9-x^2\). The \(x\) changes everything. Answer type: square root.
(e) ∫ dx/(x²−9)
Partial fractions — the denominator factors as \((x-3)(x+3)\), so no trig is needed at all. Answer type: difference of logarithms.
THE DECISION ORDER
1. Can I simplify or split the fraction? 2. Is the inside's derivative present ⟹ substitution. 3. Does the denominator factor ⟹ partial fractions. 4. Is there an irreducible quadratic or a root of one ⟹ trig substitution / arctan / arcsin form. 5. Product of unlike things ⟹ parts. Running that checklist takes fifteen seconds and saves whole pages of wrong work.
Over \([-1,1]\) this gives \(\frac{\pi}{2}\cdot\frac12\cdot2 = \frac{\pi}{2}\)… more precisely \(\frac{\arcsin(1)-\arcsin(-1)}{2} = \frac{\pi}{2}\), with the square-root terms vanishing at both ends ✓ A positive area, as the positive integrand demands.
Part II · Module 8 Integration
Partial fractions
Break one hard rational function into a sum of easy ones. This is the technique that inverts Laplace transforms and solves the logistic equation — if you skip it, half of a differential equations course becomes impossible.
①The decomposition rules
First, two preconditions: the fraction must be proper (numerator degree < denominator degree — otherwise do polynomial long division first), and the denominator must be factored.
Then each piece integrates to something you already know: \(\frac{A}{x-a}\) gives a logarithm, \(\frac{B}{(x-a)^2}\) gives a power, and \(\frac{Ax+B}{x^2+c}\) splits into a logarithm plus an arctangent.
InteractiveOne curve, two identical descriptionsreveal the decomposition step by step
Amber is the original rational function; the thin cyan curves are the individual partial fractions; the dashed white curve is their sum. When the coefficients are right, the dashed curve lies exactly on top of the amber one — that is your check, and it works for any decomposition you do by hand.
Set up: \(\dfrac{5x-4}{(x-2)(x+1)} = \dfrac{A}{x-2} + \dfrac{B}{x+1}\).
STEP 2
Multiply through by the denominator: \(5x - 4 = A(x+1) + B(x-2)\).
STEP 3
Choose \(x\) to kill terms. Set \(x=2\): \(6 = 3A\), so \(A=2\). Set \(x=-1\): \(-9 = -3B\), so \(B=3\). (This shortcut is the "cover-up method" — far faster than expanding and matching coefficients.)
Two headline applications. Logistic growth: \(\frac{dP}{dt} = kP(1-P)\) separates to \(\int\frac{dP}{P(1-P)} = \int k\,dt\), and that left integral is a partial fraction. Inverse Laplace transforms: you will constantly face \(\frac{s+3}{s^2+3s+2}\) and need to split it into \(\frac{A}{s+1} + \frac{B}{s+2}\) before reading each piece off the transform table. It is the most-used algebra skill in the back half of the course.
Improper fractions need division first
\(\frac{x^2}{x^2-1}\) cannot be decomposed as written — the degrees are equal. Divide first: \(\frac{x^2}{x^2-1} = 1 + \frac{1}{x^2-1}\), and decompose only the leftover.
②Practice
1
Evaluate \(\displaystyle\int\frac{dx}{x^2-9}\).
Hint 1Factor the denominator as \((x-3)(x+3)\), then set up \(\frac{A}{x-3} + \frac{B}{x+3}\).
Hint 2Cover-up: \(1 = A(x+3) + B(x-3)\). Try \(x=3\) and \(x=-3\).
Solution
STEP 1 · factor and set up
\[ \frac{1}{x^2-9} = \frac{1}{(x-3)(x+3)} = \frac{A}{x-3} + \frac{B}{x+3} \]
\(\frac{B}{(x-1)^2}\) is a power, not a logarithm — \(\int (x-1)^{-2}dx = -(x-1)^{-1}\). Mixing those up is the second most common error here.
3
Evaluate \(\displaystyle\int\frac{dP}{P(1-P)}\) — the integral at the heart of the logistic equation.
Hint 1Set up \(\frac{A}{P} + \frac{B}{1-P}\) and clear denominators.
Hint 2\(1 = A(1-P) + BP\). Try \(P=0\), then \(P=1\).
Hint 3Both constants are 1 — but careful with the second integral: \(\int\frac{dP}{1-P} = -\ln|1-P|\) because of the chain rule.
Solution
STEP 1 · set up
\[ \frac{1}{P(1-P)} = \frac{A}{P} + \frac{B}{1-P} \]
STEP 2 · clear and cover up
\[ 1 = A(1-P) + BP \]
\[ P=0:\ 1 = A \qquad P=1:\ 1 = B \]
STEP 3 · integrate the first piece
\[ \int\frac{dP}{P} = \ln|P| \]
STEP 4 · integrate the second, carefully
The inside \(1-P\) has derivative \(-1\), so a minus appears:
\[ \int\frac{dP}{1-P} = -\ln|1-P| \]
STEP 5 · combine
\[ \int\frac{dP}{P(1-P)} = \ln|P| - \ln|1-P| + C = \ln\left|\frac{P}{1-P}\right| + C \]
WHERE IT LEADS
Setting this equal to \(kt + C\) and solving for \(P\) produces the S-shaped logistic curve \(P = \frac{1}{1+Ae^{-kt}}\) — see problem 9 for the full derivation with numbers.
4
Find the inverse Laplace transform of \(F(s) = \dfrac{s+3}{s^2+3s+2}\) by decomposing it. (Use \(\mathcal{L}^{-1}\{\frac{1}{s-a}\} = e^{at}\).)
Hint 1Factor: \(s^2+3s+2 = (s+1)(s+2)\).
Hint 2\(s+3 = A(s+2) + B(s+1)\); cover up with \(s=-1\) and \(s=-2\).
STEP 2 · set up the decomposition
\[ \frac{s+3}{(s+1)(s+2)} = \frac{A}{s+1} + \frac{B}{s+2} \quad\Longrightarrow\quad s+3 = A(s+2) + B(s+1) \]
STEP 3 · cover up
\[ s=-1:\ 2 = A \qquad s=-2:\ 1 = -B \Rightarrow B = -1 \]
STEP 4 · write F in transform-table form
\[ F(s) = \frac{2}{s+1} - \frac{1}{s+2} \]
STEP 5 · invert term by term
\(\frac{1}{s+1} = \frac{1}{s-(-1)}\) corresponds to \(e^{-t}\), and \(\frac{1}{s+2}\) to \(e^{-2t}\):
\[ f(t) = 2e^{-t} - e^{-2t} \]
WHAT YOU JUST DID
A complete inverse Laplace transform, done with nothing but algebra. In the ODE course this is the last step of every transform-method problem — which is why partial fractions is worth being fast at.
\(B\) came out zero — but you could not have known that in advance, and omitting the term would have made the system unsolvable. Always include every required term, then let the algebra decide.
Every irreducible-quadratic term \(\frac{Ax+B}{x^2+a^2}\) splits into a logarithm plus an arctangent. Recognizing that in advance tells you what your answer should look like before you compute it.
9
Solve the logistic equation \(\dfrac{dP}{dt} = 0.5\,P\left(1 - \dfrac{P}{100}\right)\) with \(P(0) = 10\), and interpret the long-run behavior.
Hint 1Separate first: all \(P\)'s with \(dP\), all \(t\)'s with \(dt\).
Hint 2Decompose \(\frac{1}{P(1-P/100)} = \frac{1}{P} + \frac{1/100}{1-P/100}\).
Hint 3After integrating you get \(\ln\left|\frac{P}{1-P/100}\right| = 0.5t + C\); apply the initial condition before solving for \(P\).
Solution
STEP 1 · separate
\[ \frac{dP}{P\left(1-\frac{P}{100}\right)} = 0.5\,dt \]
STEP 2 · partial fractions on the left
\[ \frac{1}{P\left(1-\frac{P}{100}\right)} = \frac{A}{P} + \frac{B}{1-\frac{P}{100}} \]
Clearing: \(1 = A\left(1-\frac{P}{100}\right) + BP\). At \(P=0\): \(A=1\). At \(P=100\): \(1 = 100B \Rightarrow B = \frac{1}{100}\).
STEP 3 · integrate both sides
The second piece needs the chain rule: \(\int\frac{1/100}{1-P/100}dP = -\ln\left|1-\frac{P}{100}\right|\).
\[ \ln|P| - \ln\left|1 - \frac{P}{100}\right| = 0.5t + C \quad\Longrightarrow\quad \ln\left|\frac{P}{1-P/100}\right| = 0.5t + C \]
STEP 4 · apply the initial condition
At \(t=0\), \(P=10\), so \(1 - \frac{10}{100} = 0.9\):
\[ \ln\left(\frac{10}{0.9}\right) = C \quad\Longrightarrow\quad C = \ln(11.111) \]
STEP 5 · exponentiate and solve for P
\[ \frac{P}{1-P/100} = 11.111\,e^{0.5t} \]
Cross-multiplying and collecting \(P\):
\[ P = \frac{100}{1 + 9e^{-0.5t}} \]
STEP 6 · check and interpret
At \(t=0\): \(\frac{100}{1+9} = 10\) ✓. As \(t\to\infty\), \(e^{-0.5t}\to0\) and \(P\to100\) — the carrying capacity. The curve is S-shaped: near-exponential growth while \(P\) is small, then a slowdown as the \(\left(1-\frac{P}{100}\right)\) factor throttles it.
WHY THIS PROBLEM MATTERS
It is the standard demonstration that Part II is not busywork: separation (I4) + partial fractions (I8) + logarithm rules together solve one of the most-used models in biology, epidemiology, and marketing.
Part II · Checkpoint Quiz
Quiz B — Integration
Ten questions on the integration toolkit. The hardest exam skill is not executing a method — it is choosing the right one, so several questions ask exactly that.
Question 1
\(\displaystyle\int_0^{2\pi}\sin x\,dx = \) ?
Why
Signed area: the hump above the axis on \([0,\pi]\) exactly cancels the one below on \([\pi,2\pi]\). The total area (with absolute values) would be 4.
Question 2
If \(A(x) = \int_2^x f(t)\,dt\), then \(A'(x) = \) ?
Why
Fundamental Theorem Part 1. The lower limit is a constant, so it contributes nothing to the derivative — it only shifts \(A\) vertically.
Question 3
Which method is most efficient for \(\displaystyle\int x e^{x^2}\,dx\)?
Why
The \(x\,dx\) is precisely \(\frac12 du\), so the integral becomes \(\frac12\int e^u du = \frac12 e^{x^2}+C\). Parts would work forever without terminating — \(\int e^{x^2}dx\) has no elementary antiderivative on its own.
Question 4
For \(\displaystyle\int x^2\ln x\,dx\), LIATE says to choose:
Why
L (log) outranks A (algebraic), so the log becomes \(u\). Choice A would require you to already know \(\int\ln x\,dx\) just to find \(v\) — backwards.
\(p = \frac12 \le 1\), so it diverges at infinity. \(2\sqrt b \to \infty\). (The same integrand on \([0,1]\) converges — the two halves of the p-test point opposite ways.)
Question 6
Which substitution suits \(\displaystyle\int\frac{dx}{\sqrt{x^2-16}}\)?
Why
Pattern \(\sqrt{x^2-a^2}\) calls for secant, since \(\sec^2\theta - 1 = \tan^2\theta\) clears the root. Choice D fails because there is no \(x\,dx\) in the numerator.
Question 7
The correct partial-fraction setup for \(\dfrac{2x+1}{(x-3)(x^2+4)}\) is:
Why
An irreducible quadratic factor gets a linear numerator \(Bx+C\). And \(x^2+4\) does not factor over the reals, ruling out C.
Question 8
\(\displaystyle\int\frac{dx}{x} = \) ?
Why
The power rule fails at \(n=-1\) (choice A divides by zero), and the absolute value matters: the antiderivative must be defined on both branches, which you will need when solving \(\frac{dy}{y}\) for possibly negative \(y\).
Question 9
After the substitution \(u = x^2+1\) in \(\displaystyle\int_0^2 x(x^2+1)^3dx\), the new integral is:
Why
Both changes are required: the limits become \(u(0)=1\) and \(u(2)=5\), and \(x\,dx = \frac12 du\). Choices A and D commit the classic error of keeping \(x\)-limits on a \(u\)-integral.
Question 10
Which is true of \(\displaystyle\int f(x)\,dx\) versus \(\displaystyle\int_a^b f(x)\,dx\)?
Why
The constant cancels in \(F(b)-F(a)\), so definite integrals never carry \(+C\). And that \(+C\) on the indefinite integral is exactly what an ODE's initial condition later determines.
Part III · Module 1 Series
Convergence and divergence tests
An infinite sum is defined as the limit of its partial sums. Either that limit exists (the series converges to a number) or it doesn't (it diverges). Everything in this module is machinery for deciding which, without computing the sum.
Note the distinction that trips people up: the terms \(a_n\) form one sequence, and the partial sums \(S_N\) form a different one. Terms going to zero is necessary for convergence, but nowhere near sufficient.
InteractiveTerms vs. partial sumsadd more terms and watch
last term aN—partial sum SN—verdict —true sum —
Bars are the terms; the dots and line are the partial sums. Compare the harmonic series (\(p=1\)) with \(p = 1.05\): the terms look identical to the eye, yet one diverges and the other converges. That is precisely why you need tests instead of eyeballing.
②The two series you must know by heart
Geometric series — the only one with an easy exact sum
The harmonic series \(\sum\frac1n\) is the \(p=1\) boundary case, and it diverges — famously slowly. Its terms go to zero, and yet the sum is infinite; keep that counterexample in your pocket.
InteractiveWhich test should I use?pick a series, follow the highlighted path
recommended test —conclusion —
Read the highlighted path top to bottom. Nearly every exam question is decided in the first two boxes: always check the \(n\)th-term test first (it is free), then look for a form you already recognize before reaching for a heavier test.
③The test catalog
Test
Use when
Says
nth-term (divergence)
Always check first
If \(\lim a_n \ne 0\), it diverges. If \(\lim a_n = 0\), inconclusive — this test can never prove convergence.
Geometric
\(\sum ar^n\)
Converges iff \(|r|<1\), to \(\frac{a}{1-r}\).
p-series
\(\sum 1/n^p\)
Converges iff \(p>1\).
Integral
\(a_n = f(n)\) with \(f\) positive, decreasing, easily integrable
\(\sum a_n\) and \(\int_1^\infty f\) do the same thing.
Comparison
Terms resemble a known series
Smaller than convergent ⟹ converges. Bigger than divergent ⟹ diverges.
Limit comparison
Rational-ish terms
If \(\lim\frac{a_n}{b_n} = L\) with \(0
Ratio
Factorials or \(n\) in an exponent
\(L = \lim\left|\frac{a_{n+1}}{a_n}\right|\): converges if \(L<1\), diverges if \(L>1\), inconclusive if \(L=1\).
Root
Whole term raised to the \(n\)th power
\(L = \lim\sqrt[n]{|a_n|}\), same verdicts as the ratio test.
Alternating series
\(\sum(-1)^n b_n\) with \(b_n>0\)
Converges if \(b_n\) decreases to 0. Error is bounded by the first omitted term.
Absolute vs. conditional convergence
If \(\sum|a_n|\) converges, so does \(\sum a_n\) (absolutely convergent) — and you may rearrange it freely. If \(\sum a_n\) converges but \(\sum|a_n|\) does not, it is conditionally convergent: the alternating harmonic series is the standard example, and rearranging it can produce any sum you like. Power series always converge absolutely inside their interval, which is why the term-by-term operations in module S6 are legal there.
Worked example · three series, three tests
A
\(\sum\frac{n}{2n+1}\): terms → \(\frac12 \ne 0\), so the \(n\)th-term test kills it immediately. Diverges. (No further work needed — and this is the free check people skip.)
B
\(\sum\frac{n+1}{n^3+2}\): terms behave like \(\frac{n}{n^3} = \frac{1}{n^2}\). Limit comparison with \(b_n = 1/n^2\) gives \(L = 1\), and \(\sum 1/n^2\) converges (\(p=2\)). Converges.
The formula \(\frac{a}{1-r}\) needs \(a\) = the actual first term. Since the sum starts at \(n=1\):
\[ a = \left(\frac23\right)^1 = \frac23 \]
STEP 4 · sum it
\[ \sum_{n=1}^{\infty}\left(\frac23\right)^n = \frac{2/3}{1-2/3} = \frac{2/3}{1/3} = 2 \]
STEP 5 · check the starting index
Had the sum started at \(n=0\), the extra term \(\left(\frac23\right)^0 = 1\) would give 3 instead. The starting index changes the sum, though never the convergence — a distinction worth checking every time.
2
Test \(\displaystyle\sum_{n=1}^{\infty}\frac{n^2}{n^3+1}\) for convergence.
Hint 1Terms do go to zero, so the \(n\)th-term test is inconclusive — you need more.
Hint 2Keep only dominant powers: \(\frac{n^2}{n^3} = \frac1n\). Compare against the harmonic series.
Hint 3Limit comparison with \(b_n = 1/n\) gives \(L=1\), a finite nonzero number.
Solution
STEP 1 · run the free test first
\[ \lim_{n\to\infty}\frac{n^2}{n^3+1} = 0 \]
Inconclusive — the \(n\)th-term test can only prove divergence.
STEP 2 · guess the behavior from dominant powers
For large \(n\), \(\frac{n^2}{n^3+1} \approx \frac{n^2}{n^3} = \frac1n\). So the series should behave like the harmonic series.
Direct comparison would need an inequality, and \(\frac{n^2}{n^3+1} < \frac1n\) points the wrong way (smaller than a divergent series proves nothing). Limit comparison sidesteps the inequality entirely — which is why it is the workhorse for rational terms.
3
Test \(\displaystyle\sum_{n=1}^{\infty}\frac{n!}{2^n}\).
Hint 1Factorial ⟹ ratio test.
Hint 2\(\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{2^{n+1}}\cdot\frac{2^n}{n!} = \frac{n+1}{2}\).
Solution
STEP 1 · choose the test
A factorial in the terms makes the ratio test the natural choice — factorials telescope beautifully in a ratio.
STEP 2 · form the ratio
\[ \frac{a_{n+1}}{a_n} = \frac{(n+1)!}{2^{n+1}}\cdot\frac{2^n}{n!} \]
STEP 3 · simplify
\((n+1)! = (n+1)\cdot n!\) and \(\frac{2^n}{2^{n+1}} = \frac12\):
\[ = \frac{n+1}{2} \]
STEP 4 · take the limit
\[ L = \lim_{n\to\infty}\frac{n+1}{2} = \infty > 1 \quad\Longrightarrow\quad \textbf{diverges} \]
STEP 5 · confirm with the free test
Since \(n!\) outgrows \(2^n\), the terms themselves blow up: \(\frac{n!}{2^n}\to\infty \ne 0\), so the \(n\)th-term test also kills it. Growth hierarchy worth memorizing: \(\ln n \ll n^p \ll a^n \ll n! \ll n^n\).
4
Show that \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{\sqrt n}\) converges conditionally, and estimate the error in using \(S_{100}\).
Hint 1Check the two alternating-series conditions: \(b_n = 1/\sqrt n\) decreasing, and \(b_n \to 0\).
Hint 2For "conditionally," you must also show \(\sum|a_n| = \sum 1/\sqrt n\) diverges — what is \(p\) there?
Hint 3The alternating-series error bound is just the first term you left out: \(|S - S_N| \le b_{N+1}\).
A hundred terms and the answer is only good to one decimal place. Conditionally convergent series converge painfully slowly — and can be rearranged to sum to any value you like (Riemann's rearrangement theorem), which is why absolute convergence matters.
5
Test \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2+1}\).
Hint 1Compare with a \(p\)-series you already know.
Hint 2\(n^2 + 1 > n^2\), so \(\frac{1}{n^2+1} < \frac{1}{n^2}\) — and this time the inequality points the useful way.
Solution
STEP 1 · pick the comparison series
Dominant powers give \(\frac{1}{n^2+1} \approx \frac{1}{n^2}\), and \(\sum\frac{1}{n^2}\) converges (\(p=2>1\)).
STEP 2 · establish the inequality
For every \(n \ge 1\), enlarging the denominator shrinks the fraction:
STEP 4 · a bound on the sum
\[ \sum\frac{1}{n^2+1} < \sum\frac{1}{n^2} = \frac{\pi^2}{6} \approx 1.645 \]
(The true value is about 1.077.)
DIRECTION MATTERS
Comparison only works one way: smaller than convergent ⟹ converges, and larger than divergent ⟹ diverges. The other two combinations tell you nothing. If your inequality points the wrong way, switch to limit comparison.
6
Test \(\displaystyle\sum_{n=1}^{\infty}\frac{2^n}{n!}\), and identify what it sums to.
Hint 1Factorial ⟹ ratio test.
Hint 2\(\frac{a_{n+1}}{a_n} = \frac{2^{n+1}}{(n+1)!}\cdot\frac{n!}{2^n} = \frac{2}{n+1}\).
Hint 3Compare the terms with the Maclaurin series for \(e^x\) at \(x=2\).
Solution
STEP 1 · form the ratio
\[ \frac{a_{n+1}}{a_n} = \frac{2^{n+1}}{(n+1)!}\cdot\frac{n!}{2^n} = \frac{2}{n+1} \]
STEP 2 · take the limit
\[ L = \lim_{n\to\infty}\frac{2}{n+1} = 0 < 1 \quad\Longrightarrow\quad \textbf{converges} \]
(And \(L=0\) means it converges fast — factorials crush exponentials.)
STEP 3 · recognize the sum
The Maclaurin series \(e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}\) at \(x=2\) gives
STEP 4 · apply the verdict
\[ L = \frac34 < 1 \quad\Longrightarrow\quad \textbf{converges absolutely} \]
STEP 5 · sanity check
For large \(n\) the terms behave like \(\left(\frac34\right)^n\) — a convergent geometric series ✓. The root test is essentially detecting "eventually geometric with ratio \(L\)."
8
Use the integral test to determine whether \(\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln n}\) converges.
Hint 1Check the three conditions first: the function \(f(x) = \frac{1}{x\ln x}\) must be positive, continuous, and decreasing on \([2,\infty)\).
On \([2,\infty)\): \(f(x) = \frac{1}{x\ln x}\) is positive (both factors positive), continuous (no zeros in the denominator), and decreasing (both \(x\) and \(\ln x\) increase). The integral test applies.
STEP 2 · set up the improper integral
\[ \int_2^\infty\frac{dx}{x\ln x} = \lim_{b\to\infty}\int_2^b\frac{dx}{x\ln x} \]
STEP 3 · substitute
\[ u = \ln x,\quad du = \frac{dx}{x} \quad\Longrightarrow\quad \int\frac{du}{u} = \ln|u| = \ln(\ln x) \]
STEP 4 · evaluate the limit
\[ \lim_{b\to\infty}\left[\ln(\ln b) - \ln(\ln 2)\right] = \infty \]
STEP 5 · conclude
The integral diverges, so the series diverges too.
WHAT THE INTEGRAL TEST IS REALLY SAYING
The terms \(a_n\) are the areas of unit-width rectangles that sandwich the curve \(y = f(x)\). If the area under the curve is infinite, so is the total rectangle area, and vice versa. Note how close this is to the convergent \(\sum\frac{1}{n(\ln n)^2}\) — the boundary between convergence and divergence is astonishingly fine.
9
Does \(\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\frac{n}{n+1}\) converge? Be careful — it is alternating.
Hint 1The alternating series test has two conditions. Check the limit one first.
Hint 2\(b_n = \frac{n}{n+1} \to 1\), not 0.
Hint 3What does the \(n\)th-term test say when the terms don't approach zero?
Solution
STEP 1 · check the alternating test's second condition
\[ b_n = \frac{n}{n+1} \longrightarrow 1 \ne 0 \]
The test fails — but failing a test proves nothing on its own, so go further.
STEP 2 · apply the nth-term test to the actual terms
\[ a_n = (-1)^n\frac{n}{n+1} \quad\Longrightarrow\quad a_n \text{ oscillates between } \approx +1 \text{ and } \approx -1 \]
So \(\lim a_n\) does not exist — in particular it is not 0.
\(S_N\) bounces roughly between \(-0.5\) and \(0\) forever without settling — divergence by oscillation, not by blowing up. Set the partial-sums lab to "Σ n/(n+1)" to see the analogous non-alternating behavior.
THE HABIT
Always run the \(n\)th-term test first, even on alternating series. It is free, and it settles a surprising fraction of exam problems in one line.
Part III · Module 2 Series
Interval and radius of convergence
A power series is a function defined by an infinite sum — but only for the \(x\)-values where that sum actually converges. Finding that set is the first thing you do with any power series, including every series solution of an ODE.
There are only three possible outcomes, and a theorem guarantees it:
Converges only at \(x=a\) (radius \(R=0\));
Converges for all \(x\) (\(R=\infty\));
Converges on \(|x-a|R\), for some finite \(R>0\).
In case 3 the endpoints \(x = a\pm R\) are genuinely undecided by the ratio test and must be checked separately, one at a time, by substituting them in.
InteractiveThe convergence interval on a number linedrag \(x\), see the partial sums behave
center a —radius R —interval —at this x —
Top: the number line, with the convergence interval in green and the endpoints marked by their individual verdicts. Bottom: the partial sums at your chosen \(x\). Inside the interval they settle down fast; step just outside and they explode — and right at an endpoint they crawl, which is exactly why endpoints need their own test.
②The procedure
Worked example · full interval of convergence
Find the interval of convergence of \(\displaystyle\sum_{n=1}^{\infty}\frac{(x-2)^n}{n\,3^n}\).
Converges when that limit is \(<1\): \(|x-2| < 3\), so \(R = 3\) and the open interval is \((-1, 5)\).
STEP 3
Test \(x = 5\): the series becomes \(\sum\frac{3^n}{n3^n} = \sum\frac1n\) — the harmonic series. Diverges, so exclude 5.
STEP 4
Test \(x = -1\): the series becomes \(\sum\frac{(-3)^n}{n3^n} = \sum\frac{(-1)^n}{n}\) — alternating harmonic. Converges, so include \(-1\).
STEP 5
Interval of convergence: \([-1, 5)\). Note the asymmetry — the two endpoints really can behave differently.
The ratio test is silent at the endpoints
At \(|x-a| = R\) the ratio limit is exactly 1, which is the test's inconclusive case. Writing "converges on \([a-R, a+R]\)" without checking is guessing, and it is wrong about half the time.
Where this shows up in ODEs
When you solve \(y'' + p(x)y' + q(x)y = 0\) with a power series about \(x_0\), the resulting series is guaranteed to converge at least as far as the nearest singular point — the nearest place where \(p\) or \(q\) blows up. The radius of convergence tells you the domain on which your solution is actually valid, which is real information about the physical problem, not a technicality.
③Practice
1
Find the radius and interval of convergence of \(\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!}\).
Hint 1Ratio test: \(\left|\frac{a_{n+1}}{a_n}\right| = \frac{|x|}{n+1}\).
Hint 2What happens to \(\frac{|x|}{n+1}\) as \(n\to\infty\), for any fixed \(x\)?
Solution
STEP 1 · form the ratio of consecutive terms
\[ \left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{x^{n+1}}{(n+1)!}\cdot\frac{n!}{x^n}\right| = \frac{|x|}{n+1} \]
STEP 2 · take the limit, treating x as fixed
\[ \lim_{n\to\infty}\frac{|x|}{n+1} = 0 \qquad\text{for every real } x \]
STEP 3 · apply the ratio test
\(0 < 1\) always, so the series converges absolutely for every \(x\).
STEP 4 · state the radius and interval
\[ R = \infty, \qquad \text{interval} = (-\infty, \infty) \]
There are no endpoints to check — that only arises for finite \(R\).
WHY IT MATTERS
This series is \(e^x\). Its infinite radius is why exponentials are such well-behaved ODE solutions: the series representation is valid everywhere, so nothing you do with it can wander outside a domain of validity.
2
Find the interval of convergence of \(\displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n^2}\).
Hint 1The ratio limit is \(|x|\cdot\lim\left(\frac{n}{n+1}\right)^2 = |x|\).
Hint 2So \(R=1\) and the open interval is \((-1,1)\). Now substitute \(x=1\) and \(x=-1\) separately.
Hint 3At \(x=1\) you get a \(p\)-series with \(p=2\).
Solution
STEP 1 · ratio test
\[ \left|\frac{a_{n+1}}{a_n}\right| = |x|\cdot\frac{n^2}{(n+1)^2} \longrightarrow |x| \]
STEP 3 · test the right endpoint x = 1
\[ \sum_{n=1}^{\infty}\frac{1^n}{n^2} = \sum\frac{1}{n^2} \quad\text{— a } p\text{-series with } p=2 > 1 \ \Longrightarrow\ \textbf{converges} \]
STEP 4 · test the left endpoint x = −1
\[ \sum\frac{(-1)^n}{n^2} \]
Its absolute-value series is again \(\sum\frac1{n^2}\), which converges — so this endpoint converges absolutely.
STEP 5 · assemble the answer
\[ \text{Interval of convergence} = [-1,\,1] \]
Both endpoints included, so brackets on both sides.
3
Find the radius of convergence of \(\displaystyle\sum_{n=0}^{\infty}n!\,x^n\).
Hint 1\(\left|\frac{a_{n+1}}{a_n}\right| = (n+1)|x|\).
Hint 2For any \(x \ne 0\), that grows without bound.
Solution
STEP 1 · form the ratio
\[ \left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)!\,|x|^{n+1}}{n!\,|x|^n} = (n+1)|x| \]
STEP 2 · take the limit for fixed x ≠ 0
\[ \lim_{n\to\infty}(n+1)|x| = \infty > 1 \quad\Longrightarrow\quad \text{diverges} \]
STEP 3 · check x = 0 separately
Every term after the first is zero, so the series trivially converges there (to \(a_0 = 1\)).
STEP 4 · conclude
\[ R = 0, \qquad \text{converges only at } x = 0 \]
WHY THIS IS USELESS AS A FUNCTION
A power series with \(R=0\) defines nothing on any interval — you cannot differentiate it, integrate it, or use it to solve an ODE. When a series solution produces \(R=0\), it is a signal that the point you expanded about is singular and the method must be modified (Frobenius' method).
4
Find the interval of convergence of \(\displaystyle\sum_{n=1}^{\infty}\frac{(x+2)^n}{n\,4^n}\).
Hint 1The center is where \(x+2 = 0\), i.e. \(a = -2\).
Hint 2The ratio limit is \(\frac{|x+2|}{4}\); set it \(<1\).
Hint 3Endpoints are \(x = -6\) and \(x = 2\). Substitute each into the original series.
Solution
STEP 1 · ratio test
\[ \left|\frac{a_{n+1}}{a_n}\right| = \frac{|x+2|^{n+1}}{(n+1)4^{n+1}}\cdot\frac{n\,4^n}{|x+2|^n} = \frac{n}{n+1}\cdot\frac{|x+2|}{4} \longrightarrow \frac{|x+2|}{4} \]
STEP 2 · solve for the radius
\[ \frac{|x+2|}{4} < 1 \ \Longrightarrow\ |x+2| < 4 \ \Longrightarrow\ R = 4, \quad -6 < x < 2 \]
STEP 3 · test x = 2 (right endpoint)
Then \(x+2 = 4\) and \(\frac{4^n}{n4^n} = \frac1n\):
The factor \(n\) in the numerator is what killed both endpoints — compare with problem 2, where the \(n^2\) in the denominator saved both.
7
Find the interval of convergence of \(\displaystyle\sum_{n=1}^{\infty}\frac{(x-1)^{2n}}{n}\). (Note the even exponent.)
Hint 1The ratio gives \(|x-1|^2\), not \(|x-1|\).
Hint 2\(|x-1|^2 < 1\) is equivalent to \(|x-1| < 1\) — take the square root of both sides.
Hint 3At both endpoints the even power makes the terms identical.
Solution
STEP 1 · ratio test with the even power
\[ \left|\frac{a_{n+1}}{a_n}\right| = \frac{|x-1|^{2n+2}}{n+1}\cdot\frac{n}{|x-1|^{2n}} = \frac{n}{n+1}|x-1|^2 \longrightarrow |x-1|^2 \]
Since both sides are non-negative, taking square roots is safe.
STEP 3 · endpoints
At \(x=2\): \((x-1)^{2n} = 1^{2n} = 1\), giving \(\sum\frac1n\) — diverges. At \(x=0\): \((-1)^{2n} = 1\) as well, giving the same harmonic series — diverges.
Because only even powers appear, the two endpoints behave identically — no alternating rescue is possible. This pattern shows up in the series for \(\cos x\) and \(\frac{1}{1+x^2}\), so it is worth recognizing.
8
You seek a power-series solution of \((x-2)\,y'' + y' + y = 0\) centered at \(x_0 = 0\). Without computing anything, what radius of convergence is guaranteed, and why?
Hint 1Put the equation in standard form \(y'' + p(x)y' + q(x)y = 0\) by dividing through.
Hint 2Look for the values of \(x\) where \(p\) or \(q\) fails to be analytic — the singular points.
Hint 3The guaranteed radius is the distance from the center to the nearest singular point.
Solution
STEP 1 · put it in standard form
Divide through by the leading coefficient \((x-2)\):
\[ y'' + \frac{1}{x-2}y' + \frac{1}{x-2}y = 0 \]
STEP 2 · find the singular points
\(p(x) = q(x) = \frac{1}{x-2}\) blow up only at
\[ x = 2 \]
Everywhere else both are analytic, so \(x_0 = 0\) is an ordinary point and the method applies.
The theorem says any series solution about \(x_0=0\) converges at least on
\[ |x| < 2 \qquad\text{i.e. } R \ge 2 \]
STEP 5 · why this is genuinely useful
You learn the domain of validity of your answer before doing a single line of algebra — and if a physical problem needs \(x=3\), you know immediately to re-expand about a different center. (Over the complex numbers the same rule uses the nearest complex singularity, which is why \(\frac{1}{1+x^2}\) has \(R=1\) despite looking perfectly smooth on the real line.)
9
Find the interval of convergence of \(\displaystyle\sum_{n=1}^{\infty}\frac{(x+3)^n}{\sqrt n}\).
Hint 1The centre is where \(x+3=0\).
Hint 2The ratio limit is \(|x+3|\), since \(\sqrt{n}/\sqrt{n+1}\to1\).
Hint 3Test each endpoint separately — one is a \(p\)-series, the other alternating.
Solution
STEP 1 · ratio test
\[ \left|\frac{a_{n+1}}{a_n}\right| = |x+3|\sqrt{\frac{n}{n+1}} \longrightarrow |x+3| \]
STEP 2 · radius and centre
\[ |x+3|<1 \quad\Longrightarrow\quad a = -3,\ R = 1, \ \text{open interval } (-4,-2) \]
STEP 3 · test x = −2
Then \(x+3 = 1\):
\[ \sum\frac{1}{\sqrt n} \ \text{— a } p\text{-series with } p=\tfrac12 \le 1 \ \Longrightarrow\ \textbf{diverges} \]
A textbook case of the two endpoints behaving differently — and of why the ratio test's silence at \(|x-a|=R\) forces you to check both by hand.
Part III · Module 3 Series
Combining series & the arithmetic of series
You can add, subtract, and scale convergent series term by term — provided the indices line up. That proviso is doing more work than it looks, and it is the source of most algebra errors in power-series ODE solutions.
Two cautions attached to that innocent-looking statement:
Both series must converge and start at the same index with the same power of \(x\) in each slot. Otherwise you are adding \(a_3\) to \(b_2\).
Convergent + divergent = divergent. But divergent + divergent can be anything — \(\sum\frac1n\) and \(\sum-\frac1n\) both diverge, yet their termwise sum is 0.
InteractiveCombining two series term by termwatch the slots line up
Σa (partial) —Σb (partial) —combined —status —
The combined partial sums (cyan) are always exactly the combination of the two individual partial sums — that is the whole content of linearity. Try harmonic + alternating harmonic: one diverges, the other converges, and the sum diverges. There is no cancellation to rescue it.
②Multiplying and dividing series
Multiplication is not termwise. The product of two power series is a Cauchy product, which collects all the ways to reach each power:
Convergence: the first has \(R = \frac12\), the second \(R=\infty\). The combination converges on the smaller interval, \(|x| < \frac12\) — a sum is only as good as its worst piece.
would be finite — contradicting the divergence of the harmonic series.
STEP 3 · conclude
\[ \textbf{Diverges} \]
STEP 4 · the general rule
Convergent + divergent = divergent, always. But divergent + divergent can be anything: \(\sum\frac1n\) and \(\sum\left(-\frac1n\right)\) both diverge, yet their termwise sum is \(\sum 0 = 0\).
TRY IT
Set the combining lab to "Σ 1/n" and "Σ (1/3)ⁿ" with the a + b option — the combined partial sums track the harmonic curve almost exactly, because the geometric piece contributes at most 0.5 total.
3
Given \(e^x = \sum\frac{x^n}{n!}\) and \(\frac{1}{1-x} = \sum x^n\) (for \(|x|<1\)), find the first three terms of the product \(\frac{e^x}{1-x}\).
Hint 1Write out both series to \(x^2\) and multiply as polynomials, discarding anything of degree 3 or higher.
Hint 2\((1 + x + \frac{x^2}{2})(1 + x + x^2)\). Collect the \(x^2\) terms: \(\frac12 + 1 + 1\).
Solution
STEP 1 · write both series to the needed degree
\[ e^x = 1 + x + \frac{x^2}{2} + \cdots, \qquad \frac{1}{1-x} = 1 + x + x^2 + \cdots \]
Since only three terms are wanted, anything of degree 3 or higher can be dropped immediately.
STEP 2 · collect the constant term
\[ c_0 = (1)(1) = 1 \]
STEP 3 · collect the x term
Two ways to reach \(x^1\): \(1\cdot x\) and \(x\cdot 1\):
\[ c_1 = 1 + 1 = 2 \]
STEP 4 · collect the x² term
Three ways: \(1\cdot x^2\), \(x\cdot x\), \(\frac{x^2}{2}\cdot 1\):
\[ c_2 = 1 + 1 + \frac12 = \frac52 \]
STEP 5 · assemble and state the domain
\[ \frac{e^x}{1-x} = 1 + 2x + \frac52x^2 + \cdots, \qquad |x| < 1 \]
The radius is the smaller of the two (\(\infty\) and 1).
A NICE PATTERN
The coefficients are the partial sums \(\sum_{k=0}^{n}\frac{1}{k!}\): \(1,\ 2,\ 2.5,\ 2.667,\dots\to e\). That is the Cauchy product \(c_n = \sum_k a_kb_{n-k}\) with \(b_j = 1\), doing exactly what it should ✓
4
Evaluate the telescoping series \(\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{n} - \frac{1}{n+1}\right)\). Why can't you evaluate each piece separately?
Hint 1Write out \(S_4\) explicitly and watch what cancels.
Hint 2Find a closed form for \(S_N\), then take \(N\to\infty\).
Hint 3Both individual series are harmonic — what does that forbid?
Solution
STEP 1 · write out a partial sum
\[ S_4 = \left(1-\tfrac12\right) + \left(\tfrac12-\tfrac13\right) + \left(\tfrac13-\tfrac14\right) + \left(\tfrac14-\tfrac15\right) \]
STEP 2 · watch the interior cancel
Every negative piece is annihilated by the next positive one; only the two ends survive:
\[ S_4 = 1 - \frac15 \]
STEP 3 · general partial sum
\[ S_N = 1 - \frac{1}{N+1} \]
STEP 4 · take the limit
\[ \sum_{n=1}^{\infty}\left(\frac1n - \frac{1}{n+1}\right) = \lim_{N\to\infty}\left(1 - \frac{1}{N+1}\right) = 1 \]
STEP 5 · why you may not split
Splitting into \(\sum\frac1n - \sum\frac{1}{n+1}\) gives \(\infty - \infty\), which is meaningless. Linearity of series requires both pieces to converge; here neither does, and only their combination is finite.
WHERE TELESCOPING COMES FROM
Notice \(\frac1n - \frac{1}{n+1} = \frac{1}{n(n+1)}\) — the partial-fraction decomposition of module I8, run in reverse. Telescoping problems are usually partial-fraction problems in disguise.
5
Write \(\displaystyle\sum_{n=0}^{\infty}\frac{3^n x^n}{n!} + \sum_{n=0}^{\infty}(-1)^n x^n\) as a single power series, give the coefficient of \(x^2\), and state where it converges.
Hint 1Both sums already start at \(n=0\) and both carry \(x^n\) — the slots line up, so you can combine directly.
Hint 2Combined coefficient: \(\frac{3^n}{n!} + (-1)^n\).
Hint 3Radii: the first is \(\infty\), the second is 1. Which one wins?
Solution
STEP 1 · confirm the slots align
Same starting index (0), same power of \(x\) in each slot. No shifting needed — you may add coefficient by coefficient.
First series: \(e^{3x}\), with \(R = \infty\). Second: geometric in \(-x\), with \(R = 1\). A sum converges only where both pieces do:
\[ |x| < 1 \]
STEP 5 · identify the function
\[ \sum = e^{3x} + \frac{1}{1+x}, \qquad |x| < 1 \]
The singularity at \(x = -1\) comes entirely from the second piece and caps the whole thing — a sum is only as good as its worst-behaved part.
6
Use the Cauchy product on the series for \(e^{x}\) and \(e^{-x}\) to verify that the product's constant term is 1 and its \(x^1\) and \(x^2\) coefficients are 0.
Hint 1\(a_k = \frac{1}{k!}\) and \(b_k = \frac{(-1)^k}{k!}\).
Hint 2\(c_n = \sum_{k=0}^{n}a_kb_{n-k}\) — for each \(n\), list every way to split the power.
Hint 3The answer should be the series for \(e^x\cdot e^{-x} = e^0 = 1\).
Solution
STEP 1 · list the coefficients
\[ a_k = \frac{1}{k!}, \qquad b_k = \frac{(-1)^k}{k!} \]
Every coefficient beyond the constant cancels, giving the series \(1 + 0x + 0x^2 + \cdots = 1\) — exactly \(e^xe^{-x} = e^0 = 1\).
WHY THE CAUCHY FORMULA LOOKS LIKE THAT
To land on \(x^n\), pick \(x^k\) from the first series and \(x^{n-k}\) from the second, for every \(k\) from 0 to \(n\). Collecting those products is \(c_n = \sum_k a_kb_{n-k}\). It is just polynomial multiplication, extended to infinitely many terms.
7
Find the coefficient of \(x^3\) in the product \(\sin x \cdot \dfrac{1}{1-x}\).
Hint 1\(\sin x = x - \frac{x^3}{6} + \cdots\) (note: no even powers) and \(\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots\).
Hint 2List every pair of exponents that adds to 3.
Hint 3Only \(x\cdot x^2\) and \(-\frac{x^3}{6}\cdot 1\) contribute — the even-power terms of sine are zero.
Solution
STEP 1 · write both series
\[ \sin x = 0 + 1\cdot x + 0\cdot x^2 - \frac{1}{6}x^3 + \cdots, \qquad \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots \]
Writing the zero coefficients explicitly prevents lost terms.
STEP 2 · enumerate the ways to reach x³
\[ (k, 3-k) = (0,3),\ (1,2),\ (2,1),\ (3,0) \]
Radii: \(\infty\) for \(\sin\), 1 for the geometric series ⟹ the product series is valid for \(|x|<1\).
8
Suppose \(\sum a_n\) converges and \(\sum b_n\) diverges. For each, decide whether it must converge, must diverge, or could be either: (a) \(\sum (a_n + b_n)\), (b) \(\sum 5a_n\), (c) \(\sum (a_n b_n)\), (d) \(\sum (b_n - b_n)\).
Hint 1Linearity covers (a) and (b); it says nothing about products.
Hint 2For (c), try to build one example that converges and one that diverges.
Hint 3In (d), simplify before applying any rule.
Solution
(a) Σ(aₙ + bₙ) — must DIVERGE
If it converged, then \(\sum b_n = \sum(a_n+b_n) - \sum a_n\) would be a difference of two convergent series, hence convergent — contradiction.
(b) Σ5aₙ — must CONVERGE
Constants factor out: \(\sum 5a_n = 5\sum a_n\), a finite number times a finite number.
(c) Σ(aₙbₙ) — COULD BE EITHER
Take \(a_n = \frac{1}{n^2}\) (convergent) and \(b_n = n\) (divergent): the product is \(\sum\frac1n\), which diverges. But with \(b_n = 1\) for all \(n\) (divergent), the product is \(\sum\frac{1}{n^2}\), which converges. No general rule exists — series do not multiply termwise.
(d) Σ(bₙ − bₙ) — must CONVERGE
Simplify first: every term is exactly 0, so the sum is 0. The temptation to write \(\sum b_n - \sum b_n = \infty - \infty\) is precisely the error linearity forbids — you may only split when both pieces converge.
THE MORAL
Linearity is a statement about convergent series. Outside that hypothesis, simplify the terms first and reason about the combined series directly.
The habit worth keeping: split first, verify each piece converges, then use linearity — the order matters, as module S3 problem 4's telescoping example shows.
Part III · Module 4 Series
Adjusting the index of a series
Reindexing is a purely cosmetic operation — you are renaming the counter, not changing a single term. But it is the mechanical step that makes power-series solutions of ODEs possible, and doing it fluently is worth an hour of practice.
The bookkeeping rule in one line: whatever you add to the starting index, you subtract inside the terms (and vice versa). If both moves go the same direction, you have changed the series.
InteractiveReindexing writes the same termsslide the shift and compare rows
original —reindexed —terms match —
The two rows show the first several terms written out from each form. Whatever the shift, the rows stay identical — if they ever differ, the reindexing was done wrong. The middle option is exactly what you must shift when substituting a power series into \(y''\).
②Why ODEs force you to do this
Substitute \(y = \sum_{n=0}^{\infty}a_nx^n\) into an ODE and you get sums with mismatched powers:
The mismatch
\[ y'' = \sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}, \qquad y = \sum_{n=0}^{\infty}a_nx^{n} \]
You cannot combine those — one has \(x^{n-2}\), the other \(x^n\). Shift the first by letting \(k = n-2\) (so \(n = k+2\)):
Now both sums run from 0 and both carry \(x^k\), so you can add them slot by slot and set each coefficient to zero. That produces the recurrence relation — the entire payoff of the method.
Worked example · a recurrence relation from scratch
Find the recurrence for \(y'' - y = 0\) using \(y = \sum_{n=0}^{\infty}a_nx^n\).
STEP 1
\(y'' = \sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}\).
STEP 2
Shift: let \(n \to n+2\), which moves the start from 2 down to 0: \(y'' = \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}\).
A power series is identically zero only if every coefficient is zero, so \((n+2)(n+1)a_{n+2} = a_n\), i.e.
\[ a_{n+2} = \frac{a_n}{(n+2)(n+1)} \]
STEP 5
With \(a_0=1, a_1=0\) this generates \(1, 0, \frac12, 0, \frac{1}{24},\ldots\) — the series for \(\cosh x\). With \(a_0=0,a_1=1\) you get \(\sinh x\). Both are solutions of \(y''=y\), exactly as expected.
Watch the starting index when you differentiate
Differentiating \(\sum_{n=0}^\infty a_nx^n\) gives \(\sum_{n=1}^\infty na_nx^{n-1}\) — the \(n=0\) term is a constant and differentiates away, so the sum now starts at 1. Differentiating again starts it at 2. Forgetting to advance the lower limit leaves phantom terms with a factor of zero; harmless in that case, but the same slip in reverse loses real terms.
③Practice
1
Rewrite \(\displaystyle\sum_{n=3}^{\infty}\frac{x^{n-3}}{n}\) so that the index starts at \(n=0\).
Hint 1Let \(k = n-3\), so \(n = k+3\). Substitute everywhere \(n\) appears.
Hint 2When \(n=3\), \(k=0\) — that gives your new lower limit.
Solution
STEP 1 · define the new index
You want the exponent \(n-3\) to become the new counter, so set
\[ k = n-3 \quad\Longleftrightarrow\quad n = k+3 \]
STEP 2 · translate the lower limit
\[ n = 3 \quad\Longrightarrow\quad k = 0 \]
STEP 4 · write the result
\[ \sum_{n=3}^{\infty}\frac{x^{n-3}}{n} = \sum_{k=0}^{\infty}\frac{x^{k}}{k+3} \]
STEP 5 · verify with the first two terms
Original: \(n=3\) gives \(\frac{x^0}{3}\); \(n=4\) gives \(\frac{x^1}{4}\). New: \(k=0\) gives \(\frac{x^0}{3}\); \(k=1\) gives \(\frac{x^1}{4}\) ✓ Identical, as reindexing must be — you renamed the counter, you did not change a single term.
2
Combine \(\displaystyle\sum_{n=1}^{\infty}na_nx^{n-1} + \sum_{n=0}^{\infty}a_nx^{n}\) into a single series.
Hint 1The powers must match. Shift the first sum so its exponent reads \(x^n\).
Hint 2In the first, let \(n \to n+1\): the start moves from 1 to 0 and the term becomes \((n+1)a_{n+1}x^n\).
Solution
STEP 1 · diagnose the mismatch
The first sum carries \(x^{n-1}\), the second \(x^n\). You cannot add coefficient by coefficient until both read the same power.
STEP 2 · shift the first sum
Replace \(n\) by \(n+1\) throughout. The starting index drops by 1 (from 1 to 0) and every index inside rises by 1:
Original \(n=1\): \(1\cdot a_1x^0\). Shifted \(n=0\): \((0+1)a_1x^0\) ✓
STEP 4 · combine under one sigma
\[ \sum_{n=0}^{\infty}\Big[(n+1)a_{n+1} + a_n\Big]x^n \]
STEP 5 · what it gives you
Setting the bracket to zero (as an ODE would demand) yields \(a_{n+1} = -\frac{a_n}{n+1}\), whose solution is \(a_n = \frac{(-1)^na_0}{n!}\) — that is \(y = a_0e^{-x}\), the solution of \(y' + y = 0\). You just solved an ODE by index shifting.
3
Rewrite \(\displaystyle\sum_{n=2}^{\infty}n(n-1)a_nx^{n}\) with the summation starting at \(n=0\) instead. Why is that legal here?
Hint 1This is not a shift — check what the \(n=0\) and \(n=1\) terms would actually contribute.
Hint 2Evaluate the coefficient \(n(n-1)\) at \(n=0\) and \(n=1\).
Solution
STEP 1 · test the terms you would be adding
\[ n=0:\ 0(0-1)a_0x^0 = 0 \qquad n=1:\ 1(1-1)a_1x^1 = 0 \]
STEP 2 · conclude they are free
Both proposed new terms are exactly zero, so including them changes nothing:
No indices were changed inside the terms. This is extending the range, not shifting — the two operations have different bookkeeping and must not be confused.
STEP 4 · when the trick applies
Only when the added terms vanish identically. Here the factor \(n(n-1)\) does the work. The same trick lets you start \(\sum_{n=1}^\infty na_nx^{n-1}\) at \(n=0\), since the \(n=0\) term carries a factor of 0.
WHY IT'S USEFUL
It aligns two sums' starting indices without touching the terms — often simpler than shifting, when the powers already match.
4
Write out the first four terms of \(\displaystyle\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}\), and confirm they match the first four nonzero terms of \(\displaystyle\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}\).
Hint 1Just substitute \(n = 0,1,2,3\) into the first expression.
Hint 2Then substitute \(n = 2,3,4,5\) into the second.
Term for term identical ✓ — the shift renamed the counter without altering the series.
STEP 4 · what these terms are
This is \(y''\) when \(y = \sum a_nx^n\). Reading off the shifted form: the constant term of \(y''\) is \(2a_2\), so \(a_2 = \frac{y''(0)}{2}\) — the Taylor coefficient formula appearing naturally.
HABIT
Whenever a shift feels uncertain, write out three terms of each form. It takes thirty seconds and removes all doubt — and the index-shift lab in this module does exactly this comparison automatically.
5
Derive the recurrence relation for \(y'' + 2y = 0\) using \(y = \sum_{n=0}^{\infty}a_nx^n\), and compute \(a_2, a_3, a_4\) in terms of \(a_0, a_1\).
Hint 1Differentiate twice, then shift \(y''\) so it carries \(x^n\).
Hint 2You should reach \(\sum\left[(n+2)(n+1)a_{n+2} + 2a_n\right]x^n = 0\).
Hint 3A power series is identically zero only if every coefficient is zero.
Two free constants \(a_0, a_1\) — exactly what a second-order equation should have. The even-index chain builds \(\cos(\sqrt2 x)\) and the odd chain builds \(\sin(\sqrt2 x)\), matching the elementary solution \(y = A\cos\sqrt2x + B\sin\sqrt2x\) ✓
6
Find the recurrence for \(y' = 2xy\) with \(y = \sum a_nx^n\), and identify the solution when \(a_0 = 1\).
Hint 1\(y' = \sum_{n=1}^{\infty}na_nx^{n-1}\), and \(2xy = \sum_{n=0}^{\infty}2a_nx^{n+1}\) — note multiplying by \(x\) raises the power.
Hint 2Shift both so they carry \(x^n\). The right side becomes \(\sum_{n=1}2a_{n-1}x^n\).
Hint 3Treat the \(n=0\) coefficient separately, since the right side has no constant term.
STEP 4 · result
\[ \sum_{n=1}^{\infty}\frac{a_n}{n}x^{n+1} = \sum_{k=2}^{\infty}\frac{a_{k-1}}{k-1}x^{k} \]
STEP 5 · verify the first term
Original \(n=1\): \(a_1x^2\). New \(k=2\): \(\frac{a_1}{1}x^2\) ✓
THE RULE IN ONE LINE
The starting index and the indices inside the terms always move in opposite directions. If both move the same way, you have changed the series — the fastest way to catch the error is to compare first terms, as in step 5.
8
Derive the recurrence for \(y'' + y = 0\) with \(y = \sum a_nx^n\), then show that choosing \(a_0=1, a_1=0\) reproduces the cosine series.
Hint 1Shift \(y''\) to \(x^n\), add \(y\), and set each coefficient to zero.
Hint 2You should get \(a_{n+2} = \frac{-a_n}{(n+2)(n+1)}\).
Hint 3With \(a_1 = 0\), all odd coefficients vanish; chase the even ones and watch factorials appear.
Choosing \(a_0=0, a_1=1\) kills the even chain and builds \(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots = \sin x\). The two initial choices generate exactly the two independent solutions — which is the whole content of "second-order equations have a two-dimensional solution space."
9
Derive the recurrence relation for Airy's equation \(y'' - xy = 0\) using \(y=\sum_{n=0}^\infty a_nx^n\), and compute \(a_3\) and \(a_4\) in terms of \(a_0,a_1\).
Hint 1Shift \(y''\) so it carries \(x^n\); the term \(xy\) raises the power instead.
Hint 2\(xy = \sum_{n=0}a_nx^{n+1} = \sum_{n=1}a_{n-1}x^n\).
Hint 3Handle the \(x^0\) coefficient separately — the second sum has no constant term.
Solution
STEP 1 · expand both terms with matching powers
\[ y'' = \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n, \qquad xy = \sum_{n=1}^{\infty}a_{n-1}x^n \]
STEP 2 · match the constant term
The second sum starts at \(n=1\), so the \(x^0\) coefficient gives
\[ 2a_2 = 0 \quad\Longrightarrow\quad a_2 = 0 \]
STEP 3 · match xⁿ for n ≥ 1
\[ (n+2)(n+1)a_{n+2} - a_{n-1} = 0 \quad\Longrightarrow\quad a_{n+2} = \frac{a_{n-1}}{(n+2)(n+1)} \]
Note this links coefficients three apart, not two.
STEP 4 · generate the first few
\[ n=1:\ a_3 = \frac{a_0}{3\cdot2} = \frac{a_0}{6} \qquad n=2:\ a_4 = \frac{a_1}{4\cdot3} = \frac{a_1}{12} \]
STEP 5 · read the structure
Because the recurrence steps by 3, the coefficients split into three chains: \(a_0\to a_3\to a_6\), \(a_1\to a_4\to a_7\), and \(a_2\to a_5\to\cdots\) which is entirely zero. Two free constants \(a_0,a_1\) as a second-order equation demands ✓
WHY AIRY MATTERS
Its solutions cannot be written with elementary functions at all — the series is the definition of \(\mathrm{Ai}(x)\) and \(\mathrm{Bi}(x)\). They describe light near a caustic (the bright edge of a rainbow) and a quantum particle at a turning point. This is the power-series method earning its keep.
Part III · Module 5 Series
Power series expansion
Any well-behaved function can be replaced, near a point, by an infinite polynomial. That is a startling claim, and it is the reason series solutions to differential equations work at all.
A Maclaurin series is just the case \(a=0\). Each coefficient is built from one derivative at the center: the polynomial matches the function's value, slope, concavity, and so on, to ever higher order.
InteractivePolynomials chasing a curveraise the degree and watch the fit spread
Tₙ(x) —radius R —
The approximation is always excellent at the center and degrades as you move away. For \(e^x\), \(\sin\), and \(\cos\) the fit keeps spreading forever (\(R=\infty\)); for \(\frac{1}{1-x}\) and \(\ln(1+x)\) it slams into a wall at \(|x|=1\) — no matter how many terms you add, the polynomial never crosses that boundary. The green band marks the interval of convergence.
②The expansions worth memorizing
Function
Maclaurin series
Valid for
\(e^x\)
\(\sum\limits_{n=0}^{\infty}\dfrac{x^n}{n!} = 1 + x + \dfrac{x^2}{2!} + \cdots\)
all \(x\)
\(\sin x\)
\(\sum\limits_{n=0}^{\infty}\dfrac{(-1)^nx^{2n+1}}{(2n+1)!} = x - \dfrac{x^3}{3!} + \cdots\)
\(\sum\limits_{n=0}^{\infty}x^n = 1 + x + x^2 + \cdots\)
\(|x|<1\)
\(\ln(1+x)\)
\(\sum\limits_{n=1}^{\infty}\dfrac{(-1)^{n+1}x^n}{n} = x - \dfrac{x^2}{2} + \cdots\)
\(-1
\(\arctan x\)
\(\sum\limits_{n=0}^{\infty}\dfrac{(-1)^nx^{2n+1}}{2n+1} = x - \dfrac{x^3}{3} + \cdots\)
\(|x|\le1\)
Build new series from old ones — don't differentiate n times
Nobody computes \(\frac{d^{12}}{dx^{12}}\) of anything. To get the series for \(e^{-x^2}\), substitute \(-x^2\) into the \(e^x\) series. For \(x\sin x\), multiply the \(\sin\) series by \(x\). For \(\frac{1}{1+x^2}\), substitute \(-x^2\) into the geometric series. Substitution and multiplication do 90% of the work in practice.
Worked example · a series in three lines
Find the Maclaurin series for \(f(x) = e^{-x^2}\) and use it to estimate \(\int_0^1 e^{-x^2}dx\).
STEP 1
Start from \(e^u = \sum\frac{u^n}{n!}\) and substitute \(u = -x^2\):
Four terms give \(0.7429\); the true value is \(0.746824\ldots\) Since the series alternates, the error is smaller than the next term \(\frac{1}{216} \approx 0.0046\) ✓
STEP 4
Note that \(e^{-x^2}\) has no elementary antiderivative. The series is not a shortcut here — it is the only way to get a number.
③Practice
1
Find the Maclaurin series for \(f(x) = \dfrac{1}{1+x^2}\) and state its interval of convergence.
Hint 1Don't differentiate. Write it as \(\frac{1}{1-(-x^2)}\) and use the geometric series.
Hint 2The geometric series needs \(|{-x^2}| < 1\), i.e. \(|x| < 1\).
Solution
STEP 1 · force it into geometric form
The template is \(\frac{1}{1-r} = \sum r^n\). Here
\[ \frac{1}{1+x^2} = \frac{1}{1-(-x^2)} \quad\Longrightarrow\quad r = -x^2 \]
At \(x = \pm1\) the terms are \(\pm1\) alternating, which do not approach 0 ⟹ diverges. Interval: \((-1,1)\).
WHY THE RADIUS IS 1
\(\frac{1}{1+x^2}\) is perfectly smooth for all real \(x\), yet the series still fails past \(|x|=1\). The reason is invisible on the real line: the function blows up at the complex points \(x = \pm i\), each a distance 1 from the center. Radius of convergence always reaches to the nearest singularity, real or not.
2
Find the first four nonzero terms of the Taylor series for \(\ln x\) centered at \(a=1\).
Hint 1Substitute \(u = x-1\) into the known series for \(\ln(1+u)\).
Hint 2\(\ln(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \cdots\).
Solution
STEP 1 · rewrite around the center
"Centered at 1" means the series must be in powers of \((x-1)\). Set \(u = x-1\), so
\[ \ln x = \ln(1+u) \]
STEP 2 · quote the standard series
\[ \ln(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \frac{u^4}{4} + \cdots \]
STEP 3 · substitute back
\[ \ln x = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \cdots \]
STEP 4 · verify against the definition
Taylor's formula needs \(f^{(n)}(1)\): \(f(1) = 0\), \(f' = \frac1x \Rightarrow f'(1) = 1\), \(f'' = -\frac{1}{x^2}\Rightarrow f''(1) = -1\), so the \(x^2\) coefficient is \(\frac{-1}{2!} = -\frac12\) ✓ Same answer, far more work.
STEP 5 · interval of validity
\(\ln(1+u)\) is valid for \(-1 < u \le 1\), which translates to \(0 < x \le 2\). The failure at \(x=0\) is exactly where \(\ln\) itself blows up.
3
Use a series to evaluate \(\displaystyle\lim_{x\to0}\frac{\sin x - x}{x^3}\).
Hint 1Replace \(\sin x\) by its series and simplify the numerator before taking the limit.
Hint 2\(\sin x - x = -\frac{x^3}{6} + \frac{x^5}{120} - \cdots\) — the \(x\) terms cancel exactly.
Solution
STEP 1 · expand the numerator
\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \quad\Longrightarrow\quad \sin x - x = -\frac{x^3}{6} + \frac{x^5}{120} - \cdots \]
The leading \(x\) cancels exactly — which is why the limit is finite despite the \(0/0\) form.
STEP 2 · divide by x³
\[ \frac{\sin x - x}{x^3} = -\frac16 + \frac{x^2}{120} - \cdots \]
STEP 3 · take the limit
Every remaining term carries a positive power of \(x\), so they all vanish:
\[ \lim_{x\to0}\frac{\sin x - x}{x^3} = -\frac16 \]
STEP 4 · compare with L'Hôpital
The same limit needs three applications of L'Hôpital, each requiring you to re-check the \(0/0\) form. The series does it in one line and simultaneously tells you the next correction term.
GENERAL TECHNIQUE
For any \(\frac00\) limit at 0, expand everything to a couple of orders past the cancellation and read off the leading behavior. It also settles limits L'Hôpital cannot easily reach.
4
Find the first three nonzero terms of the Maclaurin series for \(x\sin x\), and use it to find \(f^{(4)}(0)\).
Hint 1Multiply the sine series by \(x\) — no differentiation needed.
Hint 2The Taylor coefficient of \(x^n\) is \(\frac{f^{(n)}(0)}{n!}\), so \(f^{(4)}(0) = 4!\times(\text{coefficient of }x^4)\).
Solution
STEP 1 · start from a known series
\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \]
STEP 2 · multiply by x
Multiplication by a power simply shifts every exponent up:
Computing \(f^{(4)}\) of \(x\sin x\) directly means four rounds of the product rule. The series does it with one multiplication — and the identity \(f^{(n)}(0) = n!\,c_n\) is the standard trick whenever a problem asks for a high-order derivative at a point.
5
Estimate \(e^{0.1}\) using the first four terms of the Maclaurin series, and bound the error.
Hint 1\(e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots\) with \(x=0.1\).
Hint 2The remaining tail \(\frac{x^4}{24} + \frac{x^5}{120}+\cdots\) is dominated by its first term.
Hint 3Since all terms are positive here, the estimate is an underestimate.
Solution
STEP 1 · write the partial sum
\[ e^{0.1} \approx 1 + 0.1 + \frac{0.01}{2} + \frac{0.001}{6} \]
and every later term is at least ten times smaller, so the whole tail is under about \(4.6\times10^{-6}\).
STEP 4 · state the estimate with its error
\[ e^{0.1} = 1.10517 \pm 0.000005 \]
STEP 5 · check
True value: \(1.1051709\ldots\) — the error is \(4.2\times10^{-6}\), just as predicted, and the estimate is low because every omitted term is positive ✓
WHY SMALL x MATTERS
The error scales like \(x^4\): at \(x = 0.1\) it is \(10^{-6}\), but at \(x = 1\) the same four terms are off by \(0.05\). Taylor approximations are local, and staying near the center is what buys accuracy.
6
Use a series to evaluate \(\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}\).
Hint 1\(\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots\).
Hint 2Subtracting from 1 cancels the constant term.
STEP 4 · take the limit
\[ \lim_{x\to0}\frac{1-\cos x}{x^2} = \frac12 \]
BONUS INFORMATION
The series gives more than the limit: it shows \(1-\cos x \approx \frac{x^2}{2}\) for small \(x\), the approximation behind the small-oscillation pendulum and the energy \(\frac12k x^2\) of a spring. A limit tells you one number; a series tells you the whole local behavior.
7
Estimate \(\displaystyle\int_0^{0.5}\sin\left(x^2\right)dx\) using three terms of a series, and bound the error.
Hint 1Substitute \(x^2\) into the sine series — do not attempt to integrate \(\sin(x^2)\) directly (no elementary antiderivative exists).
Hint 2\(\sin(x^2) = x^2 - \frac{x^6}{6} + \frac{x^{10}}{120} - \cdots\).
Hint 3Integrate term by term, then use the alternating-series error bound.
Solution
STEP 1 · build the series by substitution
From \(\sin u = u - \frac{u^3}{6} + \frac{u^5}{120} - \cdots\) with \(u = x^2\):
\(\sin(x^2)\) has no elementary antiderivative — the series is not a shortcut here, it is the only route to a number. This is exactly how Fresnel integrals are tabulated in optics.
8
Find the Maclaurin series for \(f(x) = \dfrac{x}{1-2x}\) and its radius of convergence.
Hint 1Handle \(\frac{1}{1-2x}\) first with the geometric template, taking \(r = 2x\).
Hint 2Then multiply the whole series by \(x\), which shifts every exponent up by one.
STEP 2 · evaluate term by term
\[ = 1 + 0.05 - 0.00125 + 0.0000625 \]
STEP 3 · sum the first three terms
\[ \sqrt{1.1} \approx 1.04875 \]
STEP 4 · bound the error
The next term is \(+0.0000625\), so the three-term estimate is low by at most about \(6.3\times10^{-5}\).
STEP 5 · check
True value \(1.0488088\ldots\), so the actual error is \(5.9\times10^{-5}\) — inside the bound ✓ Including the fourth term gives \(1.0488125\), accurate to six figures.
THE WIDER POINT
The binomial series \((1+x)^\alpha = \sum\binom{\alpha}{n}x^n\) works for any real \(\alpha\), converging for \(|x|<1\). With \(\alpha=-1\) it is the geometric series; with \(\alpha=\frac12\) it is how calculators once took square roots; and the first two terms give the engineer's \(\sqrt{1+x}\approx1+\frac x2\), used constantly in relativity and small-signal analysis.
Part III · Module 6 Series
Differentiating & integrating a series
Inside its interval of convergence, a power series behaves exactly like a polynomial: you may differentiate and integrate it one term at a time. This is the license that makes the entire power-series method for ODEs legitimate.
Both new series have the same radius of convergence \(R\) as the original. Only the endpoints can change status: differentiating tends to lose endpoints, integrating tends to gain them.
InteractiveDifferentiate or integrate, one term at a timecompare against the true function
series —equals —radius —
Amber is the true function, cyan is the truncated series after the chosen operation. Differentiating \(\sum x^n\) gives \(\sum nx^{n-1} = \frac{1}{(1-x)^2}\); integrating it gives \(-\ln(1-x)\). Both are new series obtained with zero calculus work — you simply operated on the terms.
Worked example · manufacturing three series from one
Starting from \(\dfrac{1}{1-x} = \sum_{n=0}^{\infty}x^n\) for \(|x|<1\):
Substitute \(-x\), then integrate: \(\dfrac{1}{1+x} = \sum(-1)^nx^n\), so \(\ln(1+x) = \sum_{n=0}^{\infty}\dfrac{(-1)^nx^{n+1}}{n+1} = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots\)
C
Substitute \(-x^2\), then integrate: \(\dfrac{1}{1+x^2} = \sum(-1)^nx^{2n}\), so \(\arctan x = \sum_{n=0}^{\infty}\dfrac{(-1)^nx^{2n+1}}{2n+1}\).
D
Setting \(x=1\) in C gives Leibniz's formula \(\frac{\pi}{4} = 1 - \frac13 + \frac15 - \cdots\) — an endpoint that the integrated series gained, and a nice illustration of why endpoints deserve their own check.
Where this shows up in ODEs — the whole method in one place
To solve \(y'' + xy = 0\) (Airy's equation, which has no elementary solutions at all), you assume \(y = \sum a_nx^n\), differentiate termwise twice (this module), shift indices to align the powers (S4), combine the sums (S3), set every coefficient to zero to get a recurrence, and finally check the radius of convergence of the answer (S2). Part III is not five separate topics — it is five steps of one procedure.
Inside the interval only
Term-by-term operations are valid on the open interval \(|x-a|
②Practice
1
Use termwise differentiation of the \(\sin x\) series to derive the \(\cos x\) series.
Hint 1\(\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\). Differentiate each term with the power rule.
Hint 2\(\frac{d}{dx}\frac{x^3}{3!} = \frac{3x^2}{6} = \frac{x^2}{2!}\) — the factorials collapse by exactly one step.
Solution
STEP 1 · write the series
\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \]
STEP 2 · differentiate term by term
Legal because the radius of convergence is infinite. Each term uses only the power rule:
Differentiating four times returns the sine series exactly — the same four-step cycle you saw geometrically in module D4, now visible in the coefficients.
2
Find a power series for \(\displaystyle\int_0^x\frac{\sin t}{t}\,dt\) (the sine integral, which has no elementary form).
Hint 1First divide the sine series by \(t\) term by term — every power drops by one.
Hint 2\(\frac{\sin t}{t} = 1 - \frac{t^2}{3!} + \frac{t^4}{5!} - \cdots\); note this is perfectly well behaved at \(t=0\), where it equals 1.
Hint 3Now integrate term by term from 0 to \(x\).
Solution
STEP 1 · divide the series by t
\[ \frac{\sin t}{t} = \frac{1}{t}\left(t - \frac{t^3}{3!} + \frac{t^5}{5!} - \cdots\right) = 1 - \frac{t^2}{3!} + \frac{t^4}{5!} - \cdots \]
The apparent trouble at \(t=0\) disappears — the quotient extends continuously with value 1 there.
STEP 2 · integrate term by term
\[ \int_0^x 1\,dt = x, \qquad \int_0^x\frac{t^2}{6}dt = \frac{x^3}{18}, \qquad \int_0^x\frac{t^4}{120}dt = \frac{x^5}{600} \]
The original series converges for all \(x\), and integration preserves the radius, so this holds for every \(x\).
WHY IT MATTERS
This is how "special functions" get computed in the first place. Tables and library routines for \(\operatorname{Si}\), \(\operatorname{erf}\), and Bessel functions are ultimately evaluating series exactly like this one.
3
Solve \(y' = y\), \(y(0)=1\) by assuming \(y = \sum_{n=0}^\infty a_nx^n\) and matching coefficients.
Hint 1\(y' = \sum_{n=1}^{\infty}na_nx^{n-1}\). Shift so it reads \(x^n\).
Hint 2After shifting, \(y' = \sum_{n=0}^{\infty}(n+1)a_{n+1}x^n\). Set that equal to \(\sum a_nx^n\) coefficient by coefficient.
Hint 3\((n+1)a_{n+1} = a_n\), and \(y(0)=1\) means \(a_0 = 1\). Generate \(a_1, a_2, a_3\) and look for the pattern.
STEP 5 · spot the pattern and sum
\[ a_n = \frac{1}{n!} \quad\Longrightarrow\quad y = \sum_{n=0}^{\infty}\frac{x^n}{n!} = e^x \]
WHAT JUST HAPPENED
You solved a differential equation without integrating once — and rediscovered \(e^x\) from nothing but coefficient matching. That is the power-series method in miniature, and it works on equations whose solutions have no elementary form at all.
4
Starting from \(\frac{1}{1-x} = \sum x^n\), find a closed form for \(\displaystyle\sum_{n=1}^{\infty}nx^n\), then evaluate \(\displaystyle\sum_{n=1}^{\infty}\frac{n}{2^n}\).
Hint 1Differentiating \(\sum x^n\) produces \(\sum nx^{n-1}\) — close, but the exponent is off by one.
Hint 2Multiply the differentiated series by \(x\) to fix the exponent.
You just summed a non-geometric series exactly, using calculus on a series rather than any convergence test. This "differentiate the geometric series" trick evaluates expected values in probability (the mean of a geometric distribution is exactly this computation).
5
Find the Maclaurin series for \(\dfrac{1}{(1-x)^3}\) by differentiating the geometric series twice.
Hint 1You already know \(\frac{1}{(1-x)^2} = \sum_{n=1}^\infty nx^{n-1}\) from one differentiation.
Hint 2Differentiate again: \(\frac{2}{(1-x)^3} = \sum n(n-1)x^{n-2}\).
Hint 3Divide by 2 and reindex with \(k = n-2\).
Solution
STEP 1 · first differentiation
\[ \frac{1}{(1-x)^2} = \sum_{n=1}^{\infty}nx^{n-1} \]
STEP 2 · second differentiation
Left side: \(\frac{d}{dx}(1-x)^{-2} = 2(1-x)^{-3}\). Right side, termwise:
STEP 4 · reindex with k = n − 2
\[ = \sum_{k=0}^{\infty}\frac{(k+2)(k+1)}{2}x^{k} = 1 + 3x + 6x^2 + 10x^3 + \cdots \]
STEP 5 · recognize the coefficients
\(1, 3, 6, 10, 15,\dots\) are the triangular numbers \(\binom{k+2}{2}\) — the general binomial series \((1-x)^{-m}\) in action. Radius still 1, unchanged by differentiation.
6
Integrate the series for \(\frac{1}{1+x^2}\) to obtain the arctangent series, then estimate \(\arctan(0.5)\) with three terms and bound the error.
Hint 1\(\frac{1}{1+x^2} = 1 - x^2 + x^4 - \cdots\), and \(\arctan x = \int_0^x\frac{dt}{1+t^2}\).
Hint 2Integrate term by term from 0 to \(x\).
Hint 3The result alternates, so the error is bounded by the first omitted term.
Solution
STEP 1 · start from the geometric series
\[ \frac{1}{1+t^2} = 1 - t^2 + t^4 - t^6 + \cdots, \qquad |t|<1 \]
True: \(\arctan 0.5 = 0.4636476\). Actual error \(0.00094\), inside the bound ✓
THE FAMOUS ENDPOINT
At \(x=1\) the integrated series gains its endpoint and gives Leibniz's formula \(\frac{\pi}{4} = 1 - \frac13 + \frac15 - \cdots\) — beautiful, but so slow that a thousand terms buys three decimal places. Integration can add endpoints that the original series lacked.
7
Solve \(y' = -2xy\), \(y(0)=1\) by power series, and identify the solution in closed form.
Hint 1\(y' = \sum_{n=1}na_nx^{n-1}\) and \(-2xy = \sum_{n=0}(-2a_n)x^{n+1}\) — shift both to \(x^n\).
Hint 2Matching \(x^0\) gives \(a_1 = 0\), so all odd coefficients vanish.
Hint 3The recurrence is \(a_{n+1} = \frac{-2a_{n-1}}{n+1}\); chase the even indices.
The separable method gives the same answer in two lines — but the series method also works on equations, like Airy's \(y'' + xy = 0\), where separation is impossible.
8
Differentiate the \(\sin x\) series twice, term by term, and use the result to verify that \(y = \sin x\) satisfies \(y'' + y = 0\).
Hint 1One differentiation gives the cosine series (problem 1); do it once more.
Hint 2Compare the resulting series with the original, term by term.
Solution
STEP 1 · the series and its first derivative
\[ y = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots, \qquad y' = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \]
Term-by-term differentiation is valid inside the interval of convergence, and here \(R=\infty\), so the manipulation holds for every \(x\). Had the radius been finite, the verification would only be valid inside it.
THE WHOLE OF PART III IN ONE PROBLEM
You used termwise differentiation (S6), aligned indices to compare coefficients (S4), relied on the identity theorem to conclude from matching coefficients (S3), and needed the radius of convergence to justify the whole thing (S2). That is exactly the pipeline a power-series ODE solution runs through.
9
Find a power series for \(\ln(1+x^2)\) by integrating a known series, and use it to estimate \(\int_0^{0.5}\ln(1+x^2)\,dx\).
Hint 1Start from \(\frac{d}{dx}\ln(1+x^2) = \frac{2x}{1+x^2}\), and expand \(\frac{1}{1+x^2}\) geometrically.
Hint 2Alternatively substitute \(x^2\) into the known \(\ln(1+u)\) series — much faster.
Hint 3Then integrate term by term from 0 to 0.5.
Solution
STEP 1 · build the series by substitution
From \(\ln(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \cdots\) with \(u = x^2\):
The next term is \(\frac{0.5^9}{36} = 5.4\times10^{-5}\), so the estimate is good to about that. (True value \(0.038858\), error \(5.6\times10^{-5}\) ✓)
COMPARE THE ALTERNATIVE
Direct integration needs parts: \(\int\ln(1+x^2)dx = x\ln(1+x^2) - 2x + 2\arctan x\). Doable — but the series took three lines and generalizes to integrands with no closed form at all.
Part III · Checkpoint Quiz
Quiz C — Series
Ten questions on convergence, intervals, index shifting, and power series. These are the exact skills a power-series ODE problem will demand.
A \(p\)-series converges only for \(p>1\), and \(0.9\) misses by a hair. The ratio test returns \(L=1\) here, which is inconclusive — a common trap.
Question 4
Which test is best for \(\displaystyle\sum\frac{n^3}{n!}\)?
Why
Factorials cancel beautifully in a ratio. Here \(\frac{a_{n+1}}{a_n} = \frac{(n+1)^3}{n^3(n+1)} \to 0 < 1\), so it converges. Integrating \(n!\) is not even meaningful.
Question 5
The series \(\displaystyle\sum_{n=1}^{\infty}\frac{(x-3)^n}{n\,2^n}\) has radius of convergence:
Why
The ratio test gives \(\frac{|x-3|}{2} < 1\), so \(|x-3|<2\): center 3, radius 2, open interval \((1,5)\). Endpoints still need checking separately.
Question 6
Reindexing \(\displaystyle\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}\) to start at \(n=0\) gives:
Why
Lower the start by 2, so raise every index inside by 2: \(n \to n+2\) everywhere. This exact shift appears in every second-order power-series solution.
Question 7
The Maclaurin series for \(e^{-x}\) is:
Why
Substitute \(-x\) into the \(e^x\) series: \((-x)^n = (-1)^nx^n\). The signs alternate; the whole series is not negated.
Question 8
Differentiating \(\sum_{n=0}^{\infty}c_nx^n\) term by term produces a series with:
Why
\(R\) is preserved by both differentiation and integration. Only the endpoints are in play — differentiating tends to lose them, integrating tends to gain them.
Question 9
\(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\) is:
Why
It converges by the alternating series test (to \(\ln 2\)), but the absolute-value series is the harmonic series, which diverges. That combination is exactly the definition of conditional convergence.
Question 10
In solving \(y'' + y = 0\) by power series, after substituting and shifting you set every coefficient of \(x^n\) to zero. Why is that valid?
Why
This is the identity theorem for power series — the same fact that makes Taylor coefficients unique. It is what converts one functional equation into infinitely many ordinary equations, i.e. the recurrence relation.