20 exam-grade word problems covering the core units of Calculus II. Each question includes a memory key — a tiny English cue to lock in the concept. Choose the best answer; explanations appear only when you miss one.
20 Questions
Multiple Choice
Hard / Tricky
Word Problems
01Integration Techniques
01
Integration by PartsHard
A physicist models the energy output of a damped oscillator.
Evaluate the definite integral \(\displaystyle\int_0^{\pi} x\sin(2x)\,dx\).
Many students apply integration by parts once and stop — but what is the correct final value?
KEYLIATE → choose u = x (Algebraic), dv = sin(2x)dx → apply IBP twice if needed, or use the tabular method
✗ Not quite — here's why
Let \(u = x,\; dv = \sin(2x)\,dx\). Then \(du = dx,\; v = -\tfrac{1}{2}\cos(2x)\).
IBP gives \(\left[-\tfrac{x}{2}\cos(2x)\right]_0^{\pi} + \tfrac{1}{2}\int_0^{\pi}\cos(2x)\,dx\).
First term: \(-\tfrac{\pi}{2}(1) - 0 = -\tfrac{\pi}{2}\). Second term: \(\tfrac{1}{2}\left[\tfrac{\sin(2x)}{2}\right]_0^{\pi} = 0\). Answer: \(-\dfrac{\pi}{2}\). The trap: \(\cos(2\pi)=1\), not \(-1\).
02
Trigonometric SubstitutionHard
An engineer calculates the arc length of a parabolic cable on a suspension bridge.
The integral \(\displaystyle\int \frac{x^2}{\sqrt{4-x^2}}\,dx\) arises. After applying the substitution \(x = 2\sin\theta\), which expression correctly represents the transformed integral?
KEY√(a²−x²) → x = a·sinθ · · · √(a²+x²) → x = a·tanθ · · · √(x²−a²) → x = a·secθ
✗ Let's trace through it
With \(x=2\sin\theta\): \(x^2=4\sin^2\theta\), \(\sqrt{4-x^2}=2\cos\theta\), \(dx=2\cos\theta\,d\theta\).
Substituting: \(\dfrac{4\sin^2\theta}{2\cos\theta}\cdot 2\cos\theta\,d\theta = 4\sin^2\theta\,d\theta\). Answer: B. The \(\cos\theta\) terms cancel perfectly — a key algebraic step students often miss.
03
Partial FractionsTricky
A chemistry student models a reaction where the rate depends on concentration in a specific nonlinear way.
Decompose \(\dfrac{3x+5}{(x+1)^2(x-2)}\) into partial fractions and identify the coefficient \(B\) in:
\[\frac{3x+5}{(x+1)^2(x-2)} = \frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x-2}\]
KEYRepeated factor (x+a)^n → needs n separate fractions A/(x+a), B/(x+a)², ..., up to n
✗ Here's the systematic approach
Multiply both sides by \((x+1)^2(x-2)\). Setting \(x=-1\): \(3(-1)+5 = B(-1-2)\Rightarrow 2=-3B\Rightarrow B=-\tfrac{2}{3}\).
Wait — let's recheck: \(2 = B(-3)\), so \(B = -\tfrac{2}{3}\). Setting \(x=2\): \(11=9C\Rightarrow C=\tfrac{11}{9}\). Then compare coefficients of \(x^2\) to find \(A\). Answer: B, where \(B=-\dfrac{2}{3}\).
02Improper Integrals & Convergence
04
Improper IntegralsHard
A statistician uses the Gaussian distribution in machine learning models.
The integral \(\displaystyle\int_1^{\infty} \frac{\ln x}{x^2}\,dx\) appears in a normalization problem. Does it converge, and if so, what is its value?
KEYImproper at ∞ → take limit as b→∞ · · · ln grows slower than any x^ε → integral can still converge
✗ Apply IBP first, then take the limit
Let \(u=\ln x,\;dv=x^{-2}dx\). Then \(du=\tfrac{1}{x}dx,\;v=-x^{-1}\).
\(\int_1^b \frac{\ln x}{x^2}dx = \left[-\frac{\ln x}{x}\right]_1^b + \int_1^b \frac{1}{x^2}dx = -\frac{\ln b}{b}+\left[-\frac{1}{x}\right]_1^b\).
As \(b\to\infty\): \(\frac{\ln b}{b}\to 0\) (L'Hôpital), and \(\left[-\frac{1}{b}+1\right]\to 1\). Answer: converges to \(1\).
05
p-integralsTricky
An ecology model measures total population density over an infinite spatial domain.
Determine which of the following integrals converges:
KEY∫₁^∞ 1/xᵖ dx converges iff p > 1 · · · ∫₀¹ 1/xᵖ dx converges iff p < 1
✗ Two different p-tests apply at different bounds
\(\int_0^1 x^{-1/2}dx\): type at \(x=0\), \(p=\tfrac{1}{2}<1\) → converges to \(2\).
\(\int_1^{\infty} x^{-1/2}dx\): \(p=\tfrac{1}{2}\leq 1\) → diverges.
\(\int_0^1 x^{-2}dx\): \(p=2>1\) → diverges. Answer: C. The boundary matters — same function, different behavior on \([0,1]\) vs \([1,\infty)\).
03Infinite Series & Convergence Tests
06
Ratio TestHard
A computer scientist analyzes the runtime of a recursive algorithm with factorial growth.
Apply the Ratio Test to \(\displaystyle\sum_{n=1}^{\infty} \frac{n!\,3^n}{n^n}\). What is \(\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\)?
KEYn!/nⁿ → use Stirling hint: (n/n+1)ⁿ → 1/e as n→∞, so limit = 3/e
✗ The key is the limit of (n/(n+1))^n
\(\frac{a_{n+1}}{a_n} = \frac{(n+1)!\cdot 3^{n+1}}{(n+1)^{n+1}}\cdot\frac{n^n}{n!\cdot 3^n} = \frac{3\cdot n^n}{(n+1)^n} = 3\left(\frac{n}{n+1}\right)^n\).
As \(n\to\infty\): \(\left(\frac{n}{n+1}\right)^n = \left(1-\frac{1}{n+1}\right)^n \to e^{-1}\).
So the limit is \(\frac{3}{e}\approx 1.104\)... wait, that's \(>1\)! The series diverges. Correction: Answer is A. \(3/e \approx 1.10 > 1\) → diverges by Ratio Test.
07
Alternating Series TestTricky
A signal processing engineer approximates a waveform using an alternating series.
The series \(\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n \ln n}{n}\) — which statement is correct?
Step 1 — AST: Let \(b_n = \tfrac{\ln n}{n}\). Note \(b_n\to 0\) and \(b_n\) is eventually decreasing (check derivative of \(\tfrac{\ln x}{x}\): equals \(\tfrac{1-\ln x}{x^2}<0\) for \(x>e\)). So AST says the series converges. Step 2 — Absolute convergence? \(\sum \tfrac{\ln n}{n}\geq \sum \tfrac{1}{n}\) for \(n\geq 3\) — diverges by comparison. Answer: C — conditionally convergent (converges, but not absolutely).
08
Limit Comparison TestHard
A physicist sums energy levels of an idealized quantum system.
Determine whether \(\displaystyle\sum_{n=1}^{\infty}\frac{n^2+3n}{n^4-2n+1}\) converges or diverges. Which test and comparison series is most efficient?
KEYDominant-term trick: keep highest power in numerator & denominator → compare with 1/n²
✗ Use dominant terms
Let \(b_n=\tfrac{1}{n^2}\). Then \(\lim_{n\to\infty}\frac{a_n}{b_n}=\lim\frac{n^4+3n^3}{n^4-2n+1}=1\) (a finite positive number).
Since \(\sum 1/n^2\) converges (p-series, \(p=2>1\)) and the limit is finite and positive, \(\sum a_n\) converges by LCT. Answer: B.
09
Root TestHard
An algorithm's complexity involves an expression raised to the nth power.
Apply the Root Test to \(\displaystyle\sum_{n=1}^{\infty}\left(\frac{2n+1}{3n-2}\right)^n\). What is \(\displaystyle\lim_{n\to\infty}\sqrt[n]{|a_n|}\)?
KEYRoot Test best for (f(n))^n form → take nth root, limit = L, converges if L<1
✗ Root test removes the exponent cleanly
\(\sqrt[n]{|a_n|}=\dfrac{2n+1}{3n-2}\xrightarrow{n\to\infty}\dfrac{2}{3}<1\).
Since \(L=\tfrac{2}{3}<1\), the series converges absolutely by the Root Test. Answer: B.
04Power Series & Taylor / Maclaurin
10
Radius of ConvergenceHard
A data scientist uses a power series expansion to approximate a complex activation function.
Find the radius of convergence of \(\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n(x-3)^n}{n\cdot 2^n}\) (for \(n\geq 1\)).
KEYRatio Test on power series → R = lim|aₙ/aₙ₊₁| · · · interval of convergence: check endpoints separately
✗ Apply Ratio Test to find R
\(\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{(x-3)^{n+1}}{(n+1)2^{n+1}}\cdot\frac{n\cdot2^n}{(x-3)^n}\right|=\frac{n}{n+1}\cdot\frac{|x-3|}{2}\to\frac{|x-3|}{2}\).
Series converges when \(\dfrac{|x-3|}{2}<1\), i.e. \(|x-3|<2\). So \(R=2\). Answer: B.
11
Maclaurin SeriesTricky
A numerical analyst approximates \(\cos(x^2)\) for integration.
The Maclaurin series for \(\cos(x^2)\) begins with which expression?
KEYcos u = 1 − u²/2! + u⁴/4! − · · · then substitute u = x²
✗ Substitute into known Maclaurin series
Known: \(\cos u = \sum_{n=0}^{\infty}\frac{(-1)^n u^{2n}}{(2n)!} = 1-\frac{u^2}{2!}+\frac{u^4}{4!}-\cdots\)
Replace \(u\) with \(x^2\): \(\cos(x^2) = 1-\frac{x^4}{2!}+\frac{x^8}{4!}-\cdots = 1-\frac{x^4}{2}+\frac{x^8}{24}-\cdots\) Answer: B. The trap in A: that's \(\cos(x)\), not \(\cos(x^2)\).
12
Taylor Series ErrorHard
An aerospace engineer approximates \(e^{0.1}\) using a Taylor polynomial and needs to guarantee accuracy.
Using the Taylor polynomial \(T_3(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}\) for \(e^x\) at \(a=0\), what is the maximum error bound \(|R_3(0.1)|\) using the Remainder Estimation Theorem?
KEYLagrange remainder: |Rₙ(x)| ≤ M·|x−a|^(n+1)/(n+1)! where M = max|f^(n+1)| on interval
✗ Apply the Lagrange Remainder formula
For \(e^x\), all derivatives are \(e^x\). On \([0,0.1]\), the 4th derivative satisfies \(|f^{(4)}|\leq e^{0.1}\approx 1.105\).
\(|R_3(0.1)|\leq \frac{e^{0.1}(0.1)^4}{4!}=\frac{1.105\times 0.0001}{24}\approx 4.6\times 10^{-6}\). Answer: B. Common mistake: using \(n!\) instead of \((n+1)!\) in the denominator.
05Applications of Integration
13
Arc LengthHard
A civil engineer designs a cable for a suspension bridge modeled by \(y = \cosh(x)\) from \(x=0\) to \(x=1\).
The arc length formula gives \(\displaystyle L=\int_0^1\sqrt{1+\sinh^2(x)}\,dx\). Using a hyperbolic identity, evaluate \(L\) exactly.
A ceramics artist designs a vase by rotating the region bounded by \(y = x^2\) and \(y = 4\) around the \(y\)-axis.
Using the Shell Method, which integral correctly gives the volume?
KEYShell around y-axis: V = 2π∫ x·(top−bottom) dx · · · Washer around y-axis: V = π∫(R²−r²) dy
✗ Shell Method: identify shell radius and height
Curves intersect at \(x=0\) and \(x=2\) (since \(x^2=4\Rightarrow x=2\)).
Shell: radius = \(x\) (distance to \(y\)-axis), height = \(4-x^2\) (top minus bottom).
\(V = 2\pi\int_0^2 x(4-x^2)\,dx = 2\pi\int_0^2(4x-x^3)\,dx = 2\pi\left[2x^2-\frac{x^4}{4}\right]_0^2 = 2\pi(8-4)=8\pi\). Answer: B.
15
Surface Area of RevolutionHard
A manufacturer needs the exact surface area of a metallic paraboloid formed by rotating \(y = \sqrt{x}\) for \(0\leq x\leq 3\) about the \(x\)-axis.
Which integral correctly sets up the surface area?
KEYSurface area (x-axis rotation): S = 2π∫ y·√(1+(dy/dx)²) dx · · · don't forget the 2π and the arc-length factor
✗ Use the Surface Area formula correctly
\(y=\sqrt{x}\), so \(\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}}\), and \(\left(\dfrac{dy}{dx}\right)^2=\dfrac{1}{4x}\).
Surface area: \(S=2\pi\int_0^3 y\sqrt{1+\left(\tfrac{dy}{dx}\right)^2}\,dx = 2\pi\int_0^3\sqrt{x}\cdot\sqrt{1+\tfrac{1}{4x}}\,dx\). Answer: B. Common error: using \(x\) instead of \(y\) as the radius (that would be rotation about the \(y\)-axis).
06Parametric, Polar & Differential Equations
16
Parametric CurvesHard
A drone traces a path described by the parametric equations \(x = t^2 - 2t,\; y = t^3 - 3t\).
Find the slope \(\dfrac{dy}{dx}\) at \(t = 2\), and identify whether the curve has a vertical tangent at any \(t\in[0,3]\).
KEYdy/dx = (dy/dt)/(dx/dt) · · · vertical tangent when dx/dt = 0 and dy/dt ≠ 0
✗ Compute dy/dt and dx/dt carefully
\(\tfrac{dx}{dt}=2t-2\), \(\tfrac{dy}{dt}=3t^2-3\). At \(t=2\): \(\tfrac{dx}{dt}=2,\;\tfrac{dy}{dt}=9\Rightarrow \tfrac{dy}{dx}=\tfrac{9}{2}\).
Vertical tangent: \(\tfrac{dx}{dt}=0\Rightarrow 2t-2=0\Rightarrow t=1\). Check \(\tfrac{dy}{dt}\big|_{t=1}=0\) too — this is actually a cusp or horizontal tangent, not vertical!
Wait: at \(t=1\): \(dy/dt=3(1)-3=0\) also. Indeterminate — use L'Hôpital: limit of \((6t)/(2) = 3\) → tangent exists. No true vertical tangent at \(t=1\). Answer: D is more precise. In most textbooks, answer B is accepted.
17
Polar AreaTricky
A landscape architect designs a garden plot shaped like one petal of the rose curve \(r = \cos(3\theta)\).
What is the area of one petal?
KEYPolar area: A = ½∫r² dθ · · · one petal of r=cos(nθ): find θ where r=0 to bound one petal
✗ Find the petal limits first
One petal: \(\cos(3\theta)=0\Rightarrow 3\theta=\pm\pi/2\Rightarrow \theta\in[-\pi/6,\pi/6]\).
\(A=\tfrac{1}{2}\int_{-\pi/6}^{\pi/6}\cos^2(3\theta)\,d\theta=\int_0^{\pi/6}\cos^2(3\theta)\,d\theta\) (symmetry).
Using \(\cos^2 u=\tfrac{1+\cos(2u)}{2}\): \(=\int_0^{\pi/6}\tfrac{1+\cos(6\theta)}{2}d\theta=\tfrac{1}{2}\left[\theta+\tfrac{\sin6\theta}{6}\right]_0^{\pi/6}=\tfrac{1}{2}\cdot\tfrac{\pi}{6}=\dfrac{\pi}{12}\). Answer: B.
18
Separable ODEHard
A population biologist models bacterial growth with the differential equation \(\dfrac{dy}{dt} = ky(M-y)\), where \(M\) is the carrying capacity. This is the logistic growth equation.
After separating variables and applying partial fractions, which expression is the general solution for \(y(t)\)?
An electrical engineer models current in an RL circuit with the equation \(\dfrac{dI}{dt} + \dfrac{R}{L}I = \dfrac{E_0}{L}\), where \(R, L, E_0\) are constants.
Using the integrating factor method, which expression gives the solution \(I(t)\) with initial condition \(I(0)=0\)?
KEYIntegrating factor: μ = e^∫P(t)dt · · · multiply both sides by μ → d/dt[μ·y] = μ·Q(t)
✗ Apply the integrating factor step by step
\(P(t)=R/L\Rightarrow \mu=e^{Rt/L}\). Multiply: \(\tfrac{d}{dt}\left[Ie^{Rt/L}\right]=\tfrac{E_0}{L}e^{Rt/L}\).
Integrate: \(Ie^{Rt/L}=\tfrac{E_0}{R}e^{Rt/L}+C\Rightarrow I=\tfrac{E_0}{R}+Ce^{-Rt/L}\).
Apply \(I(0)=0\): \(0=\tfrac{E_0}{R}+C\Rightarrow C=-\tfrac{E_0}{R}\).
\(I(t)=\tfrac{E_0}{R}(1-e^{-Rt/L})\). Answer: B. As \(t\to\infty\), \(I\to E_0/R\) (steady state).
20
L'Hôpital + SeriesTricky
A physicist evaluates a limit that arises in the derivation of the sinc function used in signal processing.
Evaluate \(\displaystyle\lim_{x\to 0}\frac{\sin x - x + \frac{x^3}{6}}{x^5}\) using Maclaurin series (not L'Hôpital repeatedly).
KEYsin x = x − x³/6 + x⁵/120 − · · · → cancel terms in numerator, leading term / x⁵ = 1/120
✗ Expand sin x, then cancel
\(\sin x = x - \tfrac{x^3}{6} + \tfrac{x^5}{120} - \tfrac{x^7}{5040}+\cdots\)
Numerator: \(\sin x - x + \tfrac{x^3}{6} = \left(x-\tfrac{x^3}{6}+\tfrac{x^5}{120}-\cdots\right)-x+\tfrac{x^3}{6} = \tfrac{x^5}{120}+O(x^7)\).
\(\lim_{x\to0}\dfrac{x^5/120}{x^5}=\dfrac{1}{120}\). Answer: C. Applying L'Hôpital 5 times reaches the same answer but is extremely tedious — series is far more elegant!