Calculus II Mastery Quiz: 20 multiple-choice problems with full solutions covering integration, series, and convergence tests.

CALCULUS II · MASTERY EXAM

Core Concepts
& Practice Problems

All Major Topics · Exam Style · Multiple Choice
20Questions
40Minutes
10Units
0 / 20 answered
40:00
01
Integration Techniques Integration by Parts
★ Concept + Formula to Memorize
Integration by Parts
$$\int u\,dv = uv - \int v\,du$$
LIATE rule: choose u L=Log, I=Inverse trig, A=Algebraic, T=Trig, E=Exponential
Choose $u$ = the type earlier in LIATE. $dv$ = the remaining factor. Apply repeatedly for polynomial × exponential.
📌 Worked Example
Compute $\displaystyle\int x e^x\,dx$.
Let $u=x$, $dv=e^x dx$ → $du=dx$, $v=e^x$
$= xe^x - \int e^x dx = xe^x - e^x + C = e^x(x-1)+C$
Q1 Integration by Parts
Evaluate the integral:
$$\int x\cos x\,dx$$
A$x\sin x + \cos x + C$
B$x\sin x - \cos x + C$
C$-x\cos x + \sin x + C$
D$\sin x - x\cos x + C$
02
Trigonometric Integrals Powers of Sine & Cosine
★ Concept + Formula to Memorize
Trig Power Reduction
$$\sin^2 x = \frac{1-\cos 2x}{2},\quad \cos^2 x = \frac{1+\cos 2x}{2}$$
Odd power of sin: save one sin, convert rest via cos²=1-sin² Odd power of cos: save one cos, convert rest via sin²=1-cos² Both even: use half-angle formulas
📌 Worked Example
Compute $\displaystyle\int \cos^3 x\,dx$.
$=\int(1-\sin^2 x)\cos x\,dx$. Let $u=\sin x$: $=\sin x - \tfrac{\sin^3 x}{3}+C$
Q2 Trig Integrals
Evaluate:
$$\int \sin^2 x\,dx$$
A$\dfrac{x}{2} - \dfrac{\sin 2x}{4} + C$
B$\dfrac{x}{2} + \dfrac{\sin 2x}{4} + C$
C$-\dfrac{\cos 2x}{2} + C$
D$\dfrac{1-\cos x}{2} + C$
03
Trig Substitution Eliminating Square Roots
★ Concept + Formula to Memorize
Trigonometric Substitution Table
$\sqrt{a^2-x^2}$: let $x=a\sin\theta$
$\sqrt{a^2+x^2}$: let $x=a\tan\theta$
$\sqrt{x^2-a^2}$: let $x=a\sec\theta$
After substituting, use Pythagorean identities to simplify. Always convert back to $x$ using a reference triangle.
📌 Worked Example
Compute $\displaystyle\int\frac{dx}{\sqrt{4-x^2}}$.
Let $x=2\sin\theta$: $dx=2\cos\theta\,d\theta$, $\sqrt{4-x^2}=2\cos\theta$.
$=\int d\theta = \theta+C = \arcsin\!\left(\tfrac{x}{2}\right)+C$
Q3 Trig Substitution
Evaluate:
$$\int \frac{dx}{\sqrt{9+x^2}}$$
A$\arctan\!\left(\dfrac{x}{3}\right)+C$
B$\ln\!\left|x+\sqrt{9+x^2}\right|+C$
C$\dfrac{1}{3}\ln\!\left|x+\sqrt{9+x^2}\right|+C$
D$\sinh^{-1}(3x)+C$
04
Partial Fractions Rational Function Integration
★ Concept + Formula to Memorize
Partial Fraction Decomposition
$\dfrac{P(x)}{(x-a)(x-b)} = \dfrac{A}{x-a}+\dfrac{B}{x-b}$

Repeated: $\dfrac{P(x)}{(x-a)^2} = \dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}$
Degree(P) ≥ Degree(Q): do long division first Irreducible quadratic: use Ax+B in numerator
📌 Worked Example
Compute $\displaystyle\int\frac{dx}{(x-1)(x+1)}$.
$\tfrac{1}{(x-1)(x+1)}=\tfrac{1/2}{x-1}-\tfrac{1/2}{x+1}$
$=\tfrac{1}{2}\ln|x-1|-\tfrac{1}{2}\ln|x+1|+C=\tfrac{1}{2}\ln\left|\tfrac{x-1}{x+1}\right|+C$
Q4 Partial Fractions
Evaluate:
$$\int \frac{x+5}{(x-1)(x+2)}\,dx$$
A$2\ln|x-1|-\ln|x+2|+C$
B$\ln|x-1|+2\ln|x+2|+C$
C$\ln|x-1|-\ln|x+2|+C$
D$3\ln|(x-1)(x+2)|+C$
05
Improper Integrals Convergence at Infinity
★ Concept + Formula to Memorize
Improper Integral Definition
$$\int_a^\infty f(x)\,dx = \lim_{t\to\infty}\int_a^t f(x)\,dx$$
p-integral: $\int_1^\infty x^{-p}dx$ converges iff $p>1$ Also check: discontinuities in $[a,b]$
If the limit exists and is finite → Convergent. Otherwise → Divergent.
📌 Worked Example
Does $\displaystyle\int_1^\infty \frac{1}{x^2}dx$ converge?
$=\lim_{t\to\infty}\left[-\tfrac{1}{x}\right]_1^t=\lim_{t\to\infty}\!\left(1-\tfrac{1}{t}\right)=1$. Converges to $1$.
Q5 Improper Integrals
Which of the following improper integrals converges?
A$\displaystyle\int_1^\infty \frac{1}{\sqrt{x}}\,dx$
B$\displaystyle\int_1^\infty \frac{1}{x}\,dx$
C$\displaystyle\int_1^\infty \frac{1}{x^{3/2}}\,dx$
D$\displaystyle\int_1^\infty \frac{1}{\ln x}\,dx$
Q6 Improper Integrals
Evaluate:
$$\int_0^\infty e^{-2x}\,dx$$
A$\dfrac{1}{2}$
B$2$
C$1$
DDiverges
06
Sequences Limits of Sequences
★ Concept + Formula to Memorize
Convergence of Sequences
$$\{a_n\}\text{ converges} \iff \lim_{n\to\infty}a_n = L \in \mathbb{R}$$
Squeeze Theorem applies to sequences $r^n\to 0$ if $|r|<1$; diverges if $|r|>1$ $n^{1/n}\to 1$ as $n\to\infty$
📌 Worked Example
Find $\lim_{n\to\infty}\dfrac{n^2+3n}{2n^2-1}$.
Divide numerator and denominator by $n^2$: $\dfrac{1+3/n}{2-1/n^2}\to\dfrac{1}{2}$.
Q7 Sequences
Find the limit of the sequence:
$$a_n = \left(1+\frac{2}{n}\right)^n$$
A$1$
B$e$
C$e^2$
DDiverges
07
Series Convergence Tests Ratio, Root, Integral, Comparison
★ Concept + Formula to Memorize
Major Convergence Tests
Ratio Test: $L=\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|$: $L<1$ conv., $L>1$ div., $L=1$ inconclusive
Root Test: $L=\lim_{n\to\infty}\sqrt[n]{|a_n|}$: same conclusions
p-series: $\sum\dfrac{1}{n^p}$ converges iff $p>1$
Geometric: $\sum ar^n$ converges iff $|r|<1$, sum $=\dfrac{a}{1-r}$
Divergence Test first: if $a_n\not\to 0$, diverges Alternating: check $b_n\to 0$ & decreasing
📌 Worked Example
Does $\displaystyle\sum_{n=1}^\infty \frac{n!}{2^n}$ converge?
Ratio Test: $L=\lim\dfrac{(n+1)!/2^{n+1}}{n!/2^n}=\lim\dfrac{n+1}{2}=\infty>1$. Diverges.
Q8 Series · Ratio Test
Determine convergence of:
$$\sum_{n=1}^\infty \frac{n^2}{3^n}$$
ADiverges by the Ratio Test ($L=3$)
BConverges by the Ratio Test ($L=1/3$)
CRatio Test inconclusive ($L=1$)
DDiverges by Divergence Test
Q9 Series · p-series & Comparison
Which series converges?
A$\displaystyle\sum_{n=1}^\infty \dfrac{1}{\sqrt{n}}$
B$\displaystyle\sum_{n=1}^\infty \dfrac{1}{n}$
C$\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^2}$
D$\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^{1/2}}$
08
Power Series Radius & Interval of Convergence
★ Concept + Formula to Memorize
Power Series & Radius of Convergence
$$\sum_{n=0}^\infty c_n(x-a)^n,\quad R=\frac{1}{\limsup_{n\to\infty}\sqrt[n]{|c_n|}}$$
Use Ratio Test: $R=\lim|c_n/c_{n+1}|$ Always check endpoints separately Converges absolutely for $|x-a|
📌 Worked Example
Find $R$ for $\displaystyle\sum_{n=0}^\infty \frac{x^n}{n!}$.
Ratio Test: $\lim\left|\dfrac{x^{n+1}/(n+1)!}{x^n/n!}\right|=\lim\dfrac{|x|}{n+1}=0$ for all $x$. So $R=\infty$.
Q10 Power Series
Find the radius of convergence of:
$$\sum_{n=1}^\infty \frac{(x-2)^n}{n\cdot 3^n}$$
A$R=1$
B$R=2$
C$R=3$
D$R=\infty$
09
Taylor & Maclaurin Series Representing Functions as Series
★ Concept + Formula to Memorize
Key Maclaurin Series (Memorize All!)
$e^x = \displaystyle\sum_{n=0}^\infty \frac{x^n}{n!}$ (all $x$)
$\sin x = \displaystyle\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$ (all $x$)
$\cos x = \displaystyle\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}$ (all $x$)
$\dfrac{1}{1-x} = \displaystyle\sum_{n=0}^\infty x^n$, $|x|<1$
$\ln(1+x) = \displaystyle\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}$, $|x|\le 1$
Taylor at $a$: $\sum\frac{f^{(n)}(a)}{n!}(x-a)^n$ Maclaurin: Taylor at $a=0$
📌 Worked Example
Write the Maclaurin series for $e^{-x^2}$.
Replace $x\to-x^2$ in $e^x$ series: $\displaystyle\sum_{n=0}^\infty\frac{(-1)^n x^{2n}}{n!}$
Q11 Maclaurin Series
The Maclaurin series for $\sin x$ begins:
A$x - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots$
B$x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots$
C$1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots$
D$1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots$
Q12 Taylor Series · Approximation
Using the third-degree Taylor polynomial of $e^x$ centered at $0$, approximate $e^{0.1}$:
A$1.1050$
B$1.1052$
C$1.1000$
D$1.0050$
10
Applications Arc Length · Area · Volume
★ Concept + Formula to Memorize
Geometric Applications of Integration
Arc Length: $L=\displaystyle\int_a^b\!\sqrt{1+[f'(x)]^2}\,dx$
Volume (Disk): $V=\pi\displaystyle\int_a^b [f(x)]^2\,dx$
Volume (Shell): $V=2\pi\displaystyle\int_a^b x\,f(x)\,dx$
Surface Area: $S=2\pi\displaystyle\int_a^b f(x)\sqrt{1+[f'(x)]^2}\,dx$
Washer: $\pi\int(R^2-r^2)dx$ Parametric arc: $\int\!\sqrt{(dx/dt)^2+(dy/dt)^2}\,dt$
📌 Worked Example
Volume of solid from revolving $y=\sqrt{x}$, $0\le x\le 4$, about the $x$-axis.
$V=\pi\displaystyle\int_0^4 x\,dx=\pi\left[\tfrac{x^2}{2}\right]_0^4=8\pi$
Q13 Volume · Disk Method
The region bounded by $y=x^2$, $x=0$, $x=1$, and the $x$-axis is revolved about the $x$-axis. What is the volume?
A$\dfrac{\pi}{3}$
B$\dfrac{\pi}{5}$
C$\dfrac{2\pi}{5}$
D$\pi$
Q14 Arc Length
Find the arc length of $y=\dfrac{2}{3}x^{3/2}$ from $x=0$ to $x=3$.
A$\dfrac{14}{3}$
B$\dfrac{16}{3}$
C$\dfrac{7}{3}$
D$\dfrac{14}{9}$
+
Mixed Topics Alternating Series, L'Hôpital, Parametric, Polar
Q15 Alternating Series
Which series satisfies the Alternating Series Test and therefore converges?
A$\displaystyle\sum_{n=1}^\infty \frac{(-1)^n n}{n+1}$
B$\displaystyle\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}$
C$\displaystyle\sum_{n=1}^\infty (-1)^n$
D$\displaystyle\sum_{n=1}^\infty \frac{(-1)^n n^2}{2n^2+1}$
Q16 L'Hôpital's Rule
Evaluate:
$$\lim_{x\to 0}\frac{\sin 3x}{x}$$
A$0$
B$1$
C$3$
DDoes not exist
Q17 Parametric Curves
If $x=t^2$, $y=t^3$, find $\dfrac{dy}{dx}$ in terms of $t$.
A$\dfrac{3t}{2}$
B$\dfrac{2}{3t}$
C$6t$
D$\dfrac{3}{2}t^2$
Q18 Polar Coordinates
The area enclosed by $r=2\cos\theta$ is:
A$\pi$
B$2\pi$
C$4\pi$
D$\dfrac{\pi}{2}$
Q19 Absolute vs Conditional Convergence
The series $\displaystyle\sum_{n=1}^\infty \frac{(-1)^n}{n}$ is:
AAbsolutely convergent
BConditionally convergent
CDivergent
DConvergent only at $n=1$
Q20 Integration · Substitution + Definite
Evaluate:
$$\int_0^{\pi/2} \sin x\,e^{\cos x}\,dx$$
A$e-1$
B$1-\dfrac{1}{e}$
C$e$
D$e+1$
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Complete Solutions & Explanations
Q1 — Integration by Parts
✓ Correct Answer: A — $x\sin x + \cos x + C$
Let $u=x$, $dv=\cos x\,dx$ → $du=dx$, $v=\sin x$.
$$\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x + \cos x + C$$
Q2 — Trigonometric Integrals
✓ Correct Answer: A — $\frac{x}{2} - \frac{\sin 2x}{4} + C$
Use the half-angle identity: $\sin^2 x = \frac{1-\cos 2x}{2}$.
$$\int\sin^2 x\,dx = \int\frac{1-\cos 2x}{2}\,dx = \frac{x}{2}-\frac{\sin 2x}{4}+C$$
Q3 — Trigonometric Substitution
✓ Correct Answer: B — $\ln|x+\sqrt{9+x^2}|+C$
Let $x=3\tan\theta$, $dx=3\sec^2\theta\,d\theta$, $\sqrt{9+x^2}=3\sec\theta$. Then $\int\frac{3\sec^2\theta}{3\sec\theta}d\theta=\int\sec\theta\,d\theta=\ln|\sec\theta+\tan\theta|+C$. Converting back: $\sec\theta=\frac{\sqrt{9+x^2}}{3}$, $\tan\theta=\frac{x}{3}$, so $= \ln\!\left|\frac{\sqrt{9+x^2}+x}{3}\right|+C = \ln|x+\sqrt{9+x^2}|+C$ (absorbing $\ln 3$ into $C$).
Q4 — Partial Fractions
✓ Correct Answer: A — $2\ln|x-1|-\ln|x+2|+C$
Write $\dfrac{x+5}{(x-1)(x+2)}=\dfrac{A}{x-1}+\dfrac{B}{x+2}$. Multiply: $x+5=A(x+2)+B(x-1)$.
$$x=1:\; 6=3A \Rightarrow A=2\qquad x=-2:\; 3=-3B \Rightarrow B=-1$$ $$\int\frac{x+5}{(x-1)(x+2)}dx = \int\!\left(\frac{2}{x-1}-\frac{1}{x+2}\right)dx = 2\ln|x-1|-\ln|x+2|+C$$
Verification: $\frac{d}{dx}[2\ln|x-1|-\ln|x+2|]=\frac{2}{x-1}-\frac{1}{x+2}=\frac{2(x+2)-(x-1)}{(x-1)(x+2)}=\frac{x+5}{(x-1)(x+2)}$ ✓
Q5 — Improper Integrals (p-integral)
✓ Correct Answer: C — $\int_1^\infty x^{-3/2}dx$
This is a p-integral $\int_1^\infty x^{-p}dx$ with $p=3/2>1$, so it converges. For A: $p=1/2<1$, diverges. For B: $p=1$, diverges (harmonic). For D: $1/\ln x < 1/x$ near $\infty$ is false — actually $\ln x$ grows slower than $x$, so $1/\ln x > 1/x$ for large $x$, and since $\int 1/x$ diverges, $\int 1/\ln x$ also diverges by comparison. Option C converges: $\int_1^\infty x^{-3/2}dx=\left[-2x^{-1/2}\right]_1^\infty=2$.
Q6 — Improper Integrals
✓ Correct Answer: A — $\frac{1}{2}$
$$\int_0^\infty e^{-2x}dx=\lim_{t\to\infty}\left[-\frac{e^{-2x}}{2}\right]_0^t = 0-\!\left(-\frac{1}{2}\right)=\frac{1}{2}$$
Q7 — Sequences
✓ Correct Answer: C — $e^2$
Use the fundamental limit $\lim_{n\to\infty}(1+k/n)^n=e^k$. Here $k=2$.
$$\lim_{n\to\infty}\!\left(1+\frac{2}{n}\right)^n = e^2$$
Q8 — Series · Ratio Test
✓ Correct Answer: B — Converges by the Ratio Test ($L=1/3$)
$$L=\lim_{n\to\infty}\frac{(n+1)^2/3^{n+1}}{n^2/3^n}=\lim_{n\to\infty}\frac{(n+1)^2}{3n^2}=\frac{1}{3}<1\Rightarrow\text{Converges}$$
Q9 — p-series
✓ Correct Answer: C — $\sum 1/n^2$
$p=2>1$ → convergent (Basel problem, $=\pi^2/6$). Options A and D both have $p=1/2<1$, diverge. Option B is the harmonic series, $p=1$, diverges.
Q10 — Power Series · Radius of Convergence
✓ Correct Answer: C — $R=3$
Ratio Test on $a_n=\frac{(x-2)^n}{n\cdot 3^n}$:
$$L=\lim_{n\to\infty}\left|\frac{(x-2)^{n+1}}{(n+1)3^{n+1}}\cdot\frac{n\cdot 3^n}{(x-2)^n}\right|=\frac{|x-2|}{3}$$
Series converges when $|x-2|/3 < 1$, i.e., $|x-2|<3$. So $R=3$.
Q11 — Maclaurin Series
✓ Correct Answer: B — $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$
$\sin x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$. Odd powers only, alternating signs. Option C is $\cos x$. Option D is $e^x$. Option A mixes odd and even powers incorrectly.
Q12 — Taylor Polynomial Approximation
✓ Correct Answer: B — $1.1052$
Third-degree Maclaurin: $P_3(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}$. At $x=0.1$:
$$P_3(0.1)=1+0.1+\frac{0.01}{2}+\frac{0.001}{6}=1.1051\overline{6}\approx 1.1052$$
The degree-3 Taylor polynomial gives $\approx 1.1052$ when rounded to 4 decimal places. Option B is correct. (The true value $e^{0.1}\approx 1.10517$, extremely close.)
Q13 — Volume · Disk Method
✓ Correct Answer: B — $\pi/5$
$$V=\pi\int_0^1(x^2)^2\,dx=\pi\int_0^1 x^4\,dx=\pi\left[\frac{x^5}{5}\right]_0^1=\frac{\pi}{5}$$
Q14 — Arc Length
✓ Correct Answer: A — $14/3$
$y=\frac{2}{3}x^{3/2}$, so $y'=x^{1/2}$. Arc length:
$$L=\int_0^3\sqrt{1+x}\,dx=\left[\frac{2}{3}(1+x)^{3/2}\right]_0^3=\frac{2}{3}(4)^{3/2}-\frac{2}{3}(1)=\frac{2}{3}(8)-\frac{2}{3}=\frac{16-2}{3}=\frac{14}{3}$$
Q15 — Alternating Series Test
✓ Correct Answer: B — $\sum(-1)^n/\sqrt{n}$
AST requires: (1) $b_n=1/\sqrt{n}$ is decreasing ✓, (2) $\lim b_n=0$ ✓. Option A: $b_n=n/(n+1)\to 1\ne 0$, fails. Option C: $b_n=1$, fails. Option D: $b_n\to 1/2\ne0$, fails.
Q16 — L'Hôpital's Rule
✓ Correct Answer: C — $3$
Form $0/0$. Apply L'Hôpital's Rule (or use the known limit $\lim_{x\to0}\frac{\sin x}{x}=1$):
$$\lim_{x\to 0}\frac{\sin 3x}{x}=\lim_{x\to 0}\frac{3\cos 3x}{1}=3\cos 0=3$$
Q17 — Parametric Curves
✓ Correct Answer: A — $\frac{3t}{2}$
$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{3t^2}{2t}=\frac{3t}{2}\quad(t\ne 0)$$
Differentiate both parametric equations with respect to $t$: $dx/dt=2t$, $dy/dt=3t^2$. Divide to get the slope.
Q18 — Polar Area
✓ Correct Answer: A — $\pi$
$r=2\cos\theta$ is a circle of radius $1$ centered at $(1,0)$, so area $=\pi(1)^2=\pi$. Verify via formula: $A=\frac{1}{2}\int_{-\pi/2}^{\pi/2}(2\cos\theta)^2d\theta=2\int_{-\pi/2}^{\pi/2}\cos^2\theta\,d\theta=2\cdot\frac{\pi}{2}=\pi$.
Q19 — Absolute vs Conditional Convergence
✓ Correct Answer: B — Conditionally convergent
The alternating harmonic series $\sum(-1)^n/n$ converges by AST (since $1/n\searrow 0$). But $\sum|(-1)^n/n|=\sum 1/n$ diverges (harmonic series). Therefore: conditionally convergent, not absolutely convergent.
Q20 — Substitution · Definite Integral
✓ Correct Answer: A — $e-1$
Let $u=\cos x$, $du=-\sin x\,dx$. When $x=0$: $u=1$; when $x=\pi/2$: $u=0$.
$$\int_0^{\pi/2}\sin x\,e^{\cos x}dx=-\int_1^0 e^u\,du=\int_0^1 e^u\,du=\left[e^u\right]_0^1=e-1$$