University Mathematics

Calculus II
Mastery Exam

20 exam-style problems · Full worked solutions · All major topics

40 minutes
20 questions
Multiple choice

Topics covered

Integration Techniques Integration by Parts Partial Fractions Improper Integrals Sequences & Series Power Series Taylor & Maclaurin Convergence Tests Polar Coordinates Parametric Equations Arc Length Surface Area Differential Equations
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Integration Techniques
Key Concepts & Formulas
\(\displaystyle\int u\,dv = uv - \int v\,du\)  — Integration by Parts (IBP)
\(\displaystyle\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C\)
Trig Identities: \(\sin^2 x = \frac{1-\cos 2x}{2}\), \(\cos^2 x = \frac{1+\cos 2x}{2}\)
Trig Sub: \(\sqrt{a^2-x^2}\Rightarrow x=a\sin\theta\); \(\sqrt{x^2+a^2}\Rightarrow x=a\tan\theta\)
Worked Example
\(\displaystyle\int x e^x dx\): Let \(u=x,\;dv=e^x dx\). Then \(du=dx,\;v=e^x\). Result: \(xe^x - e^x + C\).
1
Integration by Parts Medium
Evaluate \(\displaystyle\int x\cos x\,dx\).
A
\(x\sin x - \cos x + C\)
B
\(x\sin x + \cos x + C\)
C
\(-x\sin x + \cos x + C\)
D
\(\cos x - x\sin x + C\)
✓ Solution
IBP: \(u=x,\;dv=\cos x\,dx \Rightarrow du=dx,\;v=\sin x\).
\(\int x\cos x\,dx = x\sin x - \int\sin x\,dx = x\sin x + \cos x + C\).
2
Trig Integrals Medium
Evaluate \(\displaystyle\int \sin^2 x\,dx\).
A
\(\dfrac{x}{2} + \dfrac{\sin 2x}{4} + C\)
B
\(\dfrac{x}{2} - \dfrac{\sin 2x}{4} + C\)
C
\(-\dfrac{\sin 2x}{4} + C\)
D
\(\dfrac{x}{2} - \dfrac{\cos 2x}{4} + C\)
✓ Solution
Use \(\sin^2 x = \dfrac{1-\cos 2x}{2}\).
\(\int\sin^2 x\,dx = \int\dfrac{1-\cos 2x}{2}\,dx = \dfrac{x}{2} - \dfrac{\sin 2x}{4} + C\).
3
Trigonometric Substitution Hard
Evaluate \(\displaystyle\int \frac{dx}{\sqrt{4-x^2}}\).
A
\(\arcsin\!\left(\dfrac{x}{2}\right) + C\)
B
\(\dfrac{1}{2}\arcsin\!\left(\dfrac{x}{2}\right) + C\)
C
\(\arctan\!\left(\dfrac{x}{2}\right) + C\)
D
\(2\arcsin x + C\)
✓ Solution
Standard formula: \(\displaystyle\int\frac{dx}{\sqrt{a^2-x^2}}=\arcsin\!\left(\frac{x}{a}\right)+C\) with \(a=2\).
Answer: \(\arcsin\!\left(\dfrac{x}{2}\right)+C\).
Partial Fractions & Improper Integrals
Key Concepts
Partial Fractions: \(\dfrac{P(x)}{(x-a)(x-b)} = \dfrac{A}{x-a}+\dfrac{B}{x-b}\)
Improper Integral: \(\displaystyle\int_1^\infty \frac{dx}{x^p}\) converges iff \(p>1\)
\(\displaystyle\int_a^\infty f(x)\,dx = \lim_{t\to\infty}\int_a^t f(x)\,dx\)
Quick Tip
For \(\frac{1}{x^2-1}=\frac{1}{(x-1)(x+1)}\), cover-up: \(A=\frac{1}{1+1}=\frac12\), \(B=\frac{1}{-1-1}=-\frac12\).
4
Partial Fractions Hard
Evaluate \(\displaystyle\int \frac{2x+1}{x^2-x-2}\,dx\).
A
\(\ln|x-2| + \ln|x+1| + C\)
B
\(\ln|x^2-x-2| + C\)
C
\(\dfrac{5}{3}\ln|x-2| + \dfrac{1}{3}\ln|x+1| + C\)
D
\(2\ln|x-2| - \ln|x+1| + C\)
✓ Solution
Factor: \(x^2-x-2=(x-2)(x+1)\). Partial fractions: \(\dfrac{2x+1}{(x-2)(x+1)}=\dfrac{A}{x-2}+\dfrac{B}{x+1}\).
Multiply both sides by \((x-2)(x+1)\): \(2x+1=A(x+1)+B(x-2)\).
\(x=2\): \(5=3A\Rightarrow A=\frac{5}{3}\).  \(x=-1\): \(-1=-3B\Rightarrow B=\frac{1}{3}\).
\(\displaystyle\int=\frac{5}{3}\ln|x-2|+\frac{1}{3}\ln|x+1|+C\).
5
Improper Integrals Medium
Does \(\displaystyle\int_1^\infty \frac{1}{x^2}\,dx\) converge? If so, find its value.
A
Diverges
B
Converges to \(2\)
C
Converges to \(1\)
D
Converges to \(\dfrac{1}{2}\)
✓ Solution
\(\displaystyle\int_1^\infty x^{-2}\,dx = \lim_{t\to\infty}\left[-\frac{1}{x}\right]_1^t = \lim_{t\to\infty}\!\left(-\frac{1}{t}+1\right)=1\).
The integral converges to \(1\).
Sequences & Series
Convergence Tests & Series
Geometric: \(\sum_{n=0}^\infty ar^n = \dfrac{a}{1-r}\) when \(|r|<1\)
p-Series: \(\sum\frac{1}{n^p}\) converges iff \(p>1\)
Ratio Test: \(L=\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|\). Converges if \(L<1\), diverges if \(L>1\)
Alternating Series Test (Leibniz): converges if \(b_n\searrow0\)
Divergence Test: if \(\lim a_n\neq 0\), series diverges
6
Geometric Series Medium
Find the sum of the series \(\displaystyle\sum_{n=0}^{\infty}\left(\frac{2}{3}\right)^n\).
A
\(2\)
B
\(3\)
C
\(\dfrac{2}{3}\)
D
Diverges
✓ Solution
Geometric series with \(a=1\) and \(r=\frac{2}{3}\). Since \(|r|<1\), it converges.
Sum \(=\dfrac{a}{1-r}=\dfrac{1}{1-\frac{2}{3}}=\dfrac{1}{\frac{1}{3}}=3\).
7
Ratio Test Hard
Determine the convergence of \(\displaystyle\sum_{n=1}^{\infty}\frac{n!}{3^n}\).
A
Converges by Ratio Test
B
Diverges by Ratio Test
C
Converges by p-Series
D
Ratio Test is inconclusive
✓ Solution
\(\dfrac{a_{n+1}}{a_n}=\dfrac{(n+1)!}{3^{n+1}}\cdot\dfrac{3^n}{n!}=\dfrac{n+1}{3}\to\infty\) as \(n\to\infty\).
Since \(L=\infty>1\), the series diverges by the Ratio Test.
8
Alternating Series Medium
Does \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\) converge?
A
Diverges
B
Converges absolutely
C
Converges conditionally
D
Diverges by Divergence Test
✓ Solution
This is the alternating harmonic series. By the Alternating Series Test: \(b_n=\frac{1}{n}\) is decreasing and \(\lim_{n\to\infty}\frac{1}{n}=0\), so it converges.
But \(\sum\frac{1}{n}\) (harmonic series) diverges, so convergence is conditional, not absolute.
9
p-Series Medium
Which series converges?
A
\(\displaystyle\sum_{n=1}^\infty \frac{1}{n}\)
B
\(\displaystyle\sum_{n=1}^\infty \frac{1}{\sqrt{n}}\)
C
\(\displaystyle\sum_{n=1}^\infty \frac{1}{n^2}\)
D
\(\displaystyle\sum_{n=1}^\infty \frac{1}{n^{1/3}}\)
✓ Solution
p-series \(\sum\frac{1}{n^p}\) converges iff \(p>1\).
(A) \(p=1\): diverges (harmonic).  (B) \(p=\frac{1}{2}<1\): diverges.  (C) \(p=2>1\): converges.  (D) \(p=\frac{1}{3}<1\): diverges.
Power Series & Taylor Series
Power & Taylor Series
Taylor: \(f(x)=\displaystyle\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n\)
\(e^x=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}\),  \(\sin x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^n x^{2n+1}}{(2n+1)!}\),  \(\cos x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^n x^{2n}}{(2n)!}\)
Radius of Convergence: \(R=\lim_{n\to\infty}\left|\dfrac{a_n}{a_{n+1}}\right|\)
\(\dfrac{1}{1-x}=\displaystyle\sum_{n=0}^\infty x^n\) for \(|x|<1\)
10
Radius of Convergence Hard
Find the radius of convergence of \(\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n+1}\).
A
\(R=0\)
B
\(R=\infty\)
C
\(R=1\)
D
\(R=2\)
✓ Solution
Ratio Test: \(L=\lim_{n\to\infty}\left|\dfrac{x^{n+1}}{n+2}\cdot\dfrac{n+1}{x^n}\right|=|x|\cdot\lim_{n\to\infty}\dfrac{n+1}{n+2}=|x|\).
Converges when \(|x|<1\). Therefore \(R=1\).
11
Maclaurin Series Hard
Which is the Maclaurin series for \(e^{-x^2}\)?
A
\(\displaystyle\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{n!}\)
B
\(\displaystyle\sum_{n=0}^\infty \frac{x^{2n}}{n!}\)
C
\(\displaystyle\sum_{n=0}^\infty (-1)^n\frac{x^n}{n!}\)
D
\(\displaystyle\sum_{n=0}^\infty \frac{(-x^2)^n}{(2n)!}\)
✓ Solution
Start with \(e^u=\sum_{n=0}^\infty\frac{u^n}{n!}\). Substitute \(u=-x^2\):
\(e^{-x^2}=\sum_{n=0}^\infty\frac{(-x^2)^n}{n!}=\sum_{n=0}^\infty\frac{(-1)^n x^{2n}}{n!}\).
12
Taylor Polynomial Advanced
Find the third-degree Taylor polynomial for \(f(x)=\ln(1+x)\) centered at \(a=0\).
A
\(x - \dfrac{x^2}{2} + \dfrac{x^3}{3}\)
B
\(x + \dfrac{x^2}{2} + \dfrac{x^3}{3}\)
C
\(1 + x - \dfrac{x^2}{2}\)
D
\(x - x^2 + x^3\)
✓ Solution
\(f(0)=0\). \(f'(x)=\frac{1}{1+x}\Rightarrow f'(0)=1\). \(f''(x)=-\frac{1}{(1+x)^2}\Rightarrow f''(0)=-1\). \(f'''(x)=\frac{2}{(1+x)^3}\Rightarrow f'''(0)=2\).
\(T_3(x)=0+1\cdot x+\frac{-1}{2!}x^2+\frac{2}{3!}x^3 = x-\frac{x^2}{2}+\frac{x^3}{3}\).
Polar Coordinates & Parametric Equations
Polar & Parametric
Polar area: \(A=\dfrac{1}{2}\displaystyle\int_\alpha^\beta r^2\,d\theta\)
Parametric slope: \(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}\)
Parametric arc length: \(L=\displaystyle\int_a^b\!\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt\)
13
Parametric Calculus Medium
If \(x=t^2\) and \(y=t^3\), find \(\dfrac{dy}{dx}\).
A
\(\dfrac{3t}{2}\)
B
\(\dfrac{2}{3t}\)
C
\(3t^2\)
D
\(\dfrac{3t^2}{2t}=\dfrac{3t}{2}\) (same as A)
✓ Solution
\(\dfrac{dx}{dt}=2t\), \(\dfrac{dy}{dt}=3t^2\).
\(\dfrac{dy}{dx}=\dfrac{3t^2}{2t}=\dfrac{3t}{2}\).  Note: options A and D are identical — the answer is \(\dfrac{3t}{2}\).
14
Polar Area Hard
Find the area enclosed by \(r=2\cos\theta\) (one full loop).
A
\(\pi\)
B
\(2\pi\)
C
\(4\)
D
\(\dfrac{\pi}{2}\)
✓ Solution
\(r=2\cos\theta\) is a circle of radius 1. Area \(=\pi r^2=\pi\cdot 1^2=\pi\).
Via formula: \(A=\frac{1}{2}\int_{-\pi/2}^{\pi/2}(2\cos\theta)^2\,d\theta=2\int_{-\pi/2}^{\pi/2}\cos^2\theta\,d\theta=2\cdot\frac{\pi}{2}=\pi\).
Arc Length & Surface Area
Arc Length & Surface Area
Arc Length: \(L=\displaystyle\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\)
Surface of Revolution (about \(x\)-axis): \(S=2\pi\displaystyle\int_a^b y\sqrt{1+(y')^2}\,dx\)
15
Arc Length Hard
Set up the integral for the arc length of \(y=x^2\) from \(x=0\) to \(x=1\). Which is correct?
A
\(\displaystyle\int_0^1 \sqrt{1+4x^2}\,dx\)
B
\(\displaystyle\int_0^1 \sqrt{1+x^2}\,dx\)
C
\(\displaystyle\int_0^1 \sqrt{1+2x}\,dx\)
D
\(\displaystyle\int_0^1 (1+4x^2)\,dx\)
✓ Solution
\(y=x^2\Rightarrow y'=2x\Rightarrow (y')^2=4x^2\).
Arc length \(=\displaystyle\int_0^1\sqrt{1+(2x)^2}\,dx=\int_0^1\sqrt{1+4x^2}\,dx\).
16
Surface Area Advanced
The surface area generated by revolving \(y=\sqrt{x}\), \(0\le x\le 4\), about the \(x\)-axis is:
A
\(\dfrac{\pi}{6}(17\sqrt{17}-1)\)
B
\(\pi(17\sqrt{17}-1)\)
C
\(\dfrac{\pi}{3}(17\sqrt{17}-1)\)
D
\(2\pi\sqrt{17}\)
✓ Solution
\(y=\sqrt{x}\Rightarrow y'=\frac{1}{2\sqrt{x}}\Rightarrow 1+(y')^2=1+\frac{1}{4x}=\frac{4x+1}{4x}\).
\(S=2\pi\int_0^4\sqrt{x}\cdot\sqrt{\frac{4x+1}{4x}}\,dx=2\pi\int_0^4\frac{\sqrt{4x+1}}{2}\,dx=\pi\int_0^4\sqrt{4x+1}\,dx\).
Let \(u=4x+1\), \(du=4\,dx\): \(S=\frac{\pi}{4}\cdot\frac{2}{3}\left[u^{3/2}\right]_1^{17}=\frac{\pi}{6}(17^{3/2}-1)=\frac{\pi}{6}(17\sqrt{17}-1)\).
Differential Equations
ODE Methods
Separable: \(\frac{dy}{dx}=g(x)h(y)\Rightarrow\int\frac{dy}{h(y)}=\int g(x)\,dx\)
First-order linear: \(\frac{dy}{dx}+P(x)y=Q(x)\). Integrating factor: \(\mu=e^{\int P\,dx}\)
Solution: \(y=\frac{1}{\mu}\int\mu Q\,dx\)
17
Separable ODE Medium
Solve \(\dfrac{dy}{dx} = xy\) with \(y(0)=1\).
A
\(y = e^{x^2/2}\)
B
\(y = e^{x^2}\)
C
\(y = e^x\)
D
\(y = x^2 + 1\)
✓ Solution
Separate: \(\dfrac{dy}{y}=x\,dx\). Integrate: \(\ln|y|=\frac{x^2}{2}+C\). So \(y=Ae^{x^2/2}\).
Apply IC: \(y(0)=A=1\). Therefore \(y=e^{x^2/2}\).
18
Linear First-Order ODE Hard
Solve \(\dfrac{dy}{dx} + 2y = e^{-2x}\).
A
\(y = (x+C)e^{-2x}\)
B
\(y = Ce^{-2x}\)
C
\(y = xe^{-2x}+Ce^{-2x}\)
D
\(y = xe^{2x}+C\)
✓ Solution
Integrating factor: \(\mu=e^{\int 2\,dx}=e^{2x}\).
Multiply: \(e^{2x}\frac{dy}{dx}+2e^{2x}y=1\Rightarrow\frac{d}{dx}(ye^{2x})=1\).
Integrate: \(ye^{2x}=x+C\Rightarrow y=(x+C)e^{-2x}\).
Applications of Integration
Areas & Volumes
Area between curves: \(A=\displaystyle\int_a^b[f(x)-g(x)]\,dx\) where \(f\ge g\)
Disk method: \(V=\pi\displaystyle\int_a^b [f(x)]^2\,dx\)
Shell method: \(V=2\pi\displaystyle\int_a^b x\,f(x)\,dx\)
Washer method: \(V=\pi\displaystyle\int_a^b\!\left([f(x)]^2-[g(x)]^2\right)\!dx\)
19
Volumes of Revolution Hard
Find the volume of the solid formed by revolving \(y=x^2\) from \(x=0\) to \(x=1\) about the \(x\)-axis.
A
\(\dfrac{\pi}{3}\)
B
\(\dfrac{\pi}{5}\)
C
\(\dfrac{2\pi}{5}\)
D
\(\pi\)
✓ Solution
Disk method: \(V=\pi\displaystyle\int_0^1(x^2)^2\,dx=\pi\int_0^1 x^4\,dx=\pi\left[\frac{x^5}{5}\right]_0^1=\frac{\pi}{5}\).
20
Area Between Curves Advanced
Find the area enclosed between \(y=x^2\) and \(y=x\).
A
\(\dfrac{1}{6}\)
B
\(\dfrac{1}{3}\)
C
\(\dfrac{1}{2}\)
D
\(\dfrac{1}{4}\)
✓ Solution
Intersections: \(x^2=x\Rightarrow x=0,1\). On \([0,1]\), \(x\ge x^2\).
\(A=\displaystyle\int_0^1(x-x^2)\,dx=\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\).
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Answer Key & Solutions

Full worked solutions for all 20 questions