Integration by Parts

Key Formula

$$\int u\,dv = uv - \int v\,du$$

LIATE rule — choose $u$ in this priority order: Logarithmic · Inverse trig · Algebraic · Trigonometric · Exponential.

Example

Evaluate $\displaystyle\int x e^x\,dx$.

Let $u=x$, $dv=e^x dx$ → $du=dx$, $v=e^x$
$$= xe^x - \int e^x\,dx = xe^x - e^x + C = e^x(x-1)+C$$

⭐ Memorize: $\int x e^x dx = e^x(x-1)+C$ · $\int x\sin x\,dx = \sin x - x\cos x + C$

Trigonometric Integrals

Key Identities

$\sin^2 x = \frac{1-\cos 2x}{2}$, $\cos^2 x = \frac{1+\cos 2x}{2}$
$\sin^2x+\cos^2x=1$, $\sec^2x=1+\tan^2x$

For $\int \sin^m x\cos^n x\,dx$: if $m$ or $n$ is odd, factor one out and substitute; if both even, use half-angle identities.

$\displaystyle\int \sin^3 x\,dx = \int (1-\cos^2 x)\sin x\,dx$
Let $u=\cos x$ → $-\cos x + \frac{\cos^3 x}{3}+C$

Trigonometric Substitution

$\sqrt{a^2-x^2}$: let $x=a\sin\theta$
$\sqrt{a^2+x^2}$: let $x=a\tan\theta$
$\sqrt{x^2-a^2}$: let $x=a\sec\theta$

$\displaystyle\int \frac{dx}{\sqrt{4-x^2}}$: let $x=2\sin\theta$ → $dx=2\cos\theta\,d\theta$, $\sqrt{4-x^2}=2\cos\theta$
$$=\int d\theta = \theta+C = \arcsin\!\tfrac{x}{2}+C$$

Partial Fractions

$\dfrac{P(x)}{Q(x)}$: factor $Q(x)$, then decompose.
Linear: $\dfrac{A}{x-a}$; Repeated: $\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}$;
Irreducible quad: $\dfrac{Ax+B}{x^2+bx+c}$

$\displaystyle\int \frac{1}{x^2-1}dx = \int\!\left(\frac{1/2}{x-1}-\frac{1/2}{x+1}\right)\!dx = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right|+C$

Improper Integrals

$\displaystyle\int_a^\infty f(x)\,dx = \lim_{t\to\infty}\int_a^t f(x)\,dx$
p-integral: $\displaystyle\int_1^\infty \frac{1}{x^p}dx$ converges iff $p>1$

$\displaystyle\int_1^\infty \frac{1}{x^2}dx = \lim_{t\to\infty}\!\left[-\frac{1}{x}\right]_1^t = 0-(-1) = 1$

⭐ $\int_1^\infty 1/x^p$: converges $p>1$. $\int_0^1 1/x^p$: converges $p<1$.

Sequences & Series

Geometric series: $\displaystyle\sum_{n=0}^\infty ar^n = \frac{a}{1-r}$, $|r|<1$
Telescoping: partial fractions then cancel

Divergence Test

If $\lim_{n\to\infty}a_n \neq 0$, the series diverges. (If limit = 0, test is inconclusive.)

$\displaystyle\sum_{n=0}^\infty \left(\frac{2}{3}\right)^n = \frac{1}{1-2/3} = 3$

Convergence Tests

Ratio Test: $L=\lim\left|\frac{a_{n+1}}{a_n}\right|$; converges if $L<1$, diverges if $L>1$
Root Test: $L=\lim \sqrt[n]{|a_n|}$; same rule
Integral Test: $\sum a_n$ behaves like $\int f(x)dx$ (need $f$ dec., pos.)
Comparison & Limit Comparison also key

⭐ p-series $\sum 1/n^p$: converges iff $p>1$

Power Series & Radius of Convergence

$\displaystyle\sum_{n=0}^\infty c_n(x-a)^n$; Radius $R = \frac{1}{\limsup\sqrt[n]{|c_n|}}$
Use Ratio Test: $R=\lim\left|\frac{c_n}{c_{n+1}}\right|$

$\displaystyle\sum_{n=0}^\infty \frac{x^n}{n!}$: $\lim \frac{|x|^{n+1}/(n+1)!}{|x|^n/n!}=\frac{|x|}{n+1}\to 0$ for all $x$ → $R=\infty$

Taylor & Maclaurin Series

$f(x)=\displaystyle\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$

Key Series (Memorize)

$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$  (all $x$)
$\sin x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$
$\cos x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}$
$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$  ($|x|<1$)
$\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}$  ($|x|\le1, x\neq-1$)

Parametric Equations & Arc Length

$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ (tangent slope)
Arc length: $L=\displaystyle\int_a^b\!\sqrt{\!\left(\tfrac{dx}{dt}\right)^2\!+\!\left(\tfrac{dy}{dt}\right)^2}\,dt$

$x=\cos t$, $y=\sin t$ ($0\le t\le 2\pi$)
$L=\int_0^{2\pi}\!\sqrt{\sin^2 t+\cos^2 t}\,dt=2\pi$

Polar Coordinates

$x=r\cos\theta$, $y=r\sin\theta$, $r^2=x^2+y^2$
Area: $A=\dfrac{1}{2}\displaystyle\int_\alpha^\beta r^2\,d\theta$
Arc length: $L=\displaystyle\int_\alpha^\beta\!\sqrt{r^2+\left(\tfrac{dr}{d\theta}\right)^2}\,d\theta$

Area enclosed by $r=2\cos\theta$:
$A=\frac{1}{2}\int_0^\pi (2\cos\theta)^2 d\theta = 2\int_0^\pi \cos^2\theta\,d\theta = \pi$

First-Order ODE (Separable & Linear)

Separable

$\dfrac{dy}{dx}=g(x)h(y)$ → $\displaystyle\int\frac{dy}{h(y)}=\int g(x)\,dx$

Linear ($y'+P(x)y=Q(x)$)

Integrating factor: $\mu=e^{\int P(x)dx}$
Solution: $y=\dfrac{1}{\mu}\int \mu Q(x)\,dx$

$\dfrac{dy}{dx}=xy$: separate → $\dfrac{dy}{y}=x\,dx$ → $\ln|y|=\frac{x^2}{2}+C$ → $y=Ae^{x^2/2}$

20-Question Exam

Exam-style multiple choice covering all Calculus 2 topics.
40 minutes · 5 points per question

Integration by PartsQ1 / 20

Evaluate $\displaystyle\int x e^{2x}\,dx$.

$\dfrac{1}{2}e^{2x}(2x-1)+C$

$\dfrac{1}{4}e^{2x}(2x-1)+C$

$\dfrac{1}{2}xe^{2x}-\dfrac{1}{4}e^{2x}+C$

$xe^{2x}-e^{2x}+C$

Trig IntegralsQ2 / 20

Evaluate $\displaystyle\int_0^{\pi/2}\sin^2 x\,dx$.

$\dfrac{\pi}{4}$

$\dfrac{\pi}{2}$

$1$

$\dfrac{1}{2}$

Trig SubstitutionQ3 / 20

Evaluate $\displaystyle\int \frac{x^2}{\sqrt{9-x^2}}\,dx$.

$\dfrac{9}{2}\arcsin\!\frac{x}{3}-\dfrac{x}{2}\sqrt{9-x^2}+C$

$-x\sqrt{9-x^2}+\arcsin\!\frac{x}{3}+C$

$\dfrac{x}{2}\sqrt{9-x^2}+C$

$3\arcsin\!\frac{x}{3}+C$

Partial FractionsQ4 / 20

Evaluate $\displaystyle\int \frac{3x+5}{(x+1)(x+2)}\,dx$.

$2\ln|x+1|+\ln|x+2|+C$

$\ln|x+1|+2\ln|x+2|+C$

$3\ln|(x+1)(x+2)|+C$

$\ln\!\left|\frac{x+1}{x+2}\right|+C$

Improper IntegralsQ5 / 20

Determine whether $\displaystyle\int_0^1 \frac{1}{\sqrt{x}}\,dx$ converges, and if so find its value.

Diverges

Converges to $1$

Converges to $2$

Converges to $\dfrac{1}{2}$

SequencesQ6 / 20

Find $\displaystyle\lim_{n\to\infty}\frac{n^2-3n}{2n^2+1}$.

$0$

$\dfrac{1}{2}$

$1$

$\infty$

Series ConvergenceQ7 / 20

Determine whether $\displaystyle\sum_{n=1}^\infty \frac{1}{n^2}$ converges.

Which test is most direct and what is the conclusion?

Diverges by Divergence Test

Converges by p-series test ($p=2>1$)

Diverges by p-series test ($p=2>1$)

Inconclusive — must use Integral Test

Ratio TestQ8 / 20

Apply the Ratio Test to $\displaystyle\sum_{n=1}^\infty \frac{n!}{5^n}$.

$L=5$; diverges

$L=\infty$; diverges

$L=0$; converges

$L=1$; inconclusive

Power SeriesQ9 / 20

Find the radius of convergence of $\displaystyle\sum_{n=1}^\infty \frac{x^n}{n\cdot 3^n}$.

$R=1$

$R=\dfrac{1}{3}$

$R=3$

$R=9$

Maclaurin SeriesQ10 / 20

The Maclaurin series for $\sin x$ begins $x - \dfrac{x^3}{6} + \dfrac{x^5}{120}-\cdots$
What is the coefficient of $x^7$?

$\dfrac{1}{5040}$

$-\dfrac{1}{5040}$

$\dfrac{1}{720}$

$-\dfrac{1}{720}$

ParametricQ11 / 20

For the parametric curve $x=t^2$, $y=t^3$, find $\dfrac{dy}{dx}$ at $t=2$.

$3$

$\dfrac{3}{2}$

$\dfrac{3t}{2}$

$4$

Polar AreaQ12 / 20

Find the area enclosed by the cardioid $r=1+\cos\theta$.

$\pi$

$\dfrac{3\pi}{2}$

$\dfrac{3\pi}{4}$

$2\pi$

Separable ODEQ13 / 20

Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$ with $y(0)=3$.

$y=\sqrt{x^2+9}$

$y=x^2+3$

$y=3e^{x^2/2}$

$y=\sqrt{2x+9}$

Arc LengthQ14 / 20

Find the arc length of $y=\ln(\cos x)$ on $\left[0,\dfrac{\pi}{4}\right]$.

$\ln(1+\sqrt{2})$

$\ln\sqrt{2}$

$\dfrac{\pi}{4}$

$1-\dfrac{\pi}{4}$

Alternating SeriesQ15 / 20

Does $\displaystyle\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$ converge? If so, to what value?

Diverges

Converges absolutely to $\ln 2$

Converges conditionally to $\ln 2$

Converges to $\dfrac{\pi}{4}$

Taylor PolynomialQ16 / 20

Find the 2nd-degree Taylor polynomial of $f(x)=e^x$ centered at $a=1$.

$e + e(x-1)+\dfrac{e}{2}(x-1)^2$

$1+x+\dfrac{x^2}{2}$

$e\left[1+(x-1)+\dfrac{(x-1)^2}{2!}\right]$

$e^x$

Volumes of RevolutionQ17 / 20

Find the volume generated by rotating $y=x^2$ on $[0,1]$ about the $x$-axis (Disk method).

$\dfrac{\pi}{3}$

$\dfrac{\pi}{5}$

$\dfrac{2\pi}{5}$

$\pi$

Linear ODEQ18 / 20

Solve $y' + 2y = 4x$, find the general solution.

$y = 2x - 1 + Ce^{-2x}$

$y = 2x + Ce^{-2x}$

$y = x - \dfrac{1}{2} + Ce^{-2x}$

$y = 4x + Ce^{2x}$

Geometric SeriesQ19 / 20

Find the sum of $\displaystyle\sum_{n=0}^\infty \frac{(-1)^n}{3^n}$.

$\dfrac{3}{4}$

$\dfrac{1}{4}$

$\dfrac{3}{2}$

Diverges

Surface AreaQ20 / 20

The surface area generated by rotating $y=\sqrt{x}$ on $[0,1]$ about the $x$-axis is:

$\dfrac{\pi}{6}(5\sqrt{5}-1)$

$\dfrac{\pi}{3}(5\sqrt{5}-1)$

$2\pi(\sqrt{2}-1)$

$\pi\sqrt{5}$

Excellent!

20 / 20 correct · 100 pts

Correct

Incorrect

Points

0:00

Time Used

Calculus 2 — Solutions

Q1 · Integration by Parts—

Answer: B — $\dfrac{1}{4}e^{2x}(2x-1)+C$

Let $u=x$, $dv=e^{2x}dx$ → $du=dx$, $v=\frac{1}{2}e^{2x}$.
$\int xe^{2x}dx = \frac{x}{2}e^{2x}-\int\frac{1}{2}e^{2x}dx = \frac{x}{2}e^{2x}-\frac{1}{4}e^{2x}+C = \frac{1}{4}e^{2x}(2x-1)+C$.
Note: options A and B differ only in the leading factor; verify by differentiation.

Q2 · Trig Integrals—

Answer: A — $\dfrac{\pi}{4}$

Use $\sin^2 x = \frac{1-\cos 2x}{2}$:
$\int_0^{\pi/2}\frac{1-\cos 2x}{2}dx = \left[\frac{x}{2}-\frac{\sin 2x}{4}\right]_0^{\pi/2} = \frac{\pi}{4}-0 = \dfrac{\pi}{4}$.

Q3 · Trig Substitution—

Answer: A — $\dfrac{9}{2}\arcsin\!\frac{x}{3}-\dfrac{x}{2}\sqrt{9-x^2}+C$

Let $x=3\sin\theta$, $dx=3\cos\theta\,d\theta$, $\sqrt{9-x^2}=3\cos\theta$.
$\int\frac{9\sin^2\theta}{3\cos\theta}\cdot3\cos\theta\,d\theta = 9\int\sin^2\theta\,d\theta = \frac{9}{2}(\theta-\sin\theta\cos\theta)+C$.
Back-substitute: $\theta=\arcsin(x/3)$, $\sin\theta=x/3$, $\cos\theta=\frac{\sqrt{9-x^2}}{3}$.
$= \frac{9}{2}\arcsin\!\frac{x}{3} - \frac{9}{2}\cdot\frac{x}{3}\cdot\frac{\sqrt{9-x^2}}{3}+C = \frac{9}{2}\arcsin\!\frac{x}{3}-\frac{x\sqrt{9-x^2}}{2}+C$.

Q4 · Partial Fractions—

Answer: A — $2\ln|x+1|+\ln|x+2|+C$

Decompose: $\frac{3x+5}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}$.
$3x+5=A(x+2)+B(x+1)$. Set $x=-1$: $2=A$. Set $x=-2$: $-1=B(-1)$ → $B=1$.
$\int\!\left(\frac{2}{x+1}+\frac{1}{x+2}\right)dx = 2\ln|x+1|+\ln|x+2|+C$.

Q5 · Improper Integrals—

Answer: C — Converges to $2$

$\int_0^1 x^{-1/2}dx = \lim_{\varepsilon\to 0^+}\int_\varepsilon^1 x^{-1/2}dx = \lim_{\varepsilon\to 0^+}\left[2\sqrt{x}\right]_\varepsilon^1 = 2-0 = 2$.
p-test on $(0,1]$: converges since $p=1/2 < 1$.

Q6 · Sequences—

Answer: B — $\dfrac{1}{2}$

Divide numerator and denominator by $n^2$:
$\lim_{n\to\infty}\frac{1-3/n}{2+1/n^2} = \frac{1-0}{2+0} = \dfrac{1}{2}$.

Q7 · Series Convergence—

Answer: B — Converges by p-series test ($p=2>1$)

$\sum 1/n^p$ converges iff $p>1$. Here $p=2>1$, so the series converges (to $\pi^2/6$ — Basel problem). The Divergence Test only gives divergence, not convergence.

Q8 · Ratio Test—

Answer: B — $L=\infty$; diverges

$\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)!/5^{n+1}}{n!/5^n} = \frac{n+1}{5} \to \infty$ as $n\to\infty$.
Since $L=\infty > 1$, the series diverges.

Q9 · Power Series—

Answer: C — $R=3$

$c_n = \frac{1}{n\cdot3^n}$. Apply Ratio Test:
$L = \lim\left|\frac{x^{n+1}/((n+1)\cdot3^{n+1})}{x^n/(n\cdot3^n)}\right| = |x|\cdot\lim\frac{n}{3(n+1)} = \frac{|x|}{3}$.
Converges when $|x|/3 < 1$, i.e., $|x|<3$. So $R=3$.

Q10 · Maclaurin Series—

Answer: B — $-\dfrac{1}{5040}$

$\sin x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$. For $x^7$: set $2n+1=7$ → $n=3$.
Coefficient $= \frac{(-1)^3}{7!} = \frac{-1}{5040}$.

Q11 · Parametric—

Answer: A — $3$

$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{3t^2}{2t}=\dfrac{3t}{2}$.
At $t=2$: $\dfrac{3(2)}{2}=3$.

Q12 · Polar Area—

Answer: B — $\dfrac{3\pi}{2}$

$A=\frac{1}{2}\int_0^{2\pi}(1+\cos\theta)^2 d\theta = \frac{1}{2}\int_0^{2\pi}(1+2\cos\theta+\cos^2\theta)\,d\theta$.
$\int_0^{2\pi}1\,d\theta=2\pi$, $\int_0^{2\pi}\cos\theta\,d\theta=0$, $\int_0^{2\pi}\cos^2\theta\,d\theta=\pi$.
$A=\frac{1}{2}(2\pi+0+\pi)=\dfrac{3\pi}{2}$.

Q13 · Separable ODE—

Answer: A — $y=\sqrt{x^2+9}$

Separate: $y\,dy = x\,dx$ → $\frac{y^2}{2}=\frac{x^2}{2}+C_1$ → $y^2=x^2+C$.
$y(0)=3$: $9=0+C$ → $C=9$. So $y^2=x^2+9$ → $y=\sqrt{x^2+9}$ (positive root since $y(0)=3>0$).

Q14 · Arc Length—

Answer: A — $\ln(1+\sqrt{2})$

$y'=-\tan x$, so $\sqrt{1+(y')^2}=\sqrt{1+\tan^2 x}=\sec x$ (since $\sec x>0$ on $[0,\pi/4]$).
$L=\int_0^{\pi/4}\sec x\,dx = \left[\ln|\sec x+\tan x|\right]_0^{\pi/4}$
$= \ln(\sqrt{2}+1)-\ln(1+0) = \ln(1+\sqrt{2})$.

Q15 · Alternating Series—

Answer: C — Converges conditionally to $\ln 2$

By Alternating Series Test: $b_n=1/n$ is decreasing and $\to 0$, so it converges.
But $\sum 1/n$ (harmonic series) diverges → NOT absolutely convergent. Hence conditionally convergent.
Known result: $\ln 2 = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$

Q16 · Taylor Polynomial—

Answer: A or C (equivalent) — $e + e(x-1)+\dfrac{e}{2}(x-1)^2$

$f(x)=e^x$, $f'(x)=e^x$, $f''(x)=e^x$. At $a=1$: all derivatives equal $e$.
$T_2(x)=\frac{e}{0!}+\frac{e}{1!}(x-1)+\frac{e}{2!}(x-1)^2 = e\!\left[1+(x-1)+\frac{(x-1)^2}{2}\right]$.
Options A and C are the same expression written differently; the correct marked answer is A.

Q17 · Volumes of Revolution—

Answer: B — $\dfrac{\pi}{5}$

Disk method: $V=\pi\int_0^1[f(x)]^2 dx = \pi\int_0^1 x^4\,dx = \pi\left[\frac{x^5}{5}\right]_0^1 = \dfrac{\pi}{5}$.

Q18 · Linear ODE—

Answer: A — $y=2x-1+Ce^{-2x}$

Integrating factor: $\mu=e^{\int 2\,dx}=e^{2x}$.
$(e^{2x}y)'=4xe^{2x}$ → $e^{2x}y=\int 4xe^{2x}dx$.
IBP: $\int 4xe^{2x}dx = 4\cdot\frac{e^{2x}}{2}(x-\frac{1}{2})\cdot 2 = e^{2x}(2x-1)+C$.
(Verify: $\int 4xe^{2x}dx$: let $u=4x$, $dv=e^{2x}dx$ → $2xe^{2x}-e^{2x}+C$... re-check: $=4[\frac{x}{2}e^{2x}-\frac{1}{4}e^{2x}]+C=e^{2x}(2x-1)+C$.)
So $y = e^{-2x}[e^{2x}(2x-1)+C] = 2x-1+Ce^{-2x}$.

Q19 · Geometric Series—

Answer: A — $\dfrac{3}{4}$

Write as $\sum_{n=0}^\infty \left(-\frac{1}{3}\right)^n$: geometric with $a=1$, $r=-\frac{1}{3}$. Since $|r|=\frac{1}{3}<1$:
Sum $= \frac{1}{1-(-1/3)} = \frac{1}{4/3} = \dfrac{3}{4}$.

Q20 · Surface Area—

Answer: A — $\dfrac{\pi}{6}(5\sqrt{5}-1)$

$y=x^{1/2}$, $y'=\frac{1}{2\sqrt{x}}$, $(y')^2=\frac{1}{4x}$.
$S=2\pi\int_0^1 x^{1/2}\sqrt{1+\frac{1}{4x}}\,dx = 2\pi\int_0^1\sqrt{x+\frac{1}{4}}\,dx$.
$= 2\pi\left[\frac{2}{3}\left(x+\frac{1}{4}\right)^{3/2}\right]_0^1 = \frac{4\pi}{3}\left[\left(\frac{5}{4}\right)^{3/2}-\left(\frac{1}{4}\right)^{3/2}\right]$
$= \frac{4\pi}{3}\left[\frac{5\sqrt{5}}{8}-\frac{1}{8}\right] = \frac{4\pi}{3}\cdot\frac{5\sqrt{5}-1}{8} = \dfrac{\pi(5\sqrt{5}-1)}{6}$.