Unit 1 · Integration Techniques
Integration by Parts · Trigonometric Integrals · Trigonometric Substitution · Partial Fractions · Improper Integrals
IBP: ∫u dv = uv − ∫v du
∫sin²x = x/2 − sin(2x)/4
x = a sinθ for √(a²−x²)
Partial Fractions
Improper: limit at ∞ or discontinuity
Evaluate the integral $\displaystyle\int x\,e^x\,dx$.
A $xe^x - e^x + C$
B $xe^x + e^x + C$
C $x^2 e^x/2 + C$
D $e^x(x-1) + C$
✓ Correct — Integration by Parts
Let $u = x$, $dv = e^x dx$. Then $du = dx$, $v = e^x$.
$\int x\,e^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = e^x(x-1)+C$
Verification: $\frac{d}{dx}[e^x(x-1)] = e^x(x-1)+e^x = xe^x$ ✓
Note: Options A and D are equivalent: $xe^x - e^x = e^x(x-1)$.
Evaluate $\displaystyle\int x^2\cos x\,dx$.
A $x^2\sin x - 2x\cos x - 2\sin x + C$
B $x^2\sin x + 2x\cos x - 2\sin x + C$
C $x^2\sin x + 2\cos x + C$
D $x^2\cos x + 2\sin x + C$
✓ Correct — Tabular IBP
First IBP: $u=x^2$, $dv=\cos x\,dx$ → $\int x^2\cos x\,dx = x^2\sin x - 2\int x\sin x\,dx$
Second IBP: $u=x$, $dv=\sin x\,dx$ → $\int x\sin x\,dx = -x\cos x + \sin x + C$
$= x^2\sin x - 2(-x\cos x + \sin x) + C = x^2\sin x + 2x\cos x - 2\sin x + C$
Evaluate $\displaystyle\int_0^{\pi/2} \sin^2 x\,dx$.
A $\pi/4$
B $\pi/2$
C $1/2$
D $1$
✓ Correct — Power-Reduction Formula
Use $\sin^2 x = \dfrac{1-\cos 2x}{2}$:
$\int_0^{\pi/2}\frac{1-\cos 2x}{2}\,dx = \left[\frac{x}{2}-\frac{\sin 2x}{4}\right]_0^{\pi/2}$
$= \left(\frac{\pi/2}{2} - \frac{\sin\pi}{4}\right) - 0 = \frac{\pi}{4} - 0 = \frac{\pi}{4}$
Evaluate $\displaystyle\int \frac{dx}{\sqrt{4-x^2}}$.
A $\arcsin\!\left(\dfrac{x}{2}\right) + C$
B $\arctan\!\left(\dfrac{x}{2}\right) + C$
C $\dfrac{1}{2}\arcsin(x) + C$
D $2\arcsin(x) + C$
✓ Correct — Trig Substitution
Let $x = 2\sin\theta$, so $dx = 2\cos\theta\,d\theta$ and $\sqrt{4-x^2} = 2\cos\theta$.
$\int \frac{2\cos\theta\,d\theta}{2\cos\theta} = \int d\theta = \theta + C = \arcsin\!\left(\frac{x}{2}\right)+C$
Verification: $\frac{d}{dx}\!\left[\arcsin\frac{x}{2}\right] = \frac{1}{\sqrt{1-(x/2)^2}}\cdot\frac{1}{2} = \frac{1}{\sqrt{4-x^2}}$ ✓
Decompose and integrate: $\displaystyle\int \frac{dx}{x^2-1}$.
A $\dfrac{1}{2}\ln\left|\dfrac{x-1}{x+1}\right| + C$
B $\dfrac{1}{2}\ln|x^2-1| + C$
C $\ln|x^2-1| + C$
D $-\arctan(x) + C$
✓ Correct — Partial Fractions
Factor: $x^2-1=(x-1)(x+1)$. Write $\dfrac{1}{(x-1)(x+1)} = \dfrac{A}{x-1}+\dfrac{B}{x+1}$.
Solving: $A=1/2$, $B=-1/2$. Therefore:
$\int\frac{dx}{x^2-1} = \frac{1}{2}\int\frac{dx}{x-1} - \frac{1}{2}\int\frac{dx}{x+1} = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right|+C$
Determine the value of $\displaystyle\int_1^{\infty} \frac{1}{x^2}\,dx$.
A Diverges
B $2$
C $1$
D $1/2$
✓ Correct — Improper Integral (Type I)
$\int_1^{\infty}\frac{dx}{x^2} = \lim_{b\to\infty}\int_1^b x^{-2}\,dx = \lim_{b\to\infty}\left[-\frac{1}{x}\right]_1^b$
$= \lim_{b\to\infty}\left(-\frac{1}{b}+1\right) = 0+1 = 1$
The integral converges to $1$. Note: $\int_1^\infty x^{-p}dx$ converges iff $p>1$.
Unit 2 · Applications of Integration
Area Between Curves · Volumes (Disk/Washer/Shell) · Arc Length
Area = ∫[f(x)−g(x)]dx
Disk: π∫[f(x)]² dx
Shell: 2π∫x·f(x) dx
Arc: ∫√(1+[f'(x)]²) dx
Find the area of the region enclosed by $y = x^2$ and $y = x$ for $0 \le x \le 1$.
A $1/2$
B $1/3$
C $1/6$
D $2/3$
✓ Correct — Area Between Curves
On $[0,1]$: $x \ge x^2$, so the area is $\int_0^1(x-x^2)\,dx$.
$= \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2}-\frac{1}{3} = \frac{3-2}{6} = \frac{1}{6}$
The region bounded by $y=\sqrt{x}$, $y=0$, and $x=4$ is revolved about the $x$-axis. Find the volume.
A $4\pi$
B $8\pi$
C $16\pi$
D $32\pi$
✓ Correct — Disk Method
$V = \pi\displaystyle\int_0^4 (\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx$
$= \pi\left[\frac{x^2}{2}\right]_0^4 = \pi\cdot\frac{16}{2} = 8\pi$
The region bounded by $y=x^2$, $y=0$, and $x=2$ is revolved about the $y$-axis using the shell method. Which integral gives the volume?
A $2\pi\displaystyle\int_0^2 x^3\,dx$
B $\pi\displaystyle\int_0^2 x^4\,dx$
C $2\pi\displaystyle\int_0^4 y\,dy$
D $\pi\displaystyle\int_0^4 \sqrt{y}\,dy$
✓ Correct — Shell Method Setup
Shell method revolving about $y$-axis: $V = 2\pi\int_a^b x\cdot f(x)\,dx$
$V = 2\pi\int_0^2 x\cdot x^2\,dx = 2\pi\int_0^2 x^3\,dx$
$= 2\pi\left[\frac{x^4}{4}\right]_0^2 = 2\pi\cdot 4 = 8\pi$
The arc length of $y = \dfrac{x^2}{2} - \dfrac{\ln x}{4}$ on $[1,e]$ requires computing $1+(y')^2$. Which expression simplifies correctly?
A $\left(x + \dfrac{1}{4x}\right)^2$
B $\left(x - \dfrac{1}{4x}\right)^2$
C $x^2 + \dfrac{1}{16x^2}$
D $1 + x^2$
✓ Correct — Arc Length Simplification
$y' = x - \dfrac{1}{4x}$, so $1+(y')^2 = 1 + x^2 - \dfrac{1}{2} + \dfrac{1}{16x^2} = x^2 + \dfrac{1}{2} + \dfrac{1}{16x^2}$
$= \left(x+\frac{1}{4x}\right)^2$
since $\left(x+\frac{1}{4x}\right)^2 = x^2 + 2\cdot x\cdot\frac{1}{4x} + \frac{1}{16x^2} = x^2 + \frac{1}{2} + \frac{1}{16x^2}$ ✓
Unit 3 · Sequences & Series
Convergence Tests · Power Series · Taylor & Maclaurin Series
Geometric: a/(1−r), |r|<1
p-series: Σ1/nᵖ, p>1
Ratio Test: L=lim|aₙ₊₁/aₙ|
Alternating Series Test
Taylor: Σf⁽ⁿ⁾(a)/n! (x−a)ⁿ
Find the sum of the geometric series $\displaystyle\sum_{n=0}^{\infty} \frac{3}{2^n}$.
A $3$
B $6$
C $\infty$ (diverges)
D $3/2$
✓ Correct — Geometric Series Sum
Write as $\displaystyle\sum_{n=0}^\infty 3\cdot\left(\frac{1}{2}\right)^n$ with $a=3$, $r=\frac{1}{2}$.
Since $|r|=\frac{1}{2}<1$, the series converges:
$S = \frac{a}{1-r} = \frac{3}{1-\frac{1}{2}} = \frac{3}{\frac{1}{2}} = 6$
Which of the following series diverges ?
A $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}$
B $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}$
C $\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}$
D $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{\pi}}$
✓ Correct — p-Series Test
A p-series $\sum\frac{1}{n^p}$ converges if $p>1$ and diverges if $p\le 1$.
• A: $p=2>1$ → Converges ($\pi^2/6$)
• B: $p=3/2>1$ → Converges
• C: $p=1/2\le 1$ → Diverges
• D: $p=\pi>1$ → Converges
Apply the Ratio Test to $\displaystyle\sum_{n=1}^{\infty}\frac{n!}{n^n}$. What is the limiting ratio $L = \lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|$?
A $L = 1$ (test inconclusive)
B $L = 1/e < 1$ (converges)
C $L = e > 1$ (diverges)
D $L = 0 < 1$ (converges)
✓ Correct — Ratio Test with e
$\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!} = \frac{(n+1)\cdot n^n}{(n+1)^{n+1}} = \frac{n^n}{(n+1)^n} = \left(\frac{n}{n+1}\right)^n$
$L = \lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n = \lim_{n\to\infty}\frac{1}{\left(1+\frac{1}{n}\right)^n} = \frac{1}{e} < 1$
Therefore the series converges by the Ratio Test.
The series $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$ is:
A Absolutely convergent
B Conditionally convergent
C Divergent
D Convergent with sum $\pi/4$
✓ Correct — Conditional Convergence
Alternating Series Test: $b_n=1/n$ is decreasing and $\lim_{n\to\infty}1/n=0$, so $\sum(-1)^n/n$ converges .
Absolute convergence? $\sum|(-1)^n/n| = \sum 1/n$ (harmonic series) — diverges .
Therefore it is conditionally convergent . Its sum is $-\ln 2$ (not $\pi/4$).
Find the radius of convergence of $\displaystyle\sum_{n=0}^{\infty} n!\,x^n$.
A $R = \infty$
B $R = 1$
C $R = 0$
D $R = e$
✓ Correct — Radius of Convergence via Ratio Test
Apply Ratio Test with $a_n = n!\,x^n$:
$\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(n+1)!\,x^{n+1}}{n!\,x^n}\right| = (n+1)|x| \to \infty \text{ for any } x\ne 0$
The series converges only at $x=0$, so $R = 0$.
The Maclaurin series for $e^x$ is $\sum_{n=0}^\infty \frac{x^n}{n!}$. Using this, $e^{-x^2}$ equals:
A $\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{n!}$
B $\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n}}{n!}$
C $\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n x^n}{n!}$
D $\displaystyle\sum_{n=0}^{\infty}(-1)^n x^{2n}$
✓ Correct — Substitution into Known Series
Replace $x$ with $-x^2$ in $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$:
$e^{-x^2} = \sum_{n=0}^{\infty}\frac{(-x^2)^n}{n!} = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{n!}$
This is the basis for computing $\int e^{-x^2}dx$ (the error function).
The Taylor polynomial of degree 2 for $f(x)=\cos x$ centered at $a=0$ is $T_2(x)=1-\dfrac{x^2}{2}$. By the Alternating Series Estimation Theorem, the error $|\cos(0.1)-T_2(0.1)|$ satisfies:
A Error $\le \dfrac{(0.1)^4}{24} = \dfrac{1}{240{,}000}$
B Error $\le \dfrac{(0.1)^3}{6}$
C Error $\le \dfrac{(0.1)^2}{2}$
D Error = $0$ exactly
✓ Correct — Alternating Series Error Bound
The Maclaurin series for $\cos x$ alternates: $1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$
After $T_2$, the next term is $\frac{x^4}{4!}$. By the ASET, the error is bounded by this next term:
$|E| \le \frac{(0.1)^4}{4!} = \frac{0.0001}{24} = \frac{1}{240{,}000} \approx 4.17\times 10^{-6}$
Unit 4 · ODEs, Parametric & Polar
Separable ODEs · Exponential Growth/Decay · Parametric Curves · Polar Area
Separable: g(y)dy = f(x)dx
y = Ce^(kt)
Polar Area: ½∫r² dθ
Solve the IVP: $\dfrac{dy}{dx} = xy$, $y(0)=2$.
A $y = 2e^{x^2/2}$
B $y = e^{x^2}+1$
C $y = 2e^{x^2}$
D $y = 2xe^x$
✓ Correct — Separable ODE
Separate: $\dfrac{dy}{y} = x\,dx$. Integrate both sides:
$\ln|y| = \frac{x^2}{2} + C_1 \implies y = Ce^{x^2/2}$
Apply $y(0)=2$: $2 = Ce^0 = C$. Therefore $y = 2e^{x^2/2}$.
Common error: integrating $x\,dx$ as $x^2$ instead of $x^2/2$, giving choice C.
Find the area enclosed by one petal of the rose curve $r = \cos(2\theta)$.
A $\pi/8$
B $\pi/4$
C $\pi/2$
D $1/4$
✓ Correct — Polar Area Formula
One petal of $r=\cos(2\theta)$ lies where $r\ge 0$, i.e., $\theta\in[-\pi/4,\pi/4]$.
$A = \frac{1}{2}\int_{-\pi/4}^{\pi/4}\cos^2(2\theta)\,d\theta = \frac{1}{2}\int_{-\pi/4}^{\pi/4}\frac{1+\cos(4\theta)}{2}\,d\theta$
$= \frac{1}{4}\left[\theta + \frac{\sin(4\theta)}{4}\right]_{-\pi/4}^{\pi/4} = \frac{1}{4}\cdot\frac{\pi}{2} = \frac{\pi}{8}$
By the Integral Test, $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln n}$:
A Converges, since $\int_2^\infty \frac{dx}{x\ln x}$ converges
B Diverges, since $\int_2^\infty \frac{dx}{x\ln x}$ diverges
C Converges by Comparison with $\sum 1/n^2$
D Converges by the Ratio Test
✓ Correct — Integral Test (log substitution)
Let $u = \ln x$, $du = dx/x$:
$\int_2^\infty\frac{dx}{x\ln x} = \int_{\ln 2}^{\infty}\frac{du}{u} = \Big[\ln u\Big]_{\ln 2}^{\infty} = \infty$
The integral diverges , so by the Integral Test, the series also diverges .
Note: This is the "log-harmonic" series — a classic boundary case slower than $\sum 1/n^p$ for any $p>1$.
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