Calculus II Master Problem Set

Concept Review

Integration Techniques ▼

Integration by Parts: \(\int u\,dv = uv - \int v\,du\)
LIATE rule: Log, Inverse trig, Algebraic, Trig, Exponential

Trig substitution: \(a^2-x^2 \Rightarrow x=a\sin\theta\); \(a^2+x^2 \Rightarrow x=a\tan\theta\); \(x^2-a^2 \Rightarrow x=a\sec\theta\)
Partial fractions: decompose rational functions before integrating
Reduction formulas for \(\sin^n x\), \(\cos^n x\)

Quick Example

Evaluate \(\int x e^x\,dx\)

Let \(u=x\), \(dv=e^x dx\). Then \(\int xe^x dx = xe^x - e^x + C = (x-1)e^x + C\)

Sequences & Series ▼

Geometric Series: \(\displaystyle\sum_{n=0}^{\infty} ar^n = \dfrac{a}{1-r}\), \(|r|<1\)

p-Series: \(\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^p}\) converges iff \(p>1\)

Ratio Test: converges if \(\lim|a_{n+1}/a_n|<1\)
Alternating Series Test: \(\sum(-1)^n b_n\) converges if \(b_n\) decreasing → 0
Integral Test, Comparison Test, Limit Comparison Test

Quick Example

Does \(\sum_{n=1}^{\infty}\dfrac{1}{n^2}\) converge?

p-series with p = 2 > 1. Yes, it converges to \(\pi^2/6\) (Basel problem).

Power Series & Taylor Series ▼

Taylor Series: \(f(x)=\displaystyle\sum_{n=0}^{\infty}\dfrac{f^{(n)}(a)}{n!}(x-a)^n\)

Key series: \(e^x = \sum\dfrac{x^n}{n!}\), \(\sin x = \sum\dfrac{(-1)^n x^{2n+1}}{(2n+1)!}\), \(\dfrac{1}{1-x}=\sum x^n\)

Radius of convergence via Ratio Test: \(R = \lim|a_n/a_{n+1}|\)
Interval of convergence: check endpoints separately

Polar & Parametric Curves ▼

Polar Area: \(A = \dfrac{1}{2}\int_\alpha^\beta [r(\theta)]^2\,d\theta\)

Arc Length (param.): \(L = \int_a^b\!\sqrt{\left(\frac{dx}{dt}\right)^2\!+\!\left(\frac{dy}{dt}\right)^2}\,dt\)

Slope of parametric curve: \(dy/dx = (dy/dt)/(dx/dt)\)
Convert: \(x = r\cos\theta\), \(y = r\sin\theta\), \(r^2 = x^2+y^2\)

Improper Integrals & Applications ▼

Improper Integral: \(\int_a^\infty f\,dx = \lim_{t\to\infty}\int_a^t f\,dx\)

Surface Area: \(S = 2\pi\int_a^b y\sqrt{1+(y')^2}\,dx\)

\(\int_1^\infty \frac{1}{x^p}dx\) converges iff \(p>1\)
Volume by shells: \(V = 2\pi\int_a^b x\,f(x)\,dx\)
Volume by disks: \(V = \pi\int_a^b [f(x)]^2\,dx\)

Problems

Integration by Parts

Evaluate \(\displaystyle\int x\cos x\,dx\).

✓Correct!

Step 1: Let \(u = x\), \(dv = \cos x\,dx\). Then \(du = dx\), \(v = \sin x\).

Step 2: IBP formula: \(\int x\cos x\,dx = x\sin x - \int \sin x\,dx\)

Step 3: \(= x\sin x - (-\cos x) + C = x\sin x + \cos x + C\)

Check: Differentiate: \(\frac{d}{dx}(x\sin x + \cos x) = \sin x + x\cos x - \sin x = x\cos x\) ✓

Trig Substitution

Evaluate \(\displaystyle\int\frac{dx}{\sqrt{4-x^2}}\).

Step 1: Use the standard formula: \(\int\dfrac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\!\left(\dfrac{x}{a}\right)+C\)

Step 2: Here \(a^2 = 4\), so \(a = 2\).

Result: \(\sin^{-1}\!\left(\dfrac{x}{2}\right)+C\)

Verify: \(\frac{d}{dx}\sin^{-1}(x/2) = \frac{1/2}{\sqrt{1-(x/2)^2}} = \frac{1}{\sqrt{4-x^2}}\) ✓

Partial Fractions

Evaluate \(\displaystyle\int\frac{dx}{x^2-1}\).

Step 1: \(\dfrac{1}{x^2-1}=\dfrac{1}{(x-1)(x+1)}=\dfrac{1/2}{x-1}-\dfrac{1/2}{x+1}\)

Step 2: \(\int\left(\dfrac{1/2}{x-1}-\dfrac{1/2}{x+1}\right)dx = \dfrac{1}{2}\ln|x-1|-\dfrac{1}{2}\ln|x+1|+C\)

Step 3: \(= \dfrac{1}{2}\ln\left|\dfrac{x-1}{x+1}\right|+C\)

Improper Integrals

Determine whether \(\displaystyle\int_1^{\infty}\frac{dx}{x^2}\) converges. If so, find its value.

Step 1: \(\int_1^{\infty}\frac{dx}{x^2}=\lim_{t\to\infty}\int_1^t x^{-2}dx=\lim_{t\to\infty}\left[-\frac{1}{x}\right]_1^t\)

Step 2: \(=\lim_{t\to\infty}\left(-\dfrac{1}{t}+1\right)=0+1=1\)

Alternatively: p-series with \(p=2>1\), so it converges.

Geometric Series

Find the sum of the series \(\displaystyle\sum_{n=0}^{\infty}\left(\frac{2}{3}\right)^n\).

Formula: Geometric series \(\sum_{n=0}^\infty ar^n = \dfrac{a}{1-r}\) for \(|r|<1\)

Here: \(a=1\), \(r=\dfrac{2}{3}\), so \(|r|=\dfrac{2}{3}<1\) ✓

Sum: \(\dfrac{1}{1-\frac{2}{3}}=\dfrac{1}{\frac{1}{3}}=3\)

Ratio Test

Use the Ratio Test to determine the convergence of \(\displaystyle\sum_{n=1}^{\infty}\frac{n!}{n^n}\).

Step 1: Let \(a_n = n!/n^n\). Compute \(\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{(n+1)!}{(n+1)^{n+1}}\cdot\dfrac{n^n}{n!} = \dfrac{n^n}{(n+1)^n}\)

Step 2: \(= \left(\dfrac{n}{n+1}\right)^n = \left(1-\dfrac{1}{n+1}\right)^n \to e^{-1} = \dfrac{1}{e} \approx 0.368\)

Conclusion: Since \(\dfrac{1}{e} < 1\), the series converges by the Ratio Test.

Alternating Series

Which statement is correct about \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} = 1-\frac{1}{2}+\frac{1}{3}-\cdots\)?

AST Check: \(b_n = 1/n\) is positive, decreasing, and \(b_n\to 0\). So the series converges by the Alternating Series Test.

Absolute? \(\sum 1/n\) is the harmonic series — it diverges. So convergence is conditional, not absolute.

Value: This is the Taylor series for \(\ln(1+x)\) at \(x=1\): \(\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n} = \ln 2\)

Power Series — Radius of Convergence

Find the radius of convergence of \(\displaystyle\sum_{n=1}^{\infty}\frac{(x-2)^n}{n\cdot 3^n}\).

Ratio Test: Let \(a_n = \dfrac{(x-2)^n}{n\cdot 3^n}\). Then:

\(\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{(x-2)^{n+1}}{(n+1)3^{n+1}}\cdot\frac{n\cdot3^n}{(x-2)^n}\right|=\frac{n}{n+1}\cdot\frac{|x-2|}{3}\to\frac{|x-2|}{3}\)

Converges when: \(\dfrac{|x-2|}{3}<1\), i.e., \(|x-2|<3\), so \(R=3\)

Taylor Series

The Maclaurin series for \(e^{-x^2}\) begins:

Start with: \(e^u = 1 + u + \dfrac{u^2}{2!} + \dfrac{u^3}{3!} + \cdots\)

Substitute \(u = -x^2\):

\(e^{-x^2} = 1 + (-x^2) + \dfrac{(-x^2)^2}{2!} + \dfrac{(-x^2)^3}{3!} + \cdots = 1 - x^2 + \dfrac{x^4}{2} - \dfrac{x^6}{6} + \cdots\)

p-Series

Which of the following series converges?

p-Series rule: \(\sum 1/n^p\) converges iff \(p > 1\).

A: \(p=1\) — harmonic series — DIVERGES

B & D: \(p=1/2<1\) — DIVERGES

C: \(p=3/2>1\) — CONVERGES ✓

Trig Integrals

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

Identity: \(\sin^2 x = \dfrac{1-\cos 2x}{2}\)

\(\int_0^{\pi/2}\sin^2 x\,dx = \int_0^{\pi/2}\dfrac{1-\cos 2x}{2}dx = \left[\dfrac{x}{2}-\dfrac{\sin 2x}{4}\right]_0^{\pi/2}\)

\(= \left(\dfrac{\pi/2}{2}-\dfrac{\sin\pi}{4}\right) - 0 = \dfrac{\pi}{4} - 0 = \dfrac{\pi}{4}\)

Volume of Revolution

Find the volume of the solid obtained by rotating the region bounded by \(y = \sqrt{x}\), \(x = 4\), and \(y = 0\) about the \(x\)-axis.

Disk Method: \(V = \pi\int_0^4 [f(x)]^2\,dx = \pi\int_0^4 x\,dx\)

\(= \pi\left[\dfrac{x^2}{2}\right]_0^4 = \pi\cdot\dfrac{16}{2} = 8\pi\)

Polar Area

Find the area enclosed by the polar curve \(r = 2\cos\theta\).

Recognize: \(r=2\cos\theta\) is a circle of radius 1 centered at \((1,0)\). Area = \(\pi(1)^2=\pi\).

By formula (full circle from \(-\pi/2\) to \(\pi/2\)):

\(A=\dfrac{1}{2}\int_{-\pi/2}^{\pi/2}(2\cos\theta)^2d\theta = 2\int_{-\pi/2}^{\pi/2}\cos^2\theta\,d\theta = 2\cdot\dfrac{\pi}{2}=\pi\)

Arc Length (Parametric)

Find the arc length of the parametric curve \(x = t^2\), \(y = t^3\) for \(0 \le t \le 1\). What is the integrand (before integration)?

Derivatives: \(dx/dt = 2t\), \(dy/dt = 3t^2\)

Integrand: \(\sqrt{(dx/dt)^2+(dy/dt)^2}=\sqrt{4t^2+9t^4}=\sqrt{t^2(4+9t^2)}=t\sqrt{4+9t^2}\) (since \(t\ge0\))

Both A and B are equivalent — but the simplified form is \(t\sqrt{4+9t^2}\) (option A), which is the standard simplified integrand.

L'Hôpital's Rule & Limits

Evaluate \(\displaystyle\lim_{x\to 0}\frac{\sin x - x}{x^3}\).

Taylor series: \(\sin x = x - \dfrac{x^3}{6} + \dfrac{x^5}{120} - \cdots\)

\(\sin x - x = -\dfrac{x^3}{6} + O(x^5)\)

\(\dfrac{\sin x - x}{x^3} = -\dfrac{1}{6} + O(x^2) \to -\dfrac{1}{6}\) as \(x\to0\)

Via L'Hôpital (3 times): \(\dfrac{\sin x - x}{x^3}\xrightarrow{0/0}\dfrac{\cos x-1}{3x^2}\xrightarrow{0/0}\dfrac{-\sin x}{6x}\xrightarrow{0/0}\dfrac{-\cos x}{6}\to-\dfrac{1}{6}\) ✓

Shell Method

Use the shell method to find the volume of the solid generated by rotating the region bounded by \(y = x^2\), \(y = 0\), \(x = 2\) about the \(y\)-axis.

Shell Method: \(V = 2\pi\int_0^2 x\cdot f(x)\,dx = 2\pi\int_0^2 x\cdot x^2\,dx = 2\pi\int_0^2 x^3\,dx\)

\(= 2\pi\left[\dfrac{x^4}{4}\right]_0^2 = 2\pi\cdot\dfrac{16}{4} = 2\pi\cdot 4 = 8\pi\)

Convergence Tests

Which test is most direct for \(\displaystyle\sum_{n=1}^{\infty}\frac{n}{n^3+1}\)? What is the conclusion?

Limit Comparison with \(b_n = 1/n^2\):

\(\lim_{n\to\infty}\dfrac{a_n}{b_n}=\lim_{n\to\infty}\dfrac{n}{n^3+1}\cdot n^2=\lim_{n\to\infty}\dfrac{n^3}{n^3+1}=1>0\)

Since \(\sum 1/n^2\) converges (p-series, \(p=2>1\)) and the limit is finite and positive, \(\sum\dfrac{n}{n^3+1}\) also converges.

Integral Test

Using the Integral Test, determine whether \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2+1}\) converges or diverges.

Check conditions: \(f(x)=1/(x^2+1)\) is positive, continuous, and decreasing for \(x\ge1\). ✓

\(\int_1^\infty\dfrac{dx}{x^2+1}=\lim_{t\to\infty}\left[\tan^{-1}x\right]_1^t=\dfrac{\pi}{2}-\dfrac{\pi}{4}=\dfrac{\pi}{4}\)

Since the integral converges, the series also converges (Integral Test).

Surface Area of Revolution

Set up the integral for the surface area generated by rotating \(y = x^3\) on \([0,1]\) about the \(x\)-axis.

Formula: \(S=2\pi\int_a^b y\sqrt{1+(y')^2}\,dx\)

Here: \(y = x^3\), \(y' = 3x^2\), so \((y')^2 = 9x^4\)

\(S = 2\pi\int_0^1 x^3\sqrt{1+9x^4}\,dx\)

Integration — Substitution

Evaluate \(\displaystyle\int_0^1\frac{x\,dx}{\sqrt{1-x^4}}\).

Substitution: Let \(u = x^2\), so \(du = 2x\,dx\), i.e., \(x\,dx = \frac{1}{2}du\)

New limits: \(x=0\Rightarrow u=0\); \(x=1\Rightarrow u=1\)

\(\int_0^1\dfrac{x\,dx}{\sqrt{1-x^4}}=\dfrac{1}{2}\int_0^1\dfrac{du}{\sqrt{1-u^2}}=\dfrac{1}{2}\left[\sin^{-1}u\right]_0^1=\dfrac{1}{2}\cdot\dfrac{\pi}{2}=\dfrac{\pi}{4}\)

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Calculus II — 20 Multiple Choice Questions