Placement Exam C Solutions

1. Evaluate $\int \frac{1}{\sqrt{x^{2}+2}} \mathrm{~d} x$.

   Solution: $$\begin{aligned} \int \frac{1}{\sqrt{x^{2}+2}} d x & =\int \frac{\sqrt{2} \sec ^{2} \theta}{\sqrt{2+2 \tan ^{2} \theta}} d \theta \quad\left(x=\sqrt{2} \tan \theta,-\frac{\pi}{2}<\theta<\frac{\pi}{2} ; d x=\sqrt{2} \sec ^{2} \theta d \theta\right) \\ & =\int \sec \theta d \theta+C \\ & =\ln |\sec \theta+\tan \theta|+C \\ & =\ln \left|\frac{\sqrt{x^{2}+2}}{\sqrt{2}}+\frac{x}{\sqrt{2}}\right|+C \\ & =\ln \left|\sqrt{x^{2}+2}+x\right|+C_{1} \end{aligned}$$

2. Evaluate $\int \frac{16}{x^{3}-4 x} \mathrm{~d} x$.

   Solution: $$\begin{aligned} \int \frac{16}{x^{3}-4 x} d x & =\int\left(\frac{-4}{x}+\frac{2}{x+2}+\frac{2}{x-2}\right) d x \\ & =-4 \ln |x|+2 \ln |x+2|+2 \ln |x-2|+C \end{aligned}$$

3. The region under the graph of $y=\sin x, 0 \leq x \leq \pi$, is rotated 360 degrees about the $y$-axis forming a solid of revolution $S$. Find the volume of $S$.

   Solution: Via the method of cylindrical shells, we obtain the following integral yielding the volume of $S$: $$2 \pi \int_{0}^{\pi} x \sin (x) d x$$ Integrating by parts with $u=x, d v=\sin x d x, d u=d x$, and $v=-\cos x$, we obtain $$\int x \sin x d x=-x \cos x+\int \cos x d x=-x \cos x+\sin x+C$$ Thus, the volume of $S$ is $$2 \pi \int_{0}^{\pi} x \sin (x) d x=2 \pi[-x \cos x+\sin x]_{0}^{\pi}=2 \pi^{2}.$$

4. Solve the differential equation, obtaining an explicit solution: $$\frac{d y}{d x}=y^{2} x^{2}+y^{2} .$$

   Solution: Separating variables and integrating, we obtain $$\int \frac{1}{y^{2}} d y=\int\left(1+x^{2}\right) d x,$$ which yields an implicit general solution $$-\frac{1}{y}=x+\frac{x^{3}}{3}+C.$$ Solving for $y$, we obtain an explicit general solution $$y=-\frac{1}{x+\frac{x^{3}}{3}+C}.$$

5. Show that the series $$\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{2}}$$ converges.

   Solution: The series may be written $\sum_{n=2}^{\infty} f(n)$, where $f(x)=\frac{1}{x(\ln x)^{2}}$. Note $$\begin{aligned} \int_{2}^{\infty} f(x) d x & =\lim _{t \rightarrow \infty} \int_{2}^{t}(\ln x)^{-2} \frac{1}{x} d x \\ & =\lim _{t \rightarrow \infty}\left[\frac{-1}{\ln x}\right]_{2}^{t} \\ & =\lim _{t \rightarrow \infty}\left(\frac{-1}{\ln t}+\frac{1}{\ln 2}\right) \\ & =\frac{1}{\ln 2} \end{aligned}$$ Since $f$ is continuous, positive, and decreasing on $[2, \infty)$, and, by the preceding computation, the improper integral $\int_{2}^{\infty} f(x) d x$ converges, we conclude the infinite series $\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{2}}$ converges by the Integral Test.

   Remark 1. When grading this problem, consider your solution “correct” provided you found the value of the improper integral $\int_{2}^{\infty} f(x) d x$ to be $1/(\ln 2)$ and concluded that the series converges by the Integral Test.

6. Find the Maclaurin Series for $f(x)=\frac{x}{3+x^{2}}$. What is the radius of convergence?

   Solution: $$\begin{aligned} f(x) & =\frac{x}{3+x^{2}} \\ & =\frac{x}{3} \frac{1}{1+\frac{x^{2}}{3}} \\ & =\frac{x}{3} \sum_{n=0}^{\infty}(-1)^{n}\left(\frac{x^{2}}{3}\right)^{n} \quad\left(\text { for } x \text { satisfying }\left|\frac{x^{2}}{3}\right|<1\right) \\ & =\sum_{n=0}^{\infty} \frac{(-1)^{n}}{3^{n+1}} x^{2 n+1} \end{aligned}$$ Thus, the Maclaurin series of $f$ is $\sum_{n=0}^{\infty} \frac{(-1)^{n}}{3^{n+1}} x^{2 n+1}$ and, since it converges for precisely those $x$ satisfying $$|x|<\sqrt{3},$$ its radius of convergence is $\sqrt{3}$.

7. Determine the radius and interval of convergence of the power series $\sum_{n=1}^{\infty} \frac{1}{n 2^{n}}(x-1)^{n}$.

   Solution: We employ the Ratio Test. $$\begin{aligned} \lim _{n \rightarrow \infty}\left|\frac{\frac{1}{(n+1) 2^{n+1}}(x-1)^{n+1}}{\frac{1}{n 2^{n}}(x-1)^{n}}\right| & =|x-1| \lim _{n \rightarrow \infty} \frac{n}{2(n+1)} \\ & =|x-1| \cdot \frac{1}{2} \end{aligned}$$ Thus, the series converges (absolutely) for those $x$ satisfying $\frac{1}{2}|x-1|<1$, that is, for $-1<x<3$. It diverges for $|x-1|>2$. Thus, the radius of convergence of the power series is $2$ and its interval of convergence has endpoints $-1$ and $3$. At the left endpoint $x=-1$, we obtain $$\sum_{n=1}^{\infty} \frac{1}{n 2^{n}}(-2)^{n}=\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n},$$ which converges by the Alternating-Series Test (because the sequence $(1/n)$ decreases to $0$). At the right endpoint $x=3$, we obtain $$\sum_{n=1}^{\infty} \frac{1}{n 2^{n}}(2)^{n}=\sum_{n=1}^{\infty} \frac{1}{n},$$ which is the divergent harmonic series. Hence, the interval of convergence of the given power series is $[-1,3)$.

8. Suppose that $(x, y)$ has distance $r \geq 1$ from the origin. Use polar coordinates to show that $$(|x|+|y|) \ln \left(x^{2}+y^{2}\right) \leq 4 r \ln (r).$$

   Solution: Suppose that $(x, y)$ has distance $r \geq 1$ from the origin. Choose $\theta$ such that $x=r \cos \theta$ and $y=r \sin \theta$; then, we have $$\begin{aligned} (|x|+|y|) \ln \left(x^{2}+y^{2}\right) & =(|r \cos \theta|+|r \sin \theta|) \ln \left(r^{2}\left(\cos ^{2}(\theta)+\sin ^{2}(\theta)\right)\right) \\ & =(|r||\cos \theta|+|r||\sin \theta|) \ln \left(r^{2}\right) \\ & \leq(|r|+|r|) \ln \left(r^{2}\right) \quad\left(\ln \left(r^{2}\right) \geq 0 \text { because } r^{2} \geq 1\right) \\ & =4|r| \ln (r) \quad \text { (laws of logarithms). } \end{aligned}$$

9. Find an equation of the line tangent to the curve $x=1+\ln t, y=t^{2}+2$ at the point $(1,3)$.

   Solution: The slope of the tangent line is given by $\frac{y^{\prime}(1)}{x^{\prime}(1)}=\frac{2}{1}$, and thus, an equation of the tangent line is $y-3=2(x-1)$. Alternatively, one might observe that $t=e^{x-1}$, so that $y=e^{2 x-2}+2$. Thus, $\frac{d y}{d x}=2 e^{2 x-2}$, yielding the tangent-line slope $\left.\frac{d y}{d x}\right|_{x=1}=2$.

10. Find the length of the curve $x=e^{t}-t, y=4 e^{t/2}, 0 \leq t \leq 2$.

    Solution: The length of the curve is given by $$\begin{aligned} \int_{0}^{2} \sqrt{x^{\prime}(t)^{2}+y^{\prime}(t)^{2}} d t & =\int_{0}^{2} \sqrt{\left(e^{t}-1\right)^{2}+\left(2 e^{t / 2}\right)^{2}} d t \\ & =\int_{0}^{2} \sqrt{e^{2 t}-2 e^{t}+1+4 e^{t}} d t \\ & =\int_{0}^{2} \sqrt{\left(e^{t}+1\right)^{2}} d t \\ & =\int_{0}^{2}\left(e^{t}+1\right) d t \\ & =\left[e^{t}+t\right]_{0}^{2} \\ & =e^{2}+2-1 \\ & =e^{2}+1. \end{aligned}$$