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\begin{document}
\begin{center}

\textbf{College of the Holy Cross, Spring 2018} \\
\textbf{Math 134 -- Solutions for Midterm Exam 2} \\
\textbf{Friday, March 16}
\end{center}

\vskip 10pt
\noindent
I.  All parts of this problem refer to the region $R$ bounded by $y = x$, $y = x^2 + 1$, $x = 0$ and $x = 2$.  
\begin{enumerate}
\item[A.] (10)  Sketch the region $R$.  
\item[] {\it Solution:}  See Figure 1.

\begin{figure}
\begin{center}
\includegraphics[width=.3\linewidth]{IRegion.png}
\end{center}
\caption{The region for problem I.}
\end{figure}

\item[B.] (10) Compute the area of the region $R$.   
\item[] {\it Solution:}  The area is computed by 
\begin{align*}
{\rm Area} &= \int_0^2 x^2 + 1 - x\ dx\\
&= \left. \frac{x^3}{3} + x - \frac{x^2}{2}\right|_0^2\\
&= \frac{8}{3} + 2 - 2 - 0\\
&= \frac{8}{3}.
\end{align*}
\item[C.] (10) Compute the volume of the solid of revolution obtained by rotating the region $R$ about 
the $x$-axis.  
\item[] {\it Solution:}  The cross-sections by planes $x = $ constant are washers with 
inner radius $x$ and outer radius $x^2 + 1$, so the volume is 
\begin{align*} 
{\rm Volume} &= \int_0^2 \pi(x^2+1)^2 - \pi x^2\ dx\\
&= \pi \int_0^2 x^4 + x^2 + 1\ dx\\
&= \pi \left(\frac{x^5}{5} + \frac{x^3}{3} + x\right|_0^2\\
&= \pi \left(\frac{32}{5} + \frac{8}{3} + 2 - 0\right)\\
&= \frac{(96 + 40 + 30)\pi}{15}\\
&= \frac{166\pi}{15}.
\end{align*}
\item[D.] (10) Compute the volume of the solid of revolution obtained by rotating the region $R$ about
the line $x = -2$.  (Note: Any correct method is OK here.)
\item[] {\it Solution:}  The easier way to do this is to use cylindrical shells:
\begin{align*}
{\rm Volume} &= \int_0^2 2\pi (x + 2)(x^2 + 1 - x)\ dx\\
&= 2\pi \int_0^2 x^3 + x^2 - x + 2\ dx\\
&= 2\pi \left(\frac{x^4}{4} + \frac{x^3}{3} - \frac{x^2}{2} + 2x\right|_0^2\\
&= 2\pi \left( 4 + \frac{8}{3} - 2 + 4 -0\right)\\
&= \frac{52\pi}{3}.
\end{align*}
This can also be done by slices, but the inner or outer radii of the washer
cross-sections \emph{change} at $y = 1$, $y = 2$, so the volume would be computed 
as the sum of three integrals like this:
\begin{align*}{\rm Volume} &= \int_{y=0}^{y=1} \pi (2 + y)^2 - \pi(2)^2\ dy + \int_{y=1}^{y=2} \pi (2 + y)^2 - \pi(2 + \sqrt{y - 1})^2\ dy\\ 
&+ \int_{y=2}^{y=5} \pi(2 + 2)^2 - \pi(2 + \sqrt{y - 1})^2\ dy.
\end{align*}
It's a good exercise to verify that this gives the same value as the shell method, but there's a 
lot of algebra involved(!)
\end{enumerate}
\noindent
II.  Compute each of the following integrals by an appropriate method (e.g. $u$-substitution, parts, etc.)
\begin{enumerate}
\item[A.] (15) $\displaystyle\int x^2 \cos(5x^3)\ dx$
\item[] {\it Solution:}  This one is one where you just want a $u$-substitution.
Let $u = 5x^3$, then $du = 15x^2\ dx$ and the integral becomes
$$\int \cos(u) \frac{1}{15}\ du = \frac{1}{15} \sin(u) + C = \frac{1}{15}\sin(5x^3) + C.$$
\item[B.] (15) $\displaystyle\int x^2 \cos(5x) \ dx$
\item[] {\it Solution:}  For this one, we use parts (twice), letting $u = x^2$ the first time and $dv = \cos(5x)\ dx$.  Then $du = 2x\ dx$
and $v = \frac{1}{5}\sin(5x)$.  So 
$$\int x^2\cos(5x)\ dx = \frac{x^2\sin(5x)}{5} - \frac{2}{5} \int x\sin(5x)\ dx.$$
Now use parts again with $u = x$ and $dv = \sin(5x)$; the final answer is
$$\int x^2\cos(5x)\ dx = \frac{x^2\sin(5x)}{5} + \frac{2x\cos(5x)}{25} - \frac{2}{125}\sin(5x) + C.$$
\item[C.]  (15) $\displaystyle \int e^{\sqrt{x}}\ dx$  (Hint: Let $u = x^{1/2}$.)
\item[] {\it Solution:} If we do the indicated substitution first we get
$du = \frac{1}{2} x^{-1/2}\ dx$ so $dx = 2 x^{1/2} du = 2u du$.  Rewriting the given integral
in terms of $u$, we get $\int 2u e^u\ du$, which is a simple parts form:
$$2\int u e^u\ du = 2u e^u - 2e^u + C = 2\sqrt{x} e^{\sqrt{x}} - 2e^{\sqrt{x}} + C$$ 
\end{enumerate}

\noindent
III. 
\begin{enumerate}
\item[A.] (10) Identify a $u$ and a $dv$, then carry out an integration by parts to derive the
following reduction formula for exponents $k \ge 1$.  
$$\int (\ln(x))^k\ dx = x(\ln(x))^k - k \int (\ln(x))^{k-1}\ dx$$
\item[] {\it Solution:}  The appropriate choice is $u = (\ln(x))^k$ and $dv = dx$.  Then $du = k (\ln(x))^{k-1}\frac{1}{x}\ dx$ (chain rule), 
and $v = x$.  Applying the parts formula,
$$\int (\ln(x))^k\ dx = x(\ln(x))^k - k \int (\ln(x))^{k-1}\frac{1}{x} x\ dx$$
The $\frac{1}{x}$ and the $x$ cancel leaving the given form.  
\item[B.]  (5) Apply the formula in part A to integrate $\displaystyle \int (\ln(x))^2\ dx$.  (Note:  You 
can do this part even if you didn't see how to finish part A.)
\item[] {\it Solution:}  We apply the formula first with $k = 2$, then again with $k = 1$:
\begin{align*}
\int (\ln(x))^2\ dx &= x(\ln(x))^2 - 2\int \ln(x)\ dx\\
&= x(\ln(x))^2 - 2\left(x\ln(x) - \int\ dx\right)\\
&= x(\ln(x))^2 - 2x\ln(x) + 2x + C.
\end{align*}
\end{enumerate}
\end{document}

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\MyAnswer{Integral $ = $}{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}%
\end{enumerate}
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