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\begin{document}
\begin{center}
\textbf{College of the Holy Cross, Fall 2016} \\
\textbf{Math 136, section 2 -- Solutions for Midterm Exam 2} \\
\textbf{Friday, October 28}
\end{center}

\vskip 10pt
I.  
\begin{enumerate}
\item[A.] (15)  Integrate by parts:
${\displaystyle \int x^2 \sin(5x)\ dx}$
\item[] {\it Solution:}  We integrate by parts twice, each time letting $u = $ power of $x$:
\begin{eqnarray*}
\int x^2 \sin(5x)\ dx &=& \frac{-x^2 \cos(5x)}{5} + \frac{2}{5} \int x \cos(5x)\ dx\\
  &=& \frac{-x^2 \cos(5x)}{5} + \frac{2}{5} \left(\frac{x\sin(5x)}{5} - \frac{1}{5}\int \sin(5x)\ dx\right)\\
&=& \frac{-x^2\cos(5x)}{5} + \frac{2x \sin(5x)}{25} + \frac{2\cos(5x)}{125} + C.
\end{eqnarray*}
\item[B.] (10)  Integrate using appropriate trigonometric identities:
$\displaystyle \int \sec^4(4x)\tan^2(4x)\ dx$.
\item[] {\it Solution:}  With the even power of $\sec$ we can proceed like this:
$$= \int \sec^2(4x) (1 + \tan^2(4x))\tan^2(4x)\ dx = \int \tan^2(4x)\sec^2(4x)\ dx + \int\tan^4(4x)\sec^2(4x)\ dx.$$
Each of these integrals can be handled with the substitution $u = \tan(4x)$, $du = 4\sec^2(4x)\ dx$.
The integral is
$$\frac{1}{12}\tan^3(4x) + \frac{1}{20}\tan^5(4x) + C.$$
\item[C.] (20) Integrate with a trigonometric substitution:
${\displaystyle \int \frac{x^3}{\sqrt{81 - x^2}}\ dx}$.
(Partial credit points will be given as follows:  5 points for the substitution equation $x = \ldots$,
and the computation of $dx$, 5 points for the conversion to the trigonometric integral, 
5 points for the integration of the trigonometric integral, 5 points for conversion back to 
the equivalent function of $x$.)

\item[] {\it Solution:}  Let $x = 9\sin\theta$, so $dx = 9\cos\theta\ d\theta$. The
integral goes over to 
$$\int \frac{729 \sin^3\theta}{9\cos\theta}\cdot 9\cos\theta\ d\theta = 729\int \sin^3\theta d\theta.$$
Since we have the odd power of $\sin$, we can split off one power, convert the other powers
to cosines, then integrate with a $u$-substitution:
$$= 729 \int (1 - \cos^2\theta)\sin\theta\ d\theta = 
729\int\sin\theta\ d\theta - 729\int \cos^2\theta\sin\theta\ d\theta.$$
In the second, we let $u = \cos\theta$ and recognize the form as $\int u^2\ du$.  So the trig
integral equals
$$ -729\cos\theta + 243\cos^3\theta + C$$
Then from the triangle corresponding to the original $x = 9 \sin\theta$, we have
$\cos\theta = \frac{\sqrt{81 - x^2}}{9}$ and the integral expressed in terms of $x$
is
$$-81\sqrt{81 - x^2} + \frac{(81 - x^2)^{3/2}}{3} + C$$ 
\item[D.] (10)  Integrate with any applicable method we have discussed:
${\displaystyle \int x^4 \ln(x)\ dx}$

\item[] {\it Solution:}  Use integration by parts, but with $u = \ln(x)$ and $dv = x^4\ dx$.
We get
$$\int x^4 \ln(x)\ dx = \frac{x^5\ln(x)}{5} - \int \frac{1}{x}\cdot \frac{x^5}{5}\ dx
= \frac{x^5\ln(x)}{5} - \frac{x^5}{25} + C.$$

\end{enumerate}
\noindent
II.  All parts of this question refer to the region $R$ 
bounded by the graphs $y = \sqrt{x}$ and $y = x^2$, 
$x = 0$ and $x = 2$: 
\begin{center}
\includegraphics[width=2.0truein,height=2.0truein]{136P2II.png} 
\end{center}
\begin{enumerate}
\item[A.] (10)  Set up and compute integral(s) to find the area of $R$.
\item[] {\it Solution:}  Between $x = 0$ and $x = 1$, the graph $y = \sqrt{x}$ lies
above $y = x^2$, but this switches between $x = 1$ and $x = 2$.  So the area
is computed by 
$$\int_0^1 \sqrt{x} - x^2\ dx + \int_1^2 x^2 - \sqrt{x}\ dx = \left. \frac{2}{3}x^{3/2} - \frac{1}{3} x^3\right|_0^1 +  \left. \frac{1}{3}x^3  - \frac{2}{3} x^{3/2}\right|_1^2 = \frac{10-4\sqrt{2}}{3} \doteq 1.45.$$ 
\item[B.] (10)  The portion of the region $R$ between $x = 0$ and $x = 1$
is rotated about the $x$-axis to generate
a solid.  Set up and compute an integral to find its volume.
\item[] {\it Solution:}  The cross-sections by planes perpendicular to the $x$-axis
are washers with inner radius $r_{in} = x^2$ and outer radius $r_{out} = \sqrt{x}$.  
So the volume is
$$\int_0^1 \pi(\sqrt{x})^2 - \pi(x^2)^2\ dx = \pi\left(\left. \frac{x^2}{2} - \frac{x^5}{5}\right|_0^1\right) = \frac{3\pi}{10} \doteq .94$$
\item[C.] (10)  A solid has the portion of the region $R$ between $x = 1$ and $x = 2$
as base.  The cross-sections by planes perpendicular to the $x$-axis are isosceles right 
triangles with hypotenuse extending from the lower boundary to the upper boundary
of the region.  Set up and compute an integral to find the volume.  
\item[] {\it Solution:}  An isosceles right triangle with hypotenuse $h$ has legs of 
length $\frac{h}{\sqrt{2}}$ and area 
$$\frac{1}{2}\cdot\frac{h}{\sqrt{2}}\cdot\frac{h}{\sqrt{2}} = \frac{h^2}{4}.$$
Here the cross-section for a general $1 \le x \le 2$ has $h = x^2 - \sqrt{x}$, so 
our volume is computed by 
$$V = \int_1^2 \frac{(x^2 - \sqrt{x})^2}{4}\ dx$$
The Extra Credit computation -- Continuing from the integral above,
$$  = \frac{1}{4}\int_1^2 x^4 - 2x^{5/2} + x\ dx = 
\frac{1}{4}\left(\left. \frac{x^5}{5} - \frac{4}{7}x^{7/2} + \frac{x^2}{2}\right|_1^2\right) = \frac{579}{280} - 
\frac{8\sqrt{2}}{7}.$$
\end{enumerate}
\noindent
III. (15) Suppose that a region $R$ defined by $0 \le y \le f(x)$ and $a \le x \le b$
has area $A$ and lies above the $x$-axis.  When $R$ is rotated about the $x$-axis it 
sweeps out a solid with volume $V_1$.  When $R$ is rotated about the line $y = -k$, 
where $k > 0$, it sweeps out a solid with volume $V_2$.  Express $V_2$ in terms of 
$V_1, k, A$.  
\vskip 10pt
\noindent
{\it Solution:}  Since $-k < 0$, the cross-sections of the new solid are
washers with outer radius $f(x) + k$ and inner radius $k$.  The volume $V_2$ equals
$$V_2 = \int_a^b \pi(f(x) + k)^2 - \pi k^2\ dx = \int_a^b \pi (f(x))^2\ dx +2k \pi \int_a^b f(x)\ dx$$
The first integral here computes the volume $V_1$ when $R$ is rotated about the $x$-axis
and the second computes the area of the region $R$.  Hence this equals $V_1 + 2 k \pi A$.  
We have
$$V_2 = V_1 + 2 k \pi A.$$
\end{document}
