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\begin{document}
\begin{center}

\textbf{College of the Holy Cross, Fall 2016} \\
\textbf{Math 136, section 2 -- Makeup Midterm Exam 2} \\
\textbf{Monday, October 31}
\end{center}

\vfil

\noindent\textbf{Your Name}: $\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}$



\vskip 30pt
\noindent\textbf{Instructions}:  For full credit, you must show 
{\it all work} on the test pages and place your final answer
in the box provided for the problem.  Use the back of the
preceding page if you need more space for scratch work.  The numbers
next to each part of the questions are their point values.  I will
be liberal with partial credit, so even if you don't see how to complete
a problem, do as much as you can and be sure to write what you know.

\vfil

\noindent\textbf{Please do not write in the space below}

\begin{center}
\begin{tabular}{|l||r|}
\hline
Problem & Points/Poss \\
\hline\hline
I  & \hfil/ 60\  \hfil \\
\hline
II  & \hfil/ 30\  \hfil \\
\hline
III  & \hfil/ 10\  \hfil \\
\hline\hline
Total   & \hfil/100\hfil \\
\hline
\end{tabular}
\end{center}

\vfil\eject
I.  
\begin{enumerate}
\item[A.] (15)  Integrate by parts:
${\displaystyle \int x^2 \cos(3x)\ dx}$
\vskip 7.0truein
\MyAnswer{Answer:}{$\phantom{A very very very very very very very long answer here}$}%
\vfill\eject
\item[B.] (15)  Integrate using appropriate trigonometric identities:
$\displaystyle \int \sin^4(2x)\cos^5(2x)\ dx$.
\vskip 8.0truein
\MyAnswer{Answer:}{$\phantom{A very very very very very very very long answer here}$}%
\vfill\eject
\item[C.] (20) Integrate with a trigonometric substitution:
${\displaystyle \int \frac{x^3}{\sqrt{16 + x^2}}\ dx}$.
(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 $x$.)

\vskip 7.0truein
\MyAnswer{Answer:}{$\phantom{A very very very very very very very long answer here}$}%
\vfill\eject

\item[D.] (10)  Integrate with any applicable method we have discussed:
${\displaystyle \int x^3 \ln(2x)\ dx}$

\vskip 7.0truein
\MyAnswer{Answer:}{$\phantom{A very very very very very very very long answer here}$}%

\end{enumerate}
\vfill\eject
\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 $R$.
\vskip 4.0truein
\MyAnswer{Area $=$ }{$\phantom{A very long answer here}$}%
\vfill\eject
\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.
\vskip 3.0truein
\MyAnswer{Volume $=$ }{$\phantom{A very long answer here}$}%
\item[C.] (10)  A solid has the portion of the region $R$ between $x = 0$ and $x = 1$
as base.  The cross-sections by planes perpendicular to the $x$-axis are equilateral
triangles with side extending from the lower boundary to the upper boundary
of the region.  {\it Set up} an integral to find the volume. You do not need to compute this 
one, but if you do correctly, I will give 10 points Extra Credit.
\vskip 3.0truein
\MyAnswer{Volume $=$ }{$\phantom{A very long answer here}$}%
\end{enumerate}
\vfill\eject
\noindent
III. (10) 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 7.0truein 
\MyAnswer{Volume $V_2 =$ }{$\phantom{A very long answer here}$}%
\end{document}
