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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 Retest} \\
\textbf{Tuesday, November 8}
\end{center}
\vskip 10pt
\noindent
I.  
\begin{enumerate}
\item[A.] (15)  Integrate by parts: 
${\displaystyle \int \tan^{-1}(x)\ dx}$.  (Note there's really only one choice for $u$ and hence $dv$.)
\item[] {\it Solution:}  The only possible choice that leads to anything of value is
$u = \tan^{-1}(x)$ and $dv = dx$, so $du = \frac{1}{x^2 + 1}\ dx$ and $v = x$.  Then
$$\int\tan^{-1}(x)\ dx = x\tan^{-1}(x) - \int\frac{x}{x^2 + 1}\ dx.$$
This remaining integral can be done via the substitution $w = x^2 + 1$, which takes
it to $-\frac{1}{2} \int \frac{dw}{w} = -\frac{1}{2}\ln|x^2 + 1|$.  The full answer
is 
$$x\tan^{-1}(x) - \frac{1}{2}\ln|x^2+1| + C.$$
\item[B.] (15)  Integrate using appropriate trigonometric identities:
$\displaystyle \int \sin^3(6x)\cos^4(6x)\ dx$.
\item[] {\it Solution:}  With the odd power of sine, we split off one factor of $\sin(6x)$ and
use $\sin^2(6x) = 1 - \cos^2(6x)$ to convert the rest:
\begin{eqnarray*}
\int \sin^3(6x)\cos^4(6x)\ dx &=& \int \sin(6x) (1 - \cos^2(6x))\cos^4(6x)\ dx\\
&=& \int \cos^4(6x)\sin(6x)\ dx - \int \cos^6(6x)\sin(6x)\ dx \qquad (u = \cos(6x))\\
&=& -\frac{1}{6} \int u^4\ du + \frac{1}{6} \int u^6\ du\\
&=& -\frac{1}{30} u^5 + \frac{1}{42} u^7 + C\\
&=& -\frac{1}{30} \cos^5(6x) + \frac{1}{42} \cos^7(6x) + C.
\end{eqnarray*}
\item[C.] (20) Integrate with a trigonometric substitution:
${\displaystyle \int \sqrt{25 - x^2}\ dx}$.
(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 $x = 5\sin(\theta)$, then $dx = 5\cos(\theta)\ d\theta$.
The integral substitutes to 
$$\int\sqrt{25 - 25\sin^2(\theta)}\cdot 5\cos(\theta)\ d\theta = 
25\int \cos^2(\theta)\ d\theta.$$
For this we use the identity $\cos^2(\theta) = \frac{1}{2}(1 + \cos(2\theta))$, 
so the integral becomes
$$25\int \frac{1}{2}(1 + \cos(2\theta))\ d\theta = \frac{25\theta}{2} + \frac{25}{4}\sin(2\theta) + C,$$
which can also be rewritten using the double angle formula for the sine: 
$$\frac{25\theta}{2} + \frac{25}{2}\sin(\theta)\cos(\theta) + C.$$
Converting back to functions of $x$, we have $\frac{x}{5} = \sin(\theta)$, so $\theta = 
\sin^{-1}\left(\frac{x}{5}\right)$ 
and $\cos(\theta) = \frac{\sqrt{25 - x^2}}{5}$.  So after simplifying we get
$$\frac{25}{2}\sin^{-1}\left(\frac{x}{5}\right) + \frac{x}{2} \sqrt{25 - x^2} + C.$$

\item[D.] (5)  Derive the reduction formula (assume $m \ge 2$ -- this one is \emph{not}
done by parts): 
$$\int \tan^m(x)\ dx = \frac{\tan^{m-1}(x)}{m - 1} - \int \tan^{m-2}(x)\ dx.$$
\item[] {\it Solution:}  (This same problem appeared on Problem Set 5 part B.)
Break off 2 powers of $\tan(x)$ and convert that to $\sec^2(x) - 1$:
$$\int \tan^m(x)\ dx = \int \tan^{m-2}(x)(\sec^2(x) - 1)\ dx = 
\int \tan^{m-2}(x) \sec^2(x)\ dx - \int \tan^{m-2}(x)\ dx.$$
The first integral is $\int u^{m-2} du$ for $u = \tan(x)$ so we get 
$$= \frac{\tan^{m-1}(x)}{m-1} - \int \tan^{m-2}(x)\ dx,$$
which is what we wanted to show.

\item[E.] (5)  Use the formula from part D repeatedly to integrate ${\displaystyle \int \tan^4(x)\ dx}$

\item[] {\it Solution:}  Applying the formula from part D twice, with $m = 4$, then $m = 2$, we get:
\begin{eqnarray*}
\int \tan^4(x)\ dx &=& \frac{\tan^3(x)}{3} - \int \tan^2(x)\ dx\\
&=&  \frac{\tan^3(x)}{3} - \tan(x) + \int 1\ dx\\
&=&  \frac{\tan^3(x)}{3} - \tan(x) + x + C
\end{eqnarray*}
\end{enumerate}
\noindent
II.  All parts of this question refer to the region $R$ bounded by 
$y = xe^{-x}$ (in red), the $x$-axis, $x = 0$ and $x = 2$.  
\begin{center}
\includegraphics[width=2.0truein,height=2.0truein]{136RP2II.png} 
\end{center}
\begin{enumerate}
\item[A.] (10)  Set up and compute integral(s) to find the area of $R$.
\item[] {\it Solution:}  The area is 
$$A = \int_0^2 x e^{-x}\ dx.$$
Integrating by parts ($u = x, dv = e^{-x}$) we get
$$ \left.-x e^{-x} - e^{-x}\right|_0^2 =  1-3e^{-2} \doteq .594$$ 
\item[B.] (10)  The region $R$ 
is rotated about the $x$-axis to generate
a solid.  Set up and compute an integral to find its volume.
\item[] {\it Solution:}  The volume has disk cross-sections so 
$$V = \int_0^2 \pi (x e^{-x})^2\ dx = \pi \int_0^2 x^2 e^{-2x}\ dx$$
So we integrate by parts again (\emph{twice} this time with $u = x^2$, then $u = x$).
The final answer is 
$$V = \frac{\pi}{4}(1 - 13 e^{-4}).$$

\item[C.] (10)  A solid has the portion of the region $R$ 
as base.  The cross-sections by planes perpendicular to the $x$-axis are equilateral
triangles extending from the $x$-axis up to the point on the graph $y = xe^{-x}$.  
{\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.
\item[] {\it Solution:}  The area of an equilateral triangle of side $s$ is 
$\frac{s^2\sqrt{3}}{4}$, so the integral is
$$\int_0^2 \frac{\sqrt{3}}{4} (x e^{-x})^2\ dx.$$
If you're clever, you'll note that this is exactly the same integral as in part
C, up to the constant factor $\frac{\sqrt{3}}{4\pi}$.  So this volume is equal to 
$$\frac{\sqrt{3}}{16}(1 - 13e^{-4})$$
with no extra computation(!)
\end{enumerate}
\vskip 10pt
\noindent
III. (10) The area of a region in the plane is equally distributed about 
a point called its \emph{centroid}.  For a region $R$ bounded by $y = f(x) \ge 0$, the
$x$-axis, $x = a$ and $x = b$, take it as known that the $y$-coordinate of the centroid is 
computed by this ratio of two integrals:
$$\overline{y} = \frac{\int_a^b \frac{1}{2} (f(x))^2\ dx}{\int_a^b f(x)\ dx}.$$
Explain how this result shows that {\it The volume of the solid generated by rotating $R$
about the $x$-axis is equal to the product of the area of $R$ and the distance traveled by
the centroid as $R$ moves around the $x$-axis.}\footnote{This is a famous result of the 
ancient Greek mathematician Pappus of Alexandria who lived ca. 290 - 350 CE. He proved it
without integral calculus, which had not yet been invented!}
\vskip 10pt
\noindent
{\it Solution:}  The centroid travels around a circle of radius $\overline{y}$, so the 
distance it travels is the circumference, or
 $2\pi\overline{y}$.  The area of the region is $\int_a^b f(x)\ dx$.
Therefore the product of the distance traveled by the centroid and the area is:
$$2\pi\overline{y}\cdot \int_a^b f(x)\ dx$$
We can substitute from the given formula for $\overline{y}$ to get
$$2\pi\overline{y} \cdot \int_a^b f(x)\ dx = 
2\pi \frac{\int_a^b \frac{1}{2} (f(x))^2\ dx}{\int_a^b f(x)\ dx}\cdot \int_a^b f(x)\ dx = 
\int_a^b \pi (f(x))^2\ dx.$$
We know this is the volume of the solid of revolution by Cavalieri's Principle.
\end{document}
