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\begin{document}
\begin{center}

\textbf{College of the Holy Cross, Spring 2014} \\
\textbf{Math 136, section 1 -- Solutions for Midterm Exam 2} \\
\textbf{Friday, March 28}
\end{center}

\noindent
I.  For these problems, you must show \emph{all work necessary} to justify your answers, 
but you may consult the portions of the table of integrals provided as needed.  If
you use a table entry, identify it by number.  
\begin{enumerate}
\item[A.] (15) Integrate with the partial fraction method:  
${\displaystyle \int \frac{3x + 1}{x^3 + 16x}\ dx}$
\item[] {\it Solution:}  The denominator factors as $x^3 + 16x = x(x^2 + 16)$, so the 
partial fractions will be 
$$\frac{3x + 1}{x^3 + 16x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 16}$$
Clearing denominators and collecting powers of $x$, 
$$3x + 1 = (A + B)x^2 + Cx + 16A,$$
so 
$$A + B = 0, C = 3, 16A = 1$$
Hence $A = \frac{1}{16}$, $B = \frac{-1}{16}$, and $C = 3$.  Then we 
integrate as follows
\begin{eqnarray*}
\int \frac{3x + 1}{x^3 + 16x}\ dx &=& \frac{1}{16} \int \frac{dx}{x} - \frac{1}{16}\int \frac{x}{x^2 + 16}\ dx + 3\int \frac{1}{x^2 + 16}\\
&=& \frac{1}{16} \ln|x| - \frac{1}{32} \ln|x^2 + 16| + \frac{3}{4}\tan^{-1}\left(\frac{x}{4}\right) + C.
\end{eqnarray*}
(Explanation:  The first two integrals are $du/u$ forms.  The last one can be done in 
several ways, including use of \# 17 in the table of integrals with $u = x$ and $a = 4$.)

\item[B.] (15)  Integrate via a trigonometric substitution:
${\displaystyle \int x^3 \sqrt{81 - x^2}\ dx}$
\item[] {\it Solution:}  From the $\sqrt{a^2 - u^2}$ form, we want $x = 9\sin\theta$, 
so $dx = 9\cos\theta\ d\theta$ and $\sqrt{81 - x^2} = \sqrt{81(1 - \sin^2\theta)} = 9\cos\theta$.
Hence the integral becomes
\begin{eqnarray*}
\int x^3 \sqrt{81 - x^2}\ dx &=& 9^5 \int \sin^3\theta \cos^2\theta\ d\theta\\
&=& 9^5 \int \sin^2\theta\cos^2\theta \cdot \sin\theta\ d\theta\\
&=& 9^5 \int (1 - \cos^2\theta)\cos^2\theta\sin\theta\ d\theta\\
&=& 9^5\left( \int \cos^2\theta\sin\theta\ d\theta - \int \cos^4\theta\sin\theta\ d\theta\right)\\
&=& 9^5\left( \frac{-\cos^3\theta}{3} + \frac{\cos^5\theta}{5}\right) + C,
\end{eqnarray*}
(since both of the last integrals are $u^n\ du$ forms).  Converting back to $x$ we have
$\frac{x}{9} =\sin\theta$, so $\cos\theta = \frac{\sqrt{81 - x^2}}{9}$ and the integral 
simplifies to 
$$-27 (81 - x^2)^{3/2} + \frac{(81-x^2)^{5/2}}{5} + C.$$
(Note:  The trig integral $\int \sin^3\theta\cos^2\theta\ d\theta$ can also be done via the
reduction formula \# 86 (use the second form to reduce the power of $\cos\theta$), and then 
\# 67.) 
\end{enumerate}
\vskip 10pt
\noindent
II.   
\begin{enumerate}
\item[A.] (5) Use midpoint Riemann sums with $n = 2$ to approximate 
$\displaystyle \int_1^2 \sqrt{1 + \ln(x)}\ dx$.  
\item[] {\it Solution:} With $n = 2$, $\Delta x = 1/2$ and we subdivide with 
one intermediate point at $3/2$.  The midpoint of the first half is $5/4$, 
the midpoint of the second half is at $7/4$.  The midpoint sum is 
$$\frac{1}{2}\left(\sqrt{1 + \ln(5/4)} + \sqrt{1 + \ln(7/4)}\right) \doteq 1.1774$$
\item[B.] (5) Use the trapezoidal rule with $n = 2$ to approximate 
$\displaystyle \int_1^2 \sqrt{1 + \ln(x)}\ dx$. 
\item[] {\it Solution:}  We compute
$$\frac{\Delta x}{2}\left(\sqrt{1 +\ln(1)} + 2\sqrt{1 + \ln(3/2)} + \sqrt{1+\ln(2)}\right) \doteq
1.1681.$$
\item[C.] (5) The function $y = \sqrt{1 + \ln(x)}$ is plotted below.
\begin{center}
\includegraphics[width=1.5truein,height=1.5truein]{132P2I.ps} 
\end{center}
\noindent
\item[] {\it Solution:}  Since the graph is concave down on this interval $[1,2]$, 
\vskip 10pt
\item[] The midpoint approximation is an overestimate and
\item[] The trapezoidal rule approximation is an underestimate.
\end{enumerate}

\vskip 10pt
\noindent
III.  For each of the following improper integrals, set up and evaluate the appropriate limits
to determine whether the integral converges.  If so, find its value; if not, say ``does not converge.''  (Credit will be given only for the correct limit calculation.)
\begin{enumerate}
\item[A.]  (5) ${\displaystyle \int_1^\infty e^{-x}\ dx}$
\item[] This equals 
\begin{eqnarray*} 
\lim_{b \to \infty} \int_1^b e^{-x}\ dx &=& \lim_{b\to\infty} \left(-e^{-x} \right|_1^b\\
&=& \lim_{b\to\infty}\left(e^{-1} - e^{-b}\right)\\
&=& e^{-1} - 0\\
&=& e^{-1}.
\end{eqnarray*}
Note that $\lim_{b\to \infty} e^{-b} = \lim_{b\to\infty} {1\over e^b} $ does exist
and equals {\it zero} because the numerator is constant but the denominator
grows without bound as $b \to \infty$.  Since the limit is finite, the integral converges. 
\item[B.] (10) ${\displaystyle \int_{-1}^1 \frac{1}{x^2}\ dx}$
\item[] {\it Solution:}  This integral is improper because the function $f(x) = \frac{1}{x^2}$
has a discontinuity at $x = 0$, which is in the interval $[-1,1]$.  When this happens we 
must consider
$$\lim_{b\to 0^-} \int_{-1}^b \frac{1}{x^2}\ dx + \lim_{a \to 0^+} \int_a^1 \frac{1}{x^2}\ dx$$
and both limits must exist for the integral to converge.  But
$$\lim_{b\to 0^-} \int_{-1}^b \frac{1}{x^2}\ dx = \lim_{b\to 0^-} \left( \frac{-1}{x}\right|_{-1}^b =
\lim_{b\to 0^-}\left(-1 - \frac{1}{b}\right)\ d.n.e.$$
(The other integral also has an infinite limit as $a \to 0^+$.)
This integral diverges as a result.
\end{enumerate}
\noindent
IV.  The region $R$ is bounded by the graphs $y = \sqrt{x}$ and $y = x^2$.  
\begin{enumerate}
\item[A.] (5)  Sketch the region $R$.
\begin{center}
\includegraphics[width=2.0truein,height=2.0truein]{132P2IIS.ps} 
\end{center}
\vfill\eject
\item[B.] (5)  Set up and compute an integral to find the area of $R$.
\item[] {\it Solution:}  The area is 
$$\int_0^1 \sqrt{x} - x^2\ dx = \left.\frac{2}{3} x^{3/2} - \frac{x^3}{3}\right|_0^1 = \frac{1}{3}.$$
\item[C.] (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 cross-sections by planes $x = {\rm const}$ are 
``washers'' with outer radius $\sqrt{x}$ and inner radius $x^2$, so the volume is
$$V = \int_0^1 \pi (\sqrt{x})^2 - \pi (x^2)^2\ dx = \pi \left(\left. \frac{x}{2} - \frac{x^5}{5}\right|_0^1\right) = \frac{3\pi}{10}.$$
\item[D.] (10)  A solid has the region $R$ as base and 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:}  The hypotenuse of the triangle in the slice at $x$
is $\sqrt{2}$ times the side, so the two equal sides of the triangle have length 
$$\frac{\sqrt{x} - x^2}{\sqrt{2}}.$$
The area of the triangle is 
$$\frac{1}{2} \cdot \left(\frac{\sqrt{x} - x^2}{\sqrt{2}}\right)^2 = \frac{(\sqrt{x} - x^2)^2}{4}.$$
By Cavalieri's Principle, the volume is 
$${1\over 4}\int_0^1 (\sqrt{x} - x^2)^2\ dx = {1\over 4}\int_0^1 x - 2x^{5/2} + x^4\ dx 
= {1\over 4}\left( \left. \frac{x^2}{2} - \frac{4}{7} x^{7/2} + \frac{x^5}{5} \right|_0^1 \right) = \frac{9}{280}.$$
\end{enumerate}
					\noindent
					V. (10) Suppose that the 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}

