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\begin{document}
\begin{center}

\textbf{College of the Holy Cross, Spring 2008} \\
\textbf{Math 132, Solutions for Midterm Exam 2} (All Sections) \\
\textbf{Wednesday, March 26, 6 PM}
\end{center}


\noindent
I.   
\begin{enumerate}
\item[A.] (5) Use midpoint Riemann sums with $n = 4$ to approximate $\displaystyle \int_1^2 \sqrt{1 + \ln(x)}\ dx$.  

\soln On the interval $[1,2]$, with $n = 4$, we have $\Delta x = \frac{2-1}{4} = .25$, so
$x_0 = 1, x_1 = 1.25, x_2 = 1.5, x_3 = 1.75, x_4 = 2$, and the midpoints are
$m_1 = 1.125, m_2=1.375, m_3 = 1.625, m_4 = 1.875$.  The midpoint Riemann sum is
\begin{eqnarray*}
M_4 &=& (.25)\left(\sqrt{1 + \ln(1.125)} +  \sqrt{1 + \ln(1.375)} + \sqrt{1 + \ln(1.625)} + \sqrt{1 + \ln(1.875)}\right)\\
&\doteq& (.25)(1.057252589) + (1.148239405) + (1.218814102) + (1.276169526))\\
&=& 1.175118905.
\end{eqnarray*}
\item[B.] (5) Use the trapezoidal rule with $n = 4$ to approximate $\displaystyle \int_1^2 \sqrt{1 + \ln(x)}\ dx$. 

\soln By the trapezoidal rule formula, writing $f(x) = \sqrt{1 + \ln(x)}$,
\begin{eqnarray*}
T_4 &=& \frac{(.25)}{2}(f(1) + 2f(1.25) + 2f(1.5) + 2f(1.75) + f(2))\\ 
&\doteq& \frac{(.25)}{2}(1. + 2(1.105958205) + 2(1.185523137) + 2(1.248845782) + 1.301209891)\\
&=& 1.172733017.
\end{eqnarray*}
\item[C.] (5) The function $y = \sqrt{1 + \ln(x)}$ is plotted below.
\begin{center}
\includegraphics[width=2.3truein,height=2.3truein]{132P2I.ps} 
\end{center}

\soln  Since the graph is concave down on $[1,2]$,
\item[]The midpoint approximation is a overestimate.
\item[]The trapezoidal rule approximation is an underestimate.
\end{enumerate}

\vskip 10pt
\noindent
II.  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.]  (8) ${\displaystyle \int_1^\infty e^{-x}\ dx}$

\soln  The integral is improper because the interval of integration is 
infinite.  
\begin{eqnarray*}
\int_1^\infty e^{-x}\ dx &=& \lim_{b\to\infty} \int_1^b e^{-x}\ dx\\
&=& \lim_{b\to\infty} \left(\left. -e^{-x}\right|_1^b\right)\\
&=& \lim_{b\to\infty} \frac{1}{e} - \frac{1}{e^b}\\
&=& \frac{1}{e}.
\end{eqnarray*}
The integral converges; the value is $\frac{1}{e}$.
\item[B.] (8) ${\displaystyle \int_{-1}^1 \frac{1}{x^2}\ dx}$

\soln The integral is improper because the integrand has an
infinite discontinuity at $x = 0$ in the interior of the interval.
Therefore,
\begin{eqnarray*}
\int_{-1}^1 \frac{1}{x^2}\ dx &=& \lim_{b\to 0^-} \int_{-1}^b \frac{1}{x^2}\ dx + 
\lim_{a\to 0^+} \int_a^1 \frac{1}{x^2}\ dx\\
&=& \lim_{b\to 0^-} \left(\left. -\frac{1}{x}\right|_{-1}^b\right) + \lim_{a\to 0^+} \left(\left. -\frac{1}{x}\right|_a^1\right) \\
&=& \lim_{b\to 0^-} \left(- \frac{1}{b} - 1 \right)  + \lim_{a\to 0^+} \left( -1 + \frac{1}{a}\right).
\end{eqnarray*}
Neither limit exists, so this improper integral \emph{diverges}.
\item[C.] (9) $\displaystyle \int_0^1 x\ln(x)\ dx$.  

\soln 
This integral is improper because of the infinite discontinuity of $\ln(x)$ at $x = 0$.
\[
\int_0^1 x\ln(x) \ dx = \lim_{a\to 0^+} \int_a^1 x \ln(x)\ dx
\]
(if the limit exists).  To integrate, we use parts letting $u = \ln(x)$ and $dv = x\ dx$,
so $du = \frac{1}{x}\ dx$ and $v = \frac{x^2}{2}$.
Hence 
\begin{eqnarray*}
&=&\lim_{a\to 0^+}\left[ \left(\left. \frac{x^2}{2} \ln(x)\right|_a^1\right) - \int_a^1 \frac{x^2}{2}\cdot \frac{1}{x}\ dx\right]\\
&=& \lim_{a\to 0^+} \left[ -\frac{a^2}{2}\ln(a) - \frac{1}{4} + \frac{a^2}{4}\right].
\end{eqnarray*}
To evaluate the limit of the first term here, we note that this is an indeterminate
form $0\cdot \infty$.  Hence we will use L'Hopital's Rule, after rearranging:
\begin{eqnarray*}
\lim_{a\to 0^+} \frac{a^2}{2}\ln(a) &=& \lim_{a\to 0^+} \frac{\ln(a)}{\frac{2}{a^2}}\\
&=& \lim_{a\to 0^+} \frac{\frac{1}{a}}{\frac{-4}{a^3}}\\
&=& \lim_{a\to 0^+} \frac{a^2}{-4}\\
&=& 0.
\end{eqnarray*}
Hence the integral converges to $\frac{-1}{4}$ since the last term on the right above also tends to $0$
as $a \to 0^+$. 
\end{enumerate}

\vskip 10pt
\noindent
III.  The region $R$ is bounded by the graphs $y = \sqrt{x}$ and $y = x^2$.  
\begin{enumerate}
\item[A.] (5)  Sketch the region $R$.
\soln The upper curve (in green) is $y = \sqrt{x}$.  The lower (in red) is 
$y = x^2$.  The two graphs cross when $x^{1/2} = x^2$, so $x = x^4$.
This gives $x = 0$ and $x = 1$.  The region is shown here:
\begin{center}
\includegraphics[width=2.0truein,height=2.0truein]{132P2IIS.ps} 
\end{center}
\item[B.] (5)  Set up and compute an integral to find the area of $R$.

\soln
The area of $R$ is
\begin{eqnarray*}
\int_0^1 \sqrt{x} - x^2\ dx &=& \left(\left.\frac{2x^{3/2}}{3} - \frac{x^3}{3}\right|_0^1\right)\\
&=& \frac{2}{3} - \frac{1}{3}\\
&=& \frac{1}{3}.
\end{eqnarray*}
\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.

\soln
The cross-sections by planes perpendicular to the $x$-axis are washers
with outer radius $r_{out} = \sqrt{x}$ and inner radius $r_{in} = x^2$.
The volume is the integral of the cross-section area, so
\begin{eqnarray*}
V &=& \int_0^1 \pi(\sqrt{x})^2 - \pi(x^2)^2\ dx\\
&=& \pi \int_0^1 x - x^4\ dx\\
&=& \pi \left(\left. \frac{x}{2} - \frac{x^5}{5}\right|_0^1\right)\\
&=& \frac{3\pi}{10}.
\end{eqnarray*}
\item[D.] (10)  The region $R$ is rotated about the $y$-axis to generate
a solid.  Set up and compute an integral to find its volume.

\soln
On the basis of symmetry of the region $R$, we can see that this volume will be the
same as the volume in part C.  To compute this via an integral, we
need to set up in terms of $y$.  The graph $y = x^2$ is $x = \sqrt{y}$
and the graph $y = \sqrt{x}$ is $x = y^2$.  The cross-sections by planes 
perpendicular to the $y$-axis are washers
with outer radius $r_{out} = \sqrt{y}$ and inner radius $r_{in} = y^2$.
The volume is the integral of the cross-section area, so
\[
V = \int_0^1 \pi(\sqrt{y})^2 - \pi(y^2)^2\ dy = \frac{3\pi}{10}
\]
by the same computations as in part C. 
\end{enumerate}

\vskip 10pt
\noindent
IV. (15) Set up and compute the integral for the arclength of the curve
$x = 5t^2 + 2, y = t + 1$, $0 \le t \le 2$.  (You may consult the 
table of integrals for this one; if you do, say which table entry you are using.)

\soln
The arclength is 
\begin{eqnarray*}
L &=& \int_0^2 \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\ dt\\
&=& \int_0^2 \sqrt{(10t)^2 + 1}\ dt.
\end{eqnarray*}
Let $u = 10t$, then $dt = \frac{1}{10} du$, so 
\[
L = \frac{1}{10} \int_0^{20} \sqrt{u^2 + 1}\ du.
\]
This is the form from \#21 in the table of integrals, with $a = 1$.  Hence
\begin{eqnarray*}
L &=& \frac{1}{10}\left(\left. \frac{u}{2}\sqrt{u^2 + 1} + \frac{1}{2}\ln(u + \sqrt{u^2 + 1})
\right|_0^{20}\right)\\
&=& \sqrt{401} + \frac{1}{20}\ln(20 + \sqrt{401})\\
&\doteq& 20.21.
\end{eqnarray*}

\vskip 10pt
\noindent
V. (15) Find the average value of $\displaystyle{f(x) = \frac{x^3}{\sqrt{25 - x^2}}}$ on 
the interval $0 \le x \le 2$.  (Use a trigonometric substitution to compute 
the integral you need.)

\soln
The average value is computed by the integral 
\[
f_{ave} = \frac{1}{2-0} \int_0^2 \frac{x^3}{\sqrt{25 - x^2}}\ dx.
\]
To evaluate the integral, we use the trigonometric substitution
$x = 5\sin\theta$, so $dx = 5\cos\theta\,d\theta$ and 
$\sqrt{25 - x^2} = \sqrt{25 - 25\sin^2\theta} = 5\sqrt{1 - \sin^2\theta} = 5\cos\theta$. 
This gives
\begin{eqnarray*}
\int \frac{x^3}{\sqrt{25 - x^2}}\ dx &=& \int \frac{125\sin^3\theta}{5\cos\theta} \cdot 5\cos\theta\, d\theta\\
&=& 125\int \sin^3\theta\,d\theta\\
&=& 125\int \sin^2\theta \cdot \sin\theta\,d\theta\\
&=& 125\int (1 - \cos^2\theta)\sin\theta\, d\theta\\
&=& 125\int\sin\theta\,d\theta - 125\int\cos^2\theta\sin\theta\,d\theta\\
&=& -125\cos\theta + \frac{125}{3}\cos^3\theta + C.
\end{eqnarray*}
Therefore since $\cos\theta = \frac{\sqrt{25 - x^2}}{5}$ from the reference triangle (opposite = $x$,
hypotenuse = $5$, adjacent = $\sqrt{25 - x^2}$), we have
\begin{eqnarray*}
f_{ave} &=& \frac{1}{2} \left(\left. -25\sqrt{25 - x^2} + \frac{1}{3} (25 - x^2)^{3/2}\right|_0^2\right)\\
&=& \frac{1}{2} \left(-25\sqrt{21} + 7\sqrt{21} + 125 - \frac{125}{3}\right)\\
&=& \frac{1}{2} \left(\frac{250}{3} - 18\sqrt{21}\right)\\
&=& \frac{125}{3} - 9\sqrt{21}\\
&\doteq& 0.42348541.
\end{eqnarray*}
  
\end{document}

