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\markright{Math 132, Midterm 2 Solutions}

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\begin{center}

\textbf{College of the Holy Cross, Spring Semester, 2005} \\
\textbf{Math 132, Midterm 2 Solutions}\\
\textbf{Thursday, March 31}
\end{center}

\begin{enumerate}
  \item The graph of the function $f(x)$ is shown below. 
  
  \begin{center}
    \input{exam2questions_3.eepic}
  \end{center}
  
  \begin{enumerate}
    \item{} [10 points] Use the graph of $f$ to 
    compute the following approximations of $\int_0^8 f(x)\,dx$.
    
    \bigskip
    {\bf Solution:}  Since $n = 2$, $\Delta x = \frac{8 - 0}{2} = 4$.
    
    $LEFT(2):\underline{\ 28\ } = 0\cdot 4 + 7\cdot 4$
    \vfill
     $RIGHT(2):\underline{\ 68\ } = 7\cdot 4 + 10\cdot 4$
    \vfill
     $MID(2):\underline{\ 52 \ } = 4\cdot 4 + 9\cdot 4$
    \vfill
     $TRAP(2):\underline{\ 48\ } = \displaystyle{\frac{1}{2}(LEFT(2) + RIGHT(2))}$
    \vfill
     $SIMP(2):\underline{\ 50.6\ } = \displaystyle{\frac{2 MID(2) + TRAP(2)}{3}}$
    \vfill
    \item{} [6 points] For each method, decide whether the approximation 
    of $\int_0^8 f(x)\,dx$ is an {\it overestimate}, an {\it underestimate}, 
    or that this {\it cannot be determined} from the given information. 
    In the case of an overestimate or underestimate, briefly explain your 
    reasoning. 
    
    \begin{tabular}{|c|c|c|}
    \hline
    Method & Type of Estimate \qquad& Reason\qquad\qquad\qquad\qquad\qquad\qquad\qquad\ \\
    \hline
    LEFT & underestimate & $f$ is increasing \\
    \hline
    RIGHT & overestimate & $f$ is increasing \\
    \hline
    MID & underestimate & $f$ is concave down \\
    \hline
    TRAP & overestimate & $f$ is concave down \\
    \hline
    SIMP& can't determine & \\
    \hline
    \end{tabular}
    
  \end{enumerate}
  
  
  \newpage
  
  \item 
  \begin{enumerate}
  \item{} [6 points] Set up an integral that represents the arclength of the portion of the 
  graph of $y=\sin(x)$ between $x=0$ and $x=\pi$. {\bf Do not evaluate the integral.} 
  
  \bigskip
  {\bf Solution:}  Using the usual form of the arclength integral for
  the graph $y = f(x)$, we get $\frac{dy}{dx} = \cos(x)$ and
  $$L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\ dx = \int_0^\pi \sqrt{1 + \cos^2 x}\ dx$$
  
  \item{} [6 points] Approximate the integral in part (a) using a left hand sum with 
  $n=4$ subintervals.
  
  \bigskip
  {\bf Solution:}  We have $\Delta x = \frac{\pi - 0}{4} = \frac{\pi}{4}$.  Then 
  with the left hand sum approximation:
  \begin{eqnarray*}
  \int_0^\pi \sqrt{1 + \cos^2 x}\ dx & \doteq & \sqrt{1+\cos^2(0)}\cdot \frac{\pi}{4} + 
  \sqrt{1+\cos^2\left(\frac{\pi}{4}\right)}\cdot \frac{\pi}{4}\\
  & & \qquad +
  \sqrt{1+\cos^2\left(\frac{\pi}{2}\right)}\cdot \frac{\pi}{4} 
  +\sqrt{1+\cos^2\left(\frac{3\pi}{4}\right)}\cdot \frac{\pi}{4}\\
  &=& \left(\sqrt{2} + \sqrt{3/2} + 1 + \sqrt{3/2}\right)\frac{\pi}{4}\\
  &=&\left(\sqrt{2} + \sqrt{6} + 1\right)\frac{\pi}{4}\\
  &\doteq& 3.82\\
  \end{eqnarray*}
  \end{enumerate}
  
  \item Rewrite each of the following improper integrals as a limit, or limits. 
  State whether the integral converges or diverges, and compute its value 
  if it converges.
  \begin{enumerate}
    \item{} [6 points] $\displaystyle\int_{-1}^2\frac1{x^3}\,dx$
    
    \bigskip
    {\bf Solution:}  This integral is improper because the integrand
    is undefined at $x = 0$.  Since this is in the interior of the 
    interval of integration, we look at 
 $$\lim_{b\to 0^-} \int_{-1}^b x^{-3}\ dx + \lim_{a\to 0^+}\int_a^2 x^{-3}\ dx  = 
    \lim_{b\to 0^-} \frac{-1}{2} x^{-2}\Bigg|_0^b + \lim_{a\to 0^+} \frac{-1}{2} x^{-2}\Bigg|_a^2$$
    Both of these limits are undefined, so the improper integral {\it diverges}.
    \newpage
    \item{} [6 points] $\displaystyle\int_0^2\frac1{\sqrt{4-x^2}}\,dx$
    
    \bigskip
    {\bf Solution:}  This integral is improper because the integrand is
    not defined at the endpoint $x = 2$.  We have
    \begin{eqnarray*}
    \int_0^2\frac1{\sqrt{4-x^2}}\ dx & = & \lim_{b\to 2^-} \int_0^b\frac1{\sqrt{4-x^2}}\ dx\\
    & = & \lim_{b\to 2^-} \arcsin\left(\frac{x}{2}\right)\Bigg|_0^b \quad \# 28 {\rm\ in\ table}\\
    & = & \lim_{b\to 2^-} \arcsin\left(\frac{b}{2}\right)\\
    & = & \frac{\pi}{2}\\
    \end{eqnarray*}
    So the integral converges to the value $\frac{\pi}{2}$.
    \item{} [6 points] $\displaystyle\int_0^\infty e^{-4x}\,dx$
    
    \bigskip
    {\bf Solution:}  This integral is improper because of the infinite
    limit of integration.
    \begin{eqnarray*}
    \int_0^\infty e^{-4x}\,dx & = & \lim_{b\to\infty} \int_0^b e^{-4x}\,dx\\
    &=& \lim_{b\to\infty} \frac{-1}{4}e^{-4x}\Bigg|_0^b\\
    &=&\lim_{b\to\infty} \frac{1}{4}(1 - e^{-4b})\\
    &=&\frac{1}{4}\\
    \end{eqnarray*}
    So the integral converges to the value $\frac{1}{4}$.
  \end{enumerate}
  
  \item Let $R$ denote the region bounded by $y=\sin x$, $y=\cos x$, $x=0$ and 
  $x=\pi/4$. 
  \ 
  \vskip 10pt
  \ 
  \begin{center}
    \input{exam2questions_2.eepic}
  \end{center}
  \begin{enumerate}
  
    \item{} [7 points] Find the area of $R$.
    
    \bigskip
    {\bf Solution:}  Think of subdividing the area into vertical strips.
    The height of each is $\cos(x) - \sin(x)$ and the width is $dx$, so the area is given by
    $$\int_0^{\pi/4} \cos(x) - \sin(x)\ dx = (\sin(x) + \cos(x))\Bigg|_0^{\pi/4} = \sqrt{2} - 1\doteq .414$$
    \item{} [7 points] Find $\overline{x}$, 
    the $x$-coordinate of the center of mass of~$R$. (Assume the region 
    has constant mass density $\delta=1$.)
    
    \bigskip
    {\bf Solution:}  To find $\overline{x}$, we compute $Moment/Mass$.
    The mass was already computed in part a.  The moment is given by the 
    integral
    $$\int_0^{\pi/4} x(\cos(x) - \sin(x))\ dx$$
    To integrate here we use parts with $u = x$, and
    $dv = \cos(x) - \sin(x)$.  Then $du = dx$, and 
    $v = \sin(x) + \cos(x)$.  This gives
    \begin{eqnarray*}
    \int_0^{\pi/4} x(\cos(x) - \sin(x))\ dx
    &=& x(\sin(x) + \cos(x))\Bigg|_0^{\pi/4} - \int_0^{\pi/4} \sin(x) + \cos(x)\ dx\\
    &=& x(\sin(x) + \cos(x))\Bigg|_0^{\pi/4} + (\cos(x) - \sin(x))\Bigg|_0^{\pi/4}\\
    &=& \frac{\pi\sqrt{2}}{4} - 1\\
    \end{eqnarray*}
    Then the center of mass is at 
    $$\overline{x} = \frac{\frac{\pi\sqrt{2}}{4} - 1}{\sqrt{2}-1} \doteq .267$$
  \end{enumerate}
  
  \item Let $R$ be the region bounded by the curve $y=e^x$ and the lines $x=0$ and $y=e$. 
  
  \begin{center}
    \input{exam2questions_4.eepic}
  \end{center}
  
  \begin{enumerate}
    \item{} [8 points] Find the volume of the solid obtained by revolving $R$ about the $x$-axis.
    
    \bigskip
    {\bf Solution:}  Rotating around the $x$-axis, we get a solid with 
    ``washer'' cross sections because of the gap between the region $R$ 
    and the $x$-axis.  The inner radius for the cross section at $x$
    is $r_{in} = e^x$ while the outer radius is $r_{out} = e$.  The upper limit
    of integration is found by setting $e^x = e$ and solving for $x$: $x = 1$.
    The volume is 
    $$\int_0^1 \pi e^2 - \pi e^{2x}\ dx = \pi\left(e^2 x - \frac{e^{2x}}{2}\right)\Bigg|_0^1
    = \frac{\pi}{2}(e^2 + 1) \doteq 13.18$$

    \item{} [6 points] Set up an integral that represents the volume of the solid whose base is $R$ and 
    whose cross-sections perpendicular to the $x$-axis are squares with base lying in $R$. 
    {\bf Do not evaluate the integral.}
    
    \bigskip
    {\bf Solution:}  By Cavalieri's Principle, we integrate the cross
    sectional area to get the volume.  Since the cross section at $x$
    is a square with side $e - e^x$, the cross sectional area
    is $(e - e^x)^2$, and the volume is 
    $$\int_0^1 (e - e^x)^2\ dx$$
  \end{enumerate} 
  
  \item Suppose a metal rod of length $2$ meters has a mass density 
  $\delta(x)=5+1.2x^2$ kg per meter. 
  \begin{enumerate}
    \item{} [7 points] Find the total mass of the rod.
   
    \bigskip
    {\bf Solution:}  Since the rod is a straight thin (essentially
    1-dimensional) object, the mass of a small piece is approximately
    density value times the length.  Adding up these terms,
    we get:
    
    $$M = \int_0^2 5 + 1.2 x^2\ dx = (5x + .4 x^3)\Bigg|_0^2 = 13.2 {\rm kg}$$
    
    \item{} [7 points] Find the center of mass of the rod.
    
    \bigskip
    {\bf Solution:} The center of mass is at 
    \begin{eqnarray*}
    \overline{x} & = & \frac{1}{13.2} \int_0^2  x(5 + 1.2 x^2)\ dx\\
    & = & \frac{1}{13.2} \int_0^2  5x + 1.2 x^3\ dx\\
    & = & \frac{1}{13.2} \left(\frac{5x^2}{2} + .3 x^4\Bigg|_0^2\right)\\
    & = & \frac{14.8}{13.2}\\
    &\doteq & 1.12\\
    \end{eqnarray*}
    (Note:  This is to the right of the midpoint of the rod, which
    is what we expect since the density is increasing with $x$.
    There is more mass to the right of the midpoint than to the left.
     
  \end{enumerate}
  
  
  \item A 200 meter tall pyramid has a square base with side length 100 meters. 
  Its mass density (in kg/m${}^3$) at height $h$ meters above the base is given by $\delta(h)=1+0.01h$. 
  
  \begin{center}
    \input{exam2questions_5.eepic}
  \end{center}
 
  \begin{enumerate}
    \item{} [6 points] Write down a Riemann sum that approximates the total mass of the pyramid. 
    
    \bigskip
    {\bf Solution:}  Slicing the pyramid horizontally, the slice at height
    $h$ above the base is approximately a rectangular solid with length 
    and width $s(h)$ depending on $h$, and height $\Delta h$.  
    The side $s(h)$ is found by similar triangles: 
    $$s(h) = \frac{1}{2}(200-h).$$
    So we add up (density) $\cdot$ (volume) for all the slices:
    $$M \doteq \sum (1+.01 h) \left(\frac{1}{2}(200-h)\right)^2 \Delta h$$
    \item{} [6 points] Using the answer to part (a), write down an integral that equals the 
    mass of the pyramid. {\bf Do not evaluate the integral.}
    
    \bigskip
    {\bf Solution:} 
    $$M = \int_0^{200} (1 + .01h)\cdot \frac{1}{4}(200 - h)^2\ dh$$
  
  \end{enumerate}
  
\end{enumerate}
\end{document}



