%% MATH 132-01 Final Exam, Spring 2008
\documentclass[12pt]{article}
\usepackage{latexsym,amsmath,epsfig}
\newcommand{\ds}{\displaystyle}
\newcommand{\vp}{$\vphantom{\big|}$}

\newcommand{\R}{\textbf{R}}
\newcommand{\Item}{\item{$\Box$}\quad}

\newcommand{\Answer}[1]{$\phantom{.}$\hfill
        #1\quad\fboxsep 20pt \fbox{$\phantom{\Big|Answer}$}} 

\newcommand{\MyTallAnswer}[2]{$\phantom{.}$\hfill
        #1\quad\fboxsep 20pt\fbox{$\vphantom{\Big|}#2$}}

\newcommand{\MyAnswer}[2]{$\phantom{.}$\hfill
        #1\quad\fboxsep 5pt\fbox{$\vphantom{\Big|}#2$}}

\newcommand{\soln}{\paragraph{Solution:}}
\newcommand{\sep}{{\medskip\hrule\bigskip}}
\newcommand{\blank}{\underline{$\phantom{A blank}$}}

\setlength{\textwidth}{6.5in}
%\setlength{\topmargin}{-0.5in}
\setlength{\oddsidemargin}{0in}
\setlength{\textheight}{9.0in}

\begin{document}

\begin{center}

\textbf{College of the Holy Cross, Spring Semester, 2008} \\
\textbf{MATH 132, Section 01, Solutions for Final Exam} \\
\textbf{Thursday, May~8, 2:30~PM} \\
Professor Little
\end{center}
\vskip 10pt
\noindent
I.  
\begin{enumerate}
\item[(A)] (5) Compute the derivative of the function $\ds g(x) = \int_0^{3x} \frac{\sin(t)}{t}\, dt$.

\soln By the first part of the Fundamental Theorem of Calculus and the Chain Rule for derivatives:
\[
g'(x) = \frac{\sin(3x)}{3x}\cdot 3 = \frac{\sin(3x)}{x}.
\]
\item[(B)] (4) Let 
$\ds
f(x) = \begin{cases} 1 & \text{ if } 0 \le x \le 2\\
                    x-1 & \text{ if } 2 < x \le 4.\\
                   11-2x & \text{ if } 4 < x \le 6
\end{cases}$  
Sketch the graph $y = f(x)$ on the axes provided below.

\soln
\begin{center}
\includegraphics[width=2.5truein,height=2.5truein]{132FIS.ps}
\end{center}
\item[(C)] (6) Let $F(x) = \int_0^x f(t)\, dt$, where $f(t)$ is the function from part (B).  
Complete the following table of values for $F(x)$:

\soln
The value $F(x)$ represents the signed area between the graph $y = f(x)$ and the $x$-axis:
\[
\begin{array}{|c||r|r|r|r|r|r|r|}
\hline
x          & 0 & 1 & 2 & 3 & 4 & 5 & 6 \strut \\
\hline
F(x)       & \quad 0 & \qquad 1 & \qquad 2 & \qquad 3.5 & \qquad 6 & \qquad 8 & \qquad 8\strut \\
\hline
\end{array}
\]
 
\end{enumerate}
\vfill\eject

\vskip 10pt
\noindent
II.  Compute the following integrals.  Some of these may be forms covered by entries in the
table of integrals.  Half credit will be given for using a table entry; full credit only for showing
all work leading to the final answer.
\begin{enumerate}
\item[(A)] (5) $\ds \int \frac{x^4 - 3x^3 + \sqrt{x}}{x^{2/3}}\, dx$

\soln Split into separate fractions, simplify and integrate:
\begin{eqnarray*}
\int \frac{x^4 - 3x^3 + \sqrt{x}}{x^{2/3}}\, dx &=& \int x^{10/3} - 3x^{7/3} + x^{-1/6}\, dx\\
&=& \frac{3}{13} x^{13/3} - \frac{9}{10} x^{10/3} + \frac{6}{5} x^{5/6} + C.
\end{eqnarray*}
\item[(B)] (5) $\ds \int x \cos(x^2)\, dx$

\soln Use a $u$-substitution with $u = x^2$, so $du = 2x\, dx$.  This gives
\[
\int x\cos(x^2)\, dx = \frac{1}{2}\int \cos(u)\, du = \frac{1}{2}\sin(u) + C = \frac{1}{2}\sin(x^2) + C.
\]
\item[(C)] (10) $\ds \int \frac{\sec^2(5x)\, dx}{(\tan(5x) + 7)^5}$

\soln This one can also be handled by the $u$-substitution $u = \tan(5x) + 7$, for which
$du = 5\sec^2(5x)\, dx$ by the Chain Rule.  Then
\[
\int \frac{\sec^2(5x)\, dx}{(\tan(5x) + 7)^5} = 
\frac{1}{5} \int u^{-5}\, du = \frac{-1}{20} u^{-4} + C = \frac{-1}{20} (\tan(5x) + 7)^{-4} + C.
\]
\item[(D)] (10) $\ds \int_1^e x^7 \ln(x)\, dx$.  

\soln Integrate by parts with $u = \ln(x)$ and $dv = x^7\, dx$.  Then $du = \frac{1}{x}\, dx$
and $v = \frac{1}{8}x^8$ and by the integration by parts formula,
\[
\int x^7 \ln(x)\, dx = \frac{x^8}{8}\ln(x) - \int \frac{1}{8} x^7\, dx = 
\frac{x^8}{8}\ln(x) - \frac{x^8}{64} + C.
\]
For the definite integral, we apply the Evaluation Theorem to get
\[
\int_1^e x^7\ln(x)\, dx = \left. \frac{x^8}{8}\ln(x) - \frac{x^8}{64}\right|_1^e 
= \frac{e^8}{8} - \frac{e^8}{64} + \frac{1}{64} = \frac{7e^8 + 1}{64}.
\]
\item[(E)] (12.5) $\ds \int \frac{x^2}{\sqrt{25 - x^2}}\, dx$

\soln
We use the trigonometric substitution $x = 5\sin\theta$, so $dx = 5\cos\theta\,d\theta$, and
$\sqrt{25 - x^2} = \sqrt{25(1 - \sin^2\theta)} = 5\cos\theta$.  This simplifies to
\[
\int \frac{25\sin^2\theta}{5\cos\theta}\cdot 5\cos\theta\, d\theta = 
25\int \sin^2\theta\, d\theta.
\]
Now we use the half-angle identity to integrate the even power of $\sin\theta$, then
convert back to $x$:
\begin{eqnarray*}
25\int \sin^2\theta\, d\theta &=& 25\int \frac{1}{2}(1 - \cos(2\theta))\, d\theta\\
&=&  \frac{25}{2}\theta - \frac{25}{4}\sin(2\theta) + C\\
&=&  \frac{25}{2}\theta - \frac{25}{2}\sin\theta\cos\theta + C\\
&=&  \frac{25}{2}\sin^{-1}\left(\frac{x}{5}\right) - \frac{1}{2}x\sqrt{25 - x^2} + C.
\end{eqnarray*}
\item[(F)] (12.5) $\ds \int \frac{x}{(x+1)(x^2 + 1)}\, dx$
\soln
By partial fractions, 
\[
\frac{x}{(x+1)(x^2 + 1)} = \frac{A}{x+1} + \frac{Bx + C}{x^2 + 1}.
\]
Clearing denominators,
\[
x = A(x^2 + 1) + (Bx + C)(x+1),
\]
so equating coefficients, $A + B = 0$, $B + C = 1$, and $A + C = 0$.  Solving simultaneously,
\[
A = \frac{-1}{2}, \qquad B = \frac{1}{2}, \qquad C = \frac{1}{2}.
\]
Then 
\begin{eqnarray*}
\int \frac{x}{(x+1)(x^2 + 1)}\, dx &=& \int \frac{-1/2}{x+1} + \frac{x/2 + 1/2}{x^2 + 1} \, dx\\
&=& \frac{-1}{2}\ln|x+1| + \frac{1}{4}\ln(x^2 + 1) + \frac{1}{2}\tan^{-1}x + C.
\end{eqnarray*}
\end{enumerate}
\vskip 10pt
\noindent
III.  
\begin{enumerate}
\item[(A)] (5)  Use a midpoint Riemann sum with $n = 4$ to approximate 
$\ds \int_0^2 e^{x^2}\ dx$.

\soln
The midpoint Riemann sum approximation is
\[
\int_0^2 e^{x^2}\, dx \doteq \left(e^{(0.25)^2} + e^{(0.75)^2} + e^{(1.25)^2} + e^{(1.75)^2}\right)(.5)
\doteq 14.48561253.
\] 
\item[(B)] (5) Use the Trapezoidal Rule with $n = 4$ to approximate
$\ds \int_0^2 e^{x^2}\ dx$.

\soln
The Trapezoidal Rule approximation is
\[
\int_0^2 e^{x^2}\, dx \doteq \frac{.5}{2}\left(e^{0^2} + 2e^{(0.5)^2} + 2e^{(1.0)^2} + 2e^{(1.5)^2} + e^{(2.0)^2}\right) \doteq 20.64455905.
\]
\item[(C)] (5) Check the appropriate boxes:

\soln  We have $\frac{d}{dx}(e^{x^2}) = 2x e^{x^2}$, and 
\[
\frac{d^2}{dx^2}(e^{x^2}) = 2e^{x^2} + 4x^2 e^{x^2} = (2 + 4x^2)e^{x^2}.
\]
Since this is $> 0$ for all real $x$,
the midpoint approximation is an underestimate, and 
the trapezoidal rule approximation is an overestimate, because
$e^{x^2}$ is concave up on $[0,2]$.  
\end{enumerate}
\vskip 10pt
\noindent
\emph{Note:}  From this point on in the exam, if an entry from the table of integrals
applies, you may use it for full credit if you state which entry you are using and indicate
what $u$ and what constant values $a$, $b$, etc. are involved.  
\vskip 10pt
\noindent
IV.  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.''
\begin{enumerate}
\item[(A)] (6) $\ds \int_0^2 \frac{1}{{\root 3\of {x - 1}}}\, dx.$

\soln
The function $f(x) = \frac{1}{{\root 3\of {x-1}}}$ has an infinite discontinuity at
$x = 1$, which is in the middle of the interval $[0,2]$.  Hence we must identify
\[
\lim_{b\to 1^-} \int_0^b \frac{1}{{\root 3\of {x - 1}}}\, dx = \lim_{b\to 1^-} \left.\frac{3}{2}(x-1)^{2/3}\right|_0^b = \lim_{b\to 1^-} \frac{3}{2}(b - 1)^{2/3} - \frac{3}{2} = \frac{-3}{2},
\]
and
\[
\lim_{a\to 1^+} \int_a^2 \frac{1}{{\root 3\of {x - 1}}}\, dx = \lim_{a\to 1^+} \left.\frac{3}{2}(x-1)^{2/3}\right|_a^2 = \lim_{a\to 1^+} \frac{3}{2} - \frac{3}{2}(a-1)^{2/3} = \frac{3}{2}.
\]
Since both limits exist, the integral converges to $\frac{-3}{2} + \frac{3}{2} = 0$.  
\vskip 10pt
\item[(B)] (9) $\ds \int_0^\infty \frac{1}{(x^2 + 4x + 5)^{3/2}}\, dx$.

\soln
If we complete the square inside the $-3/2$ power, we have $x^2 + 4x + 5 = (x+2)^2 + 1$.
Therefore, after the substitution $u = x + 2$, $du = dx$, we have 
\[
\int_2^\infty \frac{1}{(u^2 + 1)^{3/2}}\, du = \lim_{b\to\infty} \int_2^b \frac{1}{(u^2 + 1)^{3/2}}\, du.
\]
This is the form of \# 29 in the table of integrals with $a = 1$.  Hence we must determine if
\[
 \lim_{b\to\infty} \left.\frac{u}{\sqrt{1+u^2}}\right|_2^b = 
\lim_{b\to\infty} \frac{b}{\sqrt{1+b^2}} - \frac{2}{\sqrt{5}}
\]
exists or not.  The first term can be rewritten as 
\[
\frac{b}{\sqrt{1+b^2}} = \frac{b\cdot \frac{1}{b}}{\sqrt{1+b^2}\cdot \frac{1}{b}} = 
\frac{1}{\sqrt{\frac{1}{b^2} + 1}}.
\]
Hence 
\[
\lim_{b\to\infty} \frac{1}{\sqrt{\frac{1}{b^2} + 1}} = 1,
\]
so the integral converges, to the value $1 - \frac{2}{\sqrt{5}} = \frac{\sqrt{5} - 2}{\sqrt{5}}$.
\end{enumerate}
\vskip 10pt
\noindent
V.  A region $R$ in the plane is bounded by the graphs $y = 9 - x^2$, $y = 2x$, $x = 0$ and $x = 1$.
\begin{enumerate}
\item[(A)] (5) Sketch the region on the following axes:

\soln
\begin{center}
\includegraphics[width=2.0truein,height=2.0truein]{132FVS.ps}
\end{center}
\item[(B)] (5)  Compute the area of the region $R$.

\soln
The area is given by the integral
\[
A = \int_0^1 9 - x^2 - 2x\, dx = \left. 9x - \frac{x^3}{3} - x^2\right|_0^1 = \frac{23}{3}.
\]
\item[(C)] (10) Compute the volume of the solid obtained by rotating $R$ about the $x$-axis.

\soln
The cross-sections of the solid by planes perpendicular to the $x$-axis are washers
with inner radius $r_{in} = 2x$ and outer radius $r_{out} = 9 - x^2$.  So the volume is
the integral of the area of the cross-section:
\begin{eqnarray*}
V = \int_0^1 \pi(9-x^2)^2 - \pi(2x)^2\, dx &=& \pi\int_0^1 81 - 22x^2 + x^4\, dx\\
&=& \pi\left(\left. 81x - \frac{22x^3}{3} + \frac{x^5}{5}\right|_0^1\right)\\
&=& \frac{1108\pi}{15}.
\end{eqnarray*}
\item[(D)] (5) Set up the integral(s) to compute the volume of the solid
obtained by rotating $R$ about the $y$-axis.  \emph{You do not need to compute the value.}

\soln
The cross-sections by planes perpendicular to the $y$-axis are all disks, but the function
giving the radius of the disk is given by three different formulas depending on
whether $0 \le y \le 2$, or $2 \le y\le 8$, or $8 \le y \le 9$.
\[
V = \int_0^2 \pi \left(\frac{y}{2}\right)^2\, dy + \int_2^8 \pi(1)^2\, dy + \int_8^9 \pi (\sqrt{9-y})^2\, dy.
\]

\end{enumerate}
\vskip 10pt
\noindent
VI. (10) The daily solar radiation $x$ per square meter (in hundreds of calories) in Florida
in October has a probability density function $f(x) = \frac{3}{32}(x - 2)(6 - x)$ if 
$2 \le x \le 6$, and zero otherwise.  Find the \emph{mean} daily solar radiation.

\soln
The mean is given by the integral
\[
\overline{x} = \int_2^6 x\cdot \frac{3}{32}(x-2)(6-x)\, dx = \int_2^6 \frac{3x^2}{4}-\frac{3x^3}{32}-\frac{9x}{8}\, dx = 4.
\]
\vskip 10pt
\noindent
VII.  Does each of the following infinite series converge or diverge? For full
credit, you must justify your answer completely by showing how the indicated test
applies and leads to your stated conclusion.
\begin{enumerate}
\item[(A)] (10)  $\ds \sum_{n=1}^\infty \frac{n}{e^{2n}}$ -- Integral Test

\soln
The function $f(x) = xe^{-2x}$ is continuous and positive on $[1,\infty)$.  Moreover,
$f'(x) = e^{-2x}(1 - 2x) < 0$ for all $x > 1/2$.  Therefore $f(x)$ is decreasing on $[1,\infty)$,
and the Integral Test applies.  The integral can be evaluated by parts with $u =x$, $dv = e^{-2x}$
(this is also the form of \#96 in the table of integrals with $a = -2$):
\begin{eqnarray*}
\int_1^\infty x e^{-2x}\, dx &=& \lim_{b\to\infty} \int_1^b x e^{-2x}\, dx\\
&=& \lim_{b\to\infty} \left.\frac{-x}{2}e^{-2x} - \frac{1}{4}e^{-2x}\right|_1^b\\
&=& \lim_{b\to\infty} \frac{-b}{2}e^{-2b} - \frac{1}{4}e^{-2b} + \frac{3}{4e^2}\\
&=& \frac{3}{4e^2}.
\end{eqnarray*}
Since the integral converges, the series also converges.
\item[(B)] (10) $\ds \sum_{n=1}^\infty \frac{3^n}{n^2 e^n}$ -- any applicable method.

\soln
Apply the Ratio Test:
\[
\lim_{n\to\infty} \left|\frac{3^{n+1}}{(n+1)^2 e^{n+1}}\cdot \frac{n^2 e^n}{3^n} \right|
= \lim_{n\to\infty} \frac{3}{e}\frac{(n+1)^2}{n^2} = \frac{3}{e} > 1.
\]
Therefore the series diverges. 
\end{enumerate}  
\vskip 10pt
\noindent
VIII.  
\begin{enumerate}
\item[(A)] (7.5) Using the Ratio Test and testing the endpoints, 
determine the interval of convergence for the power series $\ds \sum_{n=1}^\infty \frac{2^n x^n}{n}$.

\soln
Applying the Ratio Test,
\[
\lim_{n\to\infty} \left|\frac{2^{n+1} x^{n+1}}{n+1}\cdot \frac{n}{2^n x^n}\right|
=\lim_{n\to\infty} 2 \frac{n}{n+1} |x| = 2|x|.
\]
The condition for absolute convergence is $2|x| < 1$, so $|x| < 1/2$.  At the endpoints:
\begin{itemize}
\item $x = 1/2$:  $\sum_{n=1}^\infty \frac{1}{n}$ is the harmonic series, which diverges.
\item $x = -1/2$: $\sum_{n=1}^\infty \frac{(-1)^n}{n}$ is an alternating harmonic series
which converges by the Alternating Series Test. 
\end{itemize}
Therefore the interval of convergence is $[-1/2,1/2)$.
\item[(B)] (7.5)  Find the Taylor series for the function $f(x) = \sin(x/2)$.  Express 
your answer in summation (``sigma'') notation.

\soln
We have the standard Taylor series for the sine function:
\[
\sin(u) = \sum_{n=0}^\infty (-1)^n \frac{u^{2n+1}}{(2n+1)!},
\]
and that series converges for all real $u$ to $\sin(u)$.
Therefore, substituting $u = x/2$ gives
\[
\sin(x/2) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2^{2n+1}(2n+1)!}.
\]
\end{enumerate}

\vskip 10pt
\noindent
IX.
\begin{enumerate}
\item[(A)] (5)  A drug is administered to a patient intravenously at a constant
rate of $10$mg per hour.  The patient's body breaks down the drug and removes
it from the bloodstream at a rate proportional to the amount present.  Write 
a differential equation for the function $Q(t) = $ amount of the drug present 
(in mg) in the bloodstream at time $t$ (in hours) that describes this situation.  
You \emph{do not need to solve} the equation.

\soln
The rate of change of $Q$ is equal to \emph{the rate in minus the rate out}, so 
the differential equation is 
\[
\frac{dQ}{dt} = 10 - kQ,
\]
for some constant $k$ (the constant of proportionality).  
\item[(B)] (10) Find the general solution of
the differential equation $y' = y\sqrt{e^{2x} + 1}$.

\soln
This is a separable equation, so we separate variables and integrate to solve for $y$:
\begin{eqnarray*}
\int \frac{dy}{y} &=& \int \sqrt{e^{2x} + 1}\, dx\\
\ln|y| &=& \int\frac{\sqrt{u^2 + 1}}{u}\, du \text{ (letting $u = e^x$) }\\
\ln|y| &=& \sqrt{u^2 + 1} - \ln\left|\frac{1 + \sqrt{u^2+1}}{u}\right| + C \text{ (by \#23 in the table, $a = 1$)}\\
\ln|y| &=& \sqrt{e^{2x} + 1} - \ln\left(\frac{1 + \sqrt{e^{2x}+1}}{e^x}\right) + C\\
y &=& b \frac{e^{x + \sqrt{e^{2x} + 1}}}{1 + \sqrt{e^{2x}+1}},
\end{eqnarray*}
where $b = \pm e^C$. 
\item[(C)] (10) A population $P$ of tree frogs 
is undergoing logistic growth following the differential equation
$P' = (.1)P(1-\frac{P}{100})$, $t$ in years.  
If the initial population is $P(0) = 10$ (in thousands of individuals), how long does it
take for the population to reach $45$ thousand?

\soln
Recall that the general solution of the logistic equation $P' = kP(1 - P/M)$ 
is $P = \frac{M}{1 + be^{-kt}}$, where $b$ is an arbitrary constant.  (This can be derived
by separation of variables, and integration using partial fractions.)  Here $M = 100$ and 
$k = .1$, so 
\[
P(t) = \frac{100}{1 + be^{-(0.1)t}}.
\]
If $P(0) = 10$, then 
\[
10 = \frac{100}{1 + b},
\]
so $b = 9$.  Then we want to solve
\[
45 = \frac{100}{1 + 9e^{-(0.1)t}},
\]
so $1 + 9e^{-(0.1)t} = \frac{20}{9}$, and $e^{-(0.1)t} = \frac{11}{81}$.  Therefore,
\[
t = \frac{\ln(11/81)}{-0.1} \doteq 19.97 \doteq 20 \text{ years }. 
\]
\end{enumerate}
\end{document}
