%% MATH 136-01 Final Exam, Spring 2014
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\begin{document}

\begin{center}

\textbf{College of the Holy Cross, Spring Semester, 2014} \\
\textbf{MATH 136, Section 01, Solutions for Final Exam} \\
\textbf{Wednesday, May~14, 11:30~AM} \\
Professor Little
\end{center}
\vskip 10pt
\noindent
I.  
\begin{enumerate}
\item[(A)] (10) Let 
$\ds
f(x) = \begin{cases} 1 & \text{ if } 0 \le x \le 2\\
                    x-1 & \text{ if } 2 < x \le 4.\\
                   11-2x & \text{ if } 4 < x \le 6
\end{cases}$  
whose graph is shown here:
\soln
\begin{center}
\includegraphics[width=2.5truein,height=2.5truein]{132FIS.ps}
\end{center}
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:
\[
\begin{array}{|c||r|r|r|r|r|r|r|}
\hline
x          & 0 & 1 & 2 & 3 & 4 & 5 & 6 \strut \\
\hline
F(x)       & \quad 0 & \qquad 1 & \qquad 2 & \qquad 3.5 & \qquad 6 & \qquad 8 & \qquad 8\strut \\
\hline
\end{array}
\]
\item[(B)] (10) Compute the derivative of the function 
$\ds g(x) = \int_0^{2x} \frac{\cos(t)}{t^2}\, dt$.

\soln By the first part of the Fundamental Theorem of Calculus and the Chain Rule for derivatives:
\[
g'(x) = \frac{\cos(2x)}{(2x)^2}\cdot 2 = \frac{\cos(2x)}{2x^2}.
\] 
\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.  If you use a table entry, state which one.
\begin{enumerate}
\item[(A)] (5) $\ds \int \frac{x^4 - 3\pi^3 + \sqrt{x}}{x^{2/3}}\, dx$

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

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

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

\soln Integrate by parts with $u = \ln(x)$ and $dv = x^5\, dx$.  Then $du = \frac{1}{x}\, dx$
and $v = \frac{1}{6}x^6$ and by the integration by parts formula,
\[
\int x^5 \ln(x)\, dx = \frac{x^6}{6}\ln(x) - \int \frac{1}{6} x^6\cdot \frac{1}{x} \, dx = 
\frac{x^6}{6}\ln(x) - \frac{x^6}{36} + C.
\]
For the definite integral, we apply the Evaluation Theorem to get
\[
\int_1^e x^5\ln(x)\, dx = \left. \frac{x^6}{6}\ln(x) - \frac{x^6}{36}\right|_1^e 
= \frac{e^6}{6}\ln(e) - \frac{e^6}{36} - \frac{1}{6}\ln(1) + \frac{1}{36} = \frac{5e^6 + 1}{36}.
\]
\item[(E)] (10) $\ds \int \frac{1}{\sqrt{16 + x^2}}\, dx$

\soln
This can be done either with \# 25 in the table or 
by the trigonometric substitution $x = 4\tan\theta$.  With the latter
approach, $dx = 4\sec^2\theta\,d\theta$, and
$\sqrt{16 + x^2} = \sqrt{16(1 + \tan^2\theta)} = 4\sec\theta$.  This simplifies to
\[
\int \frac{1}{4\sec\theta}\cdot 4\sec^2\theta\, d\theta = 
\int \sec\theta\, d\theta.
\]
Now we use \# 14 in the table (or the memorized form), then convert back to $x$:
\begin{eqnarray*}
\int \sec\theta\, d\theta &=& \ln|\sec\theta + \tan\theta| + C\\
&=&  \ln\left|\frac{\sqrt{x^2 + 16}}{4} + \frac{x}{4}\right| + C.
\end{eqnarray*}
Using properties of logarithms and incorporating $-\ln(4)$ with 
the constant, this can also be written in the form given in \# 25 from the table:
$$\ln|\sqrt{16+x^2} + x| + C.$$
\item[(F)] (10) $\ds \int \frac{x}{x^3 + x^2 + x + 1}\, dx$
\soln
By partial fractions -- we start by factoring $x^3 + x^2 + x + 1 = (x+1)(x^2+1)$.  Then 
we must solve for $A,B,C$ to make:
\[
\frac{1}{(x+1)(x^2 + 1)} = \frac{A}{x+1} + \frac{Bx + C}{x^2 + 1}.
\]
Clearing denominators,
\[
1 = A(x^2 + 1) + (Bx + C)(x+1),
\]
so equating coefficients, $A + B = 0$, $B + C = 0$, and $A + C = 1$.  Solving simultaneously,
\[
A = \frac{1}{2}, \qquad B = \frac{-1}{2}, \qquad C = \frac{1}{2}.
\]
Then 
\begin{eqnarray*}
\int \frac{1}{(x+1)(x^2 + 1)}\, dx &=& \int \frac{\frac{1}{2}}{x+1} + \frac{\frac{-x}{2} + \frac{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)   Set up an integral to compute the length of the curve
$y = x^3$ from $x = 1$ to $x = 3$.

\soln
The integral is 
$$\int_1^3 \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\, dx = \int_1^3 \sqrt{1 + 9x^4}\, dx$$
\item[(B)] (10)  Use a midpoint Riemann sum with $n = 4$ to approximate your 
integral from part A.

\soln
The midpoint Riemann sum approximation is
\begin{eqnarray*}
\int_1^3 \sqrt{1 + 9x^4}\, dx &\doteq& \left(\sqrt{1 + 9(1.25)^4} + \sqrt{1 + 9(1.75)^4} + 
\sqrt{1 + 9(2.25)^4} + \sqrt{1 + 9(2.75)^4} \right)(.5) \\
&\doteq& 25.98.
\end{eqnarray*}

\item[(C)] (5) 

\soln  The midpoint approximation is an \emph{underestimate} because
$\sqrt{1 + 9x^4}$ 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
IV.  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)] (20)  Compute the area of the region $R$. 

\soln
Here is a sketch of the region:

\begin{center}
\includegraphics[width=2.0truein,height=2.0truein]{132FVS.ps}
\end{center}

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[(B)] (20) 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[(C)] {\it Extra Credit (10)} 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
V. (20) The daily solar radiation $x$ per square meter (in hundreds of calories) in Florida
in October has a probability density function $f(x) = c (x - 2)(6 - x)$ if 
$2 \le x \le 6$, and zero otherwise.  Find value of $c$ and 
the probability that the daily solar radiation
per square meter is greater than $4$.  

\soln
We must have 
$$1 = \int_2^6 c (x - 2)(6 - x)\, dx = c\left(\left. -\frac{x^3}{3} + 4x^2 - 12x\right|_2^6\right) = 
\frac{32c}{3}.$$
Therefore $c = \frac{3}{32}$.  Then the probability that $x > 4$ is given by the integral
\[
\overline{x} = \int_4^6 \frac{3}{32}(x-2)(6-x)\, dx = 
\int_4^6 \frac{3x}{4}-\frac{3x^2}{32}-\frac{9}{8}\, dx = \frac{1}{2}.
\]
\vskip 10pt
\noindent
VI.  An avian flu epidemic has broken out in Birdsburgh, a large city 
with total population 10 million.  Let $N$ be the number, in millions,
of people who have been infected, as a function of time $t$ in weeks.  The Birdsburgh
Public Health department proposes the model that the rate of change of $N$ is 
proportional to the product of the number of infected people ($N$) and 
the number of people not yet infected.   
\begin{enumerate}
\item[(A)] (10)  Write the proposed model above as a differential equation, calling
the constant of proportionality $k$.

\soln
If $N$ people have been infected ($N$ in millions), then the number who 
have not been infected is $10 - N$ (millions).  So
the differential equation is 
$$\frac{dN}{dt} = kN(10 - N).$$

\item[(B)] (5)  The function $N(t) = 10/(1 + 9999 e^{-t})$ should be a solution of your 
differential equation from part A.   What is the value of $k$? 

\soln 
For this $N$, 
\begin{equation}
\label{der}
\frac{dN}{dt} = \frac{-99990 e^{-t}}{(1 + 9999 e^{-t})^2}
\end{equation}
and
\begin{eqnarray}
\label{right}
N(10 - N) &=& \frac{10}{1 + 9999e^{-t}}\cdot \left(10 - \frac{10}{1 + 9999e^{-t}}\right) \nonumber\\  
&=& \frac{10\cdot (-99990e^{-t})}{(1 + 9999e^{-t})^2} 
\end{eqnarray}
Hence comparing $\eqref{der}$ and $\eqref{right}$, we must 
have $k = \frac{1}{10} = .1$.   (Note:  This could also be done by 
recognizing that the differential equation can be rearranged to the form
of a logistic equation with constant $10k$:  $\frac{dN}{dt} = (10k) N\left(1 - \frac{N}{10}\right)$.
The general solution of this is 
$$N = \frac{10}{1 + c e^{-(10k)t}}$$
So if the given function is a solution we must have $c = 9999$ and $10 k = 1$, so $k = .1$.)
 

\item[C)]  (10)  If the epidemic proceeds according the function given in part B,
how many weeks will pass before the number of infected people reaches 1 million?

\soln
We must solve 
$1 = \frac{10}{1 + 9999e^{-t}}$.
So $1 + 9999e^{-t} = 10$, and $t = -\ln(9/9999) = \ln(1111) \doteq 7.0$ weeks. 
Note that the units of $N$ are millions of people so the left side of the equation is
$N = 1$, not $N = 1000000$.  
\end{enumerate}

\vskip 10pt
\noindent
VII.  (25) Using the Ratio Test and testing the endpoints, 
determine the interval of convergence for the power series 
$\ds \sum_{n=1}^\infty \frac{4^n x^n}{n^2}$.

\soln
We have 
\begin{eqnarray*}
\lim_{n\to\infty} \left|\frac{4^{n+1} x^{n+1}}{(n+1)^2} \cdot \frac{n^2}{4^n x^n}\right|
&=& \lim_{n\to\infty} 4 \frac{(n+1)^2}{n^2} |x|\\
&=& 4|x|.
\end{eqnarray*}
We want $4|x| < 1$ or $|x| < \frac{1}{4}$
for absolute convergence, so the radius of convergence is $\frac{1}{4}$.  Then
at $x = \frac{1}{4}$, the series becomes
$$\sum_{n=1}^\infty \frac{1}{n^2}.$$
This is the $p$-series with $n = 2$, so it converges.
At $x = \frac{-1}{4}$, the series becomes
$$\sum_{n=1}^\infty \frac{(-1)^n}{n^2}.$$
This is an alternating series with $\frac{1}{n^2}$ decreasing to $0$ as $n \to \infty$.
So it converges by the Alternating Series Test.  The interval of convergence is 
$\left[-\frac{1}{4},\frac{1}{4}\right]$.  

\end{document}
