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\begin{document}

\begin{center}

\textbf{College of the Holy Cross, Fall 2016} \\
\textbf{MATH 136, Section 02, Final Exam} \\
\textbf{Solutions} \\
\end{center}

\noindent
I.  
\begin{enumerate}
\item[(A)] (5) Compute the derivative of the function $\ds g(x) = \int_0^{3x} \frac{\sin(t)}{t}\, dt$.
\item[] {\it Solution:} By part 1 of the FTC and the Chain Rule, this is 
$\frac{\sin(3x)}{3x}\cdot 3 = \frac{\sin(3x)}{x}$.
\item[(B)] (10) Let 
$\ds
f(x) = \begin{cases} 6x+2 & \text{ if } 0 \le x \le 1\\
                    -x+9 & \text{ if } 1 < x \le 4.\\
                   5 & \text{ if } 4 < x \le 5
\end{cases}$  
which is plotted in the graph at the top of the next page.  

Let $F(x) = \int_0^x f(t)\, dt$, where $f(t)$ is the function above.  
Complete the following table of values for $F(x)$:
\[
\begin{array}{|c||r|r|r|r|r|r|}
\hline
x          & 0 & 1 & 2 & 3 & 4 & 5 \strut \\
\hline
F(x)       & \quad 0 & 5  & 25/2 & 19 & 49/2 & 59/2 \strut \\
\hline
\end{array}
\]
\item[(C)]  (5)  On which interval(s) contained in $(0,5)$ is the graph 
$y = F(x)$ concave down?
\item[] {\it Solution:} Concave down when $F'(x) = f(x)$ is decreasing, so 
$(1,4)$. 
\end{enumerate}


\noindent
II.  
\begin{enumerate}
\item[(A)] (5)  Use a left-hand Riemann sum with $n = 4$ to approximate 
$\ds \int_0^1 e^{-x^2/2}\ dx$.
\item[] {\it Solution:}  The value is
$$e^{-0^2/2}(.25) + e^{-(.25)^2/2}(.25) + e^{-(.5)^2/2}(.25) + e^{-(.75)^2/2}(.25) \doteq .9016.$$
\item[(B)] (5) Use a midpoint Riemann sum with $n = 4$ to approximate
$\ds \int_0^1 e^{-x^2}\ dx$.
\item[] {\it Solution:} The value is
$$e^{-(.125)^2}(.25) + e^{-(.375)^2}(.25) + e^{-(.625)^2}(.25) + e^{-(.875)^2}(.25) \doteq 
.8572$$
\item[(C)] (10) Check the appropriate boxes and fill in the blank:
\item[] The left-hand sum is an overestimate because: $e^{-x^2/2}$ is decreasing
on $[0,1]$.  (The first derivative is $-x e^{-x^2/2}$, which is negative for 
all $x > 0$.)
\item[] The midpoint approximation is a overestimate,  
because $y = e^{-x^2/2}$ is concave down on $[0,1]$.  (The second derivative
is $(x^2-1)e^{-x^2/2}$, which is negative for all $x \in (-1,1)$.)
\end{enumerate}

\begin{figure}
\begin{center}
\includegraphics[width=2.5truein,height=2.5truein]{136FI.png}
\caption{Figure for problem I.}
\end{center}
\end{figure}

\noindent
III.  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) $\ds \int \frac{x^{2/3} - x^4 + \sqrt{x}}{x^{2/3}}\, dx$
\item[] {\it Solution:}  This equals
$$\int 1 - x^{10/3} + x^{-1/6}\ dx = x - \frac{3}{13}x^{13/3} + \frac{6}{5} x^{5/6} + C.$$
\item[(B)] (5) $\ds \int x \cos(x^2)\, dx$
\item[] {\it Solution:} Let $u = x^2$, then $du = 2x\ dx$, so the integral is
$$\frac{1}{2} \int \cos(u)\ du = \frac{1}{2}\sin(u) + C = \frac{1}{2}\sin(x^2) + C.$$
\item[(C)] (10) $\ds \int \frac{\sec^2(5x)\, dx}{(\tan(5x) + 7)^5}$
\item[] {\it Solution:} Let $u = \tan(5x) + 7$.  Then $du = 5\sec^2(5x)$.  The integral
equals
$$\frac{1}{5} \int u^{-5}\ du = \frac{-1}{20} u^{-4} + C = \frac{-1}{20(\tan(5x)+7)^4} + C.$$ 
\item[(D)] (10) $\ds \int_1^{e^2} x^3 \ln(x)\, dx$.  
\item[] {\it Solution:}  Integrate by parts with $u = \ln(x)$, $dv = x^3\ dx$.  Then
$du = \frac{1}{x}\ dx$ and $v = \frac{x^4}{4}$, so 
\begin{eqnarray*}
\int_1^{e^2} x^3\ln(x)\ dx &=&
\left. \frac{x^4\ln(x)}{4}\right|_1^{e^2} - \int_1^{e^2} \frac{x^3}{4}\ dx\\
&=& \frac{e^{8}}{2} - \left. \frac{x^4}{16}\right|_1^{e^2}\\
&=& \frac{e^{8}}{2} - \frac{e^{8}}{16} + \frac{1}{16}\\
&=& \frac{7 e^{8} + 1}{16}.
\end{eqnarray*}
\item[(E)] (15) $\ds \int \frac{x^2}{\sqrt{16 - x^2}}\, dx$
\item[] {\it Solution:}  By trigonometric substitution with $x = 4\sin\theta$, so 
$dx = 4\cos\theta\ d\theta$.  We get
$$\int \frac{16 \sin^2\theta\cdot 4\cos\theta}{\sqrt{16 - 16\sin^2\theta}}\ d\theta = 
16 \int \sin^2\theta\ d\theta.$$
Using the double angle formula $\sin^2\theta = \frac{1}{2}(1 - \cos(2\theta))$, 
this becomes
$$= 8 \int 1 - \cos(2\theta)\ d\theta = 8\theta - 4\sin(2\theta) + C =
8\theta - 8\sin\theta\cos\theta + C.$$
Then converting back from $x = 4\sin\theta$, 
$\theta = \sin^{-1}\left(\frac{x}{4}\right)$ and $\cos\theta = \frac{\sqrt{16 - x^2}}{4}$.
So the final answer is
$$ = 8\sin^{-1}\left(\frac{x}{4}\right) - \frac{x\sqrt{16-x^2}}{2} + C.$$
\item[(F)] (15) $\ds \int \frac{x + 1}{(x - 3)(x^2 + 1)}\, dx$
\item[] {\it Solution:}  By partial fractions:
$$\frac{x+1}{(x - 3)(x^2+1)} = \frac{A}{x-3} + \frac{Bx + C}{x^2 + 1},$$
so 
$$x + 1 = A(x^2 + 1) + (B x + C)(x - 3) = (A + B)x^2 + (C - 3B)x + A - 3C$$
Hence
$$A + B = 0, -3B + C = 1, A - 3C = 1$$
From the second equation $C = 1 + 3B$, and the last equation becomes $A - 9B = 4$.
Solving that with the first equation yields $A = \frac{2}{5}$, $B = \frac{-2}{5}$
and $C = \frac{-1}{5}$.  Then 
$$\int \frac{x+1}{(x-3)(x^2+1)}\ dx = \int\frac{2/5}{x - 3}\ dx + \int \frac{(-2/5)x}{x^2+1}\ dx + \int \frac{-1/5}{x^2 + 1}\ dx.
$$
The final answer is 
$$\frac{2}{5}\ln|x - 3| - \frac{1}{5}\ln(x^2 + 1) - \frac{1}{5}\tan^{-1}(x) + C.$$
\end{enumerate}
\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)] (10) $\ds \int_1^3 \frac{1}{{\root 3\of {x - 1}}}\, dx.$
\item[] {\it Solution:}  Here the discontinuity is at the left endpoint
of the interval, so we want
\begin{eqnarray*}
\lim_{a\to 1^+} \int_a^3 (x - 1)^{-1/3}\ dx &=& 
\lim_{a \to 1^{+}}\left. \frac{3}{2}(x - 1)^{2/3}\right|_a^3\\
&=& \lim_{a\to 1^+} \frac{3}{2}(2^{2/3} - (a-1)^{2/3})\\
&=& \frac{3}{2}2^{2/3}.
\end{eqnarray*}
Therefore the integral converges to $\frac{3}{2} 2^{2/3}$. 
\item[(B)] (10) $\ds \int_0^\infty \frac{1}{x^2 + 4x + 5}\, dx$.  (Hint: There are 
two ways to do this.  You can compute the integral if you complete
the square.  Or you can use $\frac{1}{x^2 + 4x + 5} < \frac{1}{x^2}$ for all $x > 0$..)
\item[] {\it Solution:}  This integral is improper because of the
infinite limit of integration.  Completing the square, we get
\begin{eqnarray*}
\lim_{b\to\infty} \int_0^b \frac{1}{(x + 2)^2 + 1}\ dx
&=& \lim_{b\to\infty} \left.\tan^{-1}(x + 2)\right|_0^b\\
&=& \lim_{b\to\infty} \tan^{-1}(b+2) - \tan^{-1}(2)\\
&=& \frac{\pi}{2} - \tan^{-1}(2).
\end{eqnarray*}
This integral also converges, to $\frac{\pi}{2} - \tan^{-1}(2) \doteq .4636$.  
\end{enumerate}
\vskip 10pt
\noindent
V.  A region $R$ in the plane is bounded by the graphs $y = x^2$, $y = x + 6$.
See Figure 2. 
\begin{figure}
\begin{center}
\includegraphics[width=1.6truein,height=1.6truein]{136FV.png}
\caption{Figure for problem V.}
\end{center}
\end{figure}
\begin{enumerate}
\item[(A)] (10)  Compute the area of the region $R$.
\item[] {\it Solution:}  The parabola crosses the line where $x^2 - x - 6 = (x - 3)(x + 2) = 0$, so $x = -2,3$.  The area is
\begin{eqnarray*}
\int_{-2}^3 x + 6 - x^2\ dx &=& \left. \frac{x^2}{2} + 6x - \frac{x^3}{3}\right|_{-2}^3\\
&=& \frac{9}{2} + 18 - 9 - 2 + 12 - \frac{8}{3}\\
&=& \frac{125}{6}.
\end{eqnarray*}
\item[(B)] (10) Compute the volume of the solid obtained by rotating $R$ about the $x$-axis.
\item[] {\it Solution:}  The cross-sections by planes perpendicular to the $x$-axis
 are washers with inner radius $r_{in} = x^2$ and outer radius $r_{out} = x + 6$.
Therefore the integral that computes the volume is 
\begin{eqnarray*}
V &=& \int_{-2}^3 \pi(x + 6)^2 - \pi(x^2)^2\ dx\\
&=& \pi \int_{-2}^3 x^2 + 12x + 36 - x^4\ dx\\
&=& \pi \left. \frac{x^3}{3} + 6x^2 + 36x - \frac{x^5}{5}\right|_{-2}^3\\
&=& \pi \left(9 + 54 + 108 - \frac{243}{5} + \frac{8}{3} - 24 + 72 - \frac{32}{5}\right)\\
&=& \frac{500\pi}{3}.
\end{eqnarray*}
\item[(C)] (10) Set up the integral(s) to compute the volume of the solid
obtained by rotating $R$ about the line $y = 12$.  \emph{You do not need to compute the value.}
\item[] {\it Solution:}  Also washer cross-sections.  The integral is 
$$V = \int_{-2}^3 \pi (12 - x^2)^2 - \pi(6 - x)^2\ dx.$$
\end{enumerate}
\vskip 10pt
VI.
\begin{enumerate}
\item[(A)] (10)  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 $y(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.
\item[] {\it Solution:}  The equation is $\ds \frac{dy}{dt} = 10 - ky$.
The idea is similar to the derivation of the mixing problem differential
equation we did in class.  The $\frac{dy}{dt}$ is the rate of change of the
amount of the drug, the 10 represents the inflow of the drug into the blood
stream via the IV, and the $-ky$ represents the drug being removed at a rate
proportional to the amount.  I wrote the constant of proportionality here
as $-k$ to emphasize that this is the term representing the outflow.  The
value of $k$ itself would be $> 0$.
\item[(B)] (10) Find the general solution of
the differential equation $\frac{dy}{dx} = xy\sqrt{x^2 + 1}$.
\item[] {\it Solution:}  This is a separable equation.  After separating variables
$$\int \frac{dy}{y} = \int x (x^2 + 1)^{1/2}\ dx.$$
So 
\begin{eqnarray*}
\ln|y| &=& \frac{1}{3}(x^2 + 1)^{3/2} + C\\
y &=& D e^{\frac{1}{3}(x^2 + 1)^{3/2}},
\end{eqnarray*}
where the arbitrary constant $D = \pm e^C$
\item[(C)] (10) Let $y(t)$ represent the population of a colony of tree frogs 
that is undergoing logistic growth following the differential equation
$\frac{dy}{dt} = (.1)y(1-\frac{y}{100})$, $t$ in years.  
If the initial population is $y(0) = 10$, how long does it
take for the population to reach $45$?
\item[] {\it Solution:}  The general solution of this logistic equation is
$$y = \frac{100}{1 + d e^{-(0.1)t}}$$
From the initial condition 
$$10 = \frac{100}{1 + d} \Rightarrow d = 9$$
Then we solve for $t$:
$$45 = \frac{100}{1 + 9 e^{-(0.1)t}}$$
which gives 
$$t = -10 \ln(11/81) \doteq 19.96$$
or about 20 years.  
\end{enumerate}
\vskip 10pt
\noindent
VII. 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)] (5)  Use the Integral Test to determine if
$\ds \sum_{n=1}^\infty \frac{n}{e^{2n}}$ converges.
\item[] {\it Solution:}  We need to decide whether $\int_1^\infty x e^{-2x}\ dx$
converges.  Integrating by parts with $u = x$, $dv = e^{-2x}\ dx$, we have
\begin{eqnarray*}
\lim_{b\to\infty} \left. - \frac{x}{2 e^{2x}} - \frac{1}{4 e^{2x}}\right|_1^b
&=& \lim_{b\to\infty} \left(- \frac{b}{2 e^{2b}} - \frac{1}{4e^{-2b}} +  \frac{3}{4e^2}\right)\\
&=& \frac{3}{4e^2}.
\end{eqnarray*}
(We used L'Hopital's Rule here to see $\lim_{b\to \infty} \frac{b}{2 e^{2b}} = 0$.)
Since the improper integral converges, the series does too.
\item[(B)] (15) Using the Ratio Test and testing convergence at the endpoints, 
determine the interval of convergence for the power series 
$$\sum_{n=1}^\infty \frac{x^n}{n 3^n}.$$
\item[] {\it Solution:} The Ratio Test gives
$$\lim_{n\to\infty} \left|\frac{x^{n+1}}{(n+1)3^{n+1}}\cdot \frac{n3^n}{x^n}\right|
= \lim_{n\to\infty} \frac{1}{3} \frac{n}{n+1} |x| = \frac{1}{3}|x|.$$
This says the series converges absolutely when $|x|/3 < 1$, or $|x| < 3$.
It diverges if $|x| = 3$.  Substituting $x = 3$ into the series
we have $\sum_{n=1}^\infty \frac{1}{n}$, the divergent harmonic series.
On the other hand $x = -3$ gives $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$, 
which converges by the Alternating Series Test.  Thus the full interval 
of convergence is $[-3,3)$, or $-3 \le x < 3$.  
\end{enumerate}
\vskip 10pt
\noindent
Extra Credit. (10) In economics, the \emph{multiplier effect} refers to the fact
that when there is an injection of money to consumers in an economy, the consumers
spend a certain proportion of it. That amount recirculates through the enconomy and
adds additional income, which comes back to the consumers and they
spend the same percentage.  The process repeats \emph{indefinitely} circulating
additional money through the economy.  Suppose that in order to stimulate 
the economy, the government cuts taxes by \$ 50 billion, thereby injecting
that much money back to consumers.  If consumers 
save 10\% of the money they get and spend the other 90\%, what is the total
additional spending circulated through the economy by the tax cut?
\vskip 10pt
\noindent
{\it Solution:} 
The total is the sum of the geometric series
$$50 + 50(.9) + 50(.9)^2 + 50(.9)^3 + \cdots = \frac{50}{1 - .9} = 500$$
(units are billions of dollars).   
\end{document}
