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\begin{document}
\begin{center}
\textbf{College of the Holy Cross, Fall 2016} \\
\textbf{Math 136 -- Solutions for Midterm Exam 3} \\
\textbf{December 1}
\end{center}
\vskip 10pt
\noindent
I.  [20~points] Integrate with the partial fraction method:  
${\displaystyle \int \frac{2x^2 + 3}{x^3 + 9x}\ dx}$
\vskip 10pt
\noindent
{\it Solution:}  The denominator factors as $x^3 + 9x = x(x^2 + 9)$, where
$x^2 + 9 = 0$ has no real roots.  Therefore, the partial fractions will have
the form:
$$\frac{2x^2 + 3}{x^3 + 9x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 9},$$
so 
$$2x^2 + 3 = A(x^2 + 9) + (Bx + C)x = (A + B)x^2 + Cx + 9A.$$
Equating coefficients of like powers of $x$, we get $9A = 3$, so $A = \frac{1}{3}$,
$C = 0$, and $A + B = 2$, so $B = \frac{5}{3}$.  Then we integrate:
\begin{eqnarray*}
\int \frac{2x^2 + 3}{x^3 + 9x}\ dx &=& \frac{1}{3}\int \frac{1}{x}\ dx 
+ \frac{5}{3} \int \frac{x}{x^2 + 9}\\
&=& \frac{1}{3}\ln|x| + \frac{5}{6}\ln|x^2 + 9| + C.
\end{eqnarray*}
\vskip 10pt
\noindent
II.  For each integral, say why the integral is \emph{improper}, then set up and 
evaluate the appropriate limits to determine whether the integral converges.  If so, find its value; if not, say ``does not converge.''  (Full credit will be given only for the correct limit calculation.)
\begin{enumerate}
\item[A.]  [15~points] ${\displaystyle \int_2^\infty x e^{-x}\ dx}$. 
\item[]{\it Solution:}  This is improper because of the infinite upper limit
of integration.  To determine whether the integral converges, we compute
$$\lim_{b\to\infty} \int_2^b x e^{-x}\ dx,$$
which is done by parts:  $u = x$, so $du = dx$ and $dv = e^{-x}\ dx$, so $v = -e^{-x}$:
\begin{eqnarray*}
&=& \lim_{b\to\infty} \left(\left.-xe^{-x}\right|_2^b + \int_2^b e^{-x}\ dx\right)\\
&=& \lim_{b\to \infty} \left(-\frac{b}{e^b} + \frac{2}{e^2} - e^b + \frac{1}{e^2}\right)\\
&=& \frac{3}{2e},
\end{eqnarray*}
(because $\lim_{b\to\infty} \frac{b}{e^b} = \lim_{b\to\infty}\frac{1}{e^b} = 0$
by L'Hopital's Rule).  
The integral \emph{converges} to $\frac{3}{2e}$.
\item[B.] [15~points] ${\displaystyle \int_{0}^3 \frac{1}{x^2 + 4x}\ dx}$\\
\item[] {\it Solution:} This is improper because $\frac{1}{x^2 + 4x}$ has a discontinuity (a vertical asymptote)
at $x = 0$, which is the left hand end-point of the interval $[0,3]$.  It also has
a second vertical asyptote at $x = -4$, but that is not relevant for this question.
To determine whether the integral converges we need to find
$$\lim_{a\to 0^+} \int_a^3 \frac{1}{x^2 + 4x}\ dx.$$
This integral is one we can do with partial fractions:
$$\frac{1}{x^2 + 4x} = \frac{A}{x} + \frac{B}{x + 4}$$
which yields $1 = A(x + 4) + Bx$, so $A + B = 0$ and $4A = 1$.  Therefore
$A = \frac{1}{4}$ and $B = -\frac{1}{4}$.  As a result, we have
\begin{eqnarray*} 
\lim_{a\to 0^+} \int_a^3 \frac{1}{x^2 + 4x}\ dx &=&
\lim_{a\to 0^+} \int_a^3 \frac{1}{4x} - \frac{1}{4(x+4)}\ dx\\
&=& \lim_{a\to 0^+} \left(\left.\frac{1}{4} \ln|x| - \frac{1}{4}\ln|x+4|\right|_a^3\right)\\
&=& \lim_{a\to 0^+}\frac{1}{4}\left(\ln(3) - \ln(a) - \ln(7) + \ln(a+4)\right)
\end{eqnarray*}
This does not exist because $\ln_{a\to 0^+} \ln(a)$ does not exist (the natural
log has a vertical asymptote at $0$ from the right).  Therefore the integral
{\it does not exist} (or diverges).   
\end{enumerate}
\vskip 10pt
\noindent
III.  Both parts of this problem deal with the differential equation $\frac{dy}{dx} = \frac{y}{\sqrt{x}}$.
\begin{enumerate}
\item[A.] [15~points] Find the general solution $y(x)$ of the equation 
by separating variables and integrating.
\item[] {\it Solution:}  We have 
$$\int \frac{1}{y}\ dy = \int x^{-1/2}\ dx$$
so 
$$\ln(y) = 2x^{1/2} + C$$
and hence
$$y = D e^{2 x^{1/2}}$$
where $D = \pm e^C$ is an arbitrary constant.
\item[B.] [5~points] Find the particular solution $y(x)$ satisfying the initial 
condition $y(1) = 4$ and compute the exact value of $y(2)$.  
\item[] {\it Solution:}  The equation $y(1) = 4$ says 
$4 = De^2$, so $D = 4e^{-2}$.  Then $y(2) = 4e^{-2} e^{2\sqrt{2}} = 4 e^{2\sqrt{2} - 2}.$
\end{enumerate}
\vskip 10pt
\noindent
IV. [10~points]  A population $y$ satisfies the logistic growth equation
$$\frac{dy}{dt} = (.3)y\left(1 - \frac{y}{100}\right)$$
and the initial condition $y(0) = 3$.  
What is the population level when the population is \emph{growing the fastest}?
\vskip 10pt
\noindent
{\it Solution 1:}  $\frac{dy}{dt}$ is the rate of change of the population.
So we are looking for $y$ that maximizes 
$$(.3)y\left(1 - \frac{y}{100}\right) = (.3)\left(y - \frac{y^2}{100}\right).$$
The graph of this function is a parabola opening down.  The maximum
occurs where the derivative of this quadratic function of $y$
is zero: 
$$(.3)\left(1 - \frac{y}{50}\right) = 0,$$
so $y = 50$.  
\vskip 10pt 
\noindent
{\it Solution 2:}  The general solution of this logistic equation is 
$$y = \frac{100}{1 + d e^{-.3t}}.$$ 
With the initial condition $y(0) = 3$, we have $3 = \frac{100}{1 + d}$, so
$d = 97/3$.  Then the rate of change of $y$ as a function of $t$ is given by:
$$\frac{dy}{dt} = \frac{-970 e^{-.3t}}{(1 + \frac{97}{3}e^{-.3t})^2}.$$
To determine when this is a maximum, we can differentiate again with the 
quotient rule.  After simplification:
$$\frac{d^2y}{dt^2} = \frac{(25043e^{-0.3t} -7857)e^{-0.3t}}{(1 + \frac{97}{3}e^{-.3t})^3}$$
This is zero when 
$$25043e^{-0.3t} -7857 = 0, \quad {\rm or} \quad t \doteq 11.59.$$
Then $y(11.59) \doteq 50$.  
\vskip 10pt
\noindent
V.  
\begin{enumerate}
\item[A.] [10~points] A function $y = f(x)$ is plotted below.
\begin{center}
\includegraphics[width=1.5truein,height=1.5truein]{1363V.png} 
\end{center}
\noindent
Check the appropriate boxes for statements about $\int_0^4 f(x)\ dx$:
\item[] {\it Solution:}  The graph is concave up, which shows that
the midpoint approximation is an {\it underestimate} and 
the trapezoidal rule approximation is an {\it overestimate}.
\item[B.] [10~points]  Suppose you know that the function $f(x)$ plotted
here is $g''(x)$ for some other function $g(x)$.  How big would you 
need to take $N$ to get a midpoint rule approximation to $\int_0^4 g(x)\ dx$
with error $< 10^{-5}$?
\item[] {\it Solution:}  From the error bound for the midpoint rule, we 
have
$$|{\rm Error}(M_N)| \le \frac{K_2\cdot (4-0)^3}{24 N^2},$$
where $K_2 = \max_{x\in [0,3]} |g''(x)|$.
From the graph we can take $K_2 = 1.5$ and then we want
$$\frac{96}{24 N^2} < 10^{-5} = .00001.$$
This is equivalent to
$$N^2 > \frac{96}{24\cdot 0.00001},$$
or 
$$N > \sqrt{\frac{96}{24\cdot 0.00001}} = 2000.$$
\end{enumerate}
\end{document}

