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\begin{center}

\textbf{College of the Holy Cross, Spring Semester, 2018} \\
\textbf{MATH 134 Solutions for Practice Final Exam} \\
\textbf{May 3, 2018} \\
\end{center}
\vskip 10pt
\noindent
I.  
\begin{enumerate}
\item[(A)] 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.png}
\end{center}
Let $F(x) = \int_{-1}^x f(t)\, dt$, where $f(t)$ is the function from part (B).  (There
is a part of the graph to the left of $x = 0$ that you are not seeing.  
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.
By the interval union property for integrals 
$$\int_{-1}^x f(x)\ dx = \int_{-1}^0 f(x)\ dx + \int_0^x f(x)\ dx$$
and we can tell $\int_{-1}^0 f(x)\ dx = 3$ from the first entry in the table.  Then
to find the others we just compute areas and add that 3:
\[
\begin{array}{|c||r|r|r|r|r|r|r|}
\hline
x          & 0 & 1 & 2 & 3 & 4 & 5 & 6 \strut \\
\hline
F(x)       & \quad 3 & \qquad 4 & \qquad 5 & \qquad 6.5 & \qquad 9 & \qquad 11 & \qquad 11\strut \\
\hline
\end{array}
\]
\item[(B)] 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}

\vskip 10pt
\noindent
II.  Compute the following integrals.  
\begin{enumerate}
\item[(A)] $\ds \int \frac{x^4 - 3\pi^3 + \sqrt{x}}{x^{2/3}}\, dx$

\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)] $\ds \int x^3 e^{x^2}\, dx$  (integrate by parts)

\soln The correct choice is $u = x^2$ and $dv = x e^{x^2}\ dx$ (note that you 
need the $x$ with the exponential in order to integrate!)  Hence $du = 2x\ dx$
and $v = \frac{1}{2} e^{x^2}$.  Then by the parts formula
\[
\int x^3 e^{x^2}\, dx = \frac{1}{2} x^2 e^{x^2} - \int x e^{x^2}\ dx = 
 \frac{1}{2} x^2 e^{x^2} -  \frac{1}{2} e^{x^2} + C.
\]
\item[(C)] $\ds \int \frac{\csc^2(5x)\, dx}{\cot(5x) + 7}$

\soln This one can 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)] $\ds \int_1^e x^5 \ln(x)\, dx$.  

\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 Fundamental 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)] $\ds \int \frac{1}{\sqrt{16 + x^2}}\, dx$

\soln
This can be done
by the trigonometric substitution $x = 4\tan\theta$, so $dx = 4\sec^2\theta\,d\theta$, 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 formula ST4 from our table, 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:
$$\ln|\sqrt{16+x^2} + x| + C.$$
\item[(F)]  $\ds \int \frac{x}{(x^2 + 1)(x + 1)}\, dx$
\soln
Using partial fractions, 
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)] Use a midpoint Riemann sum with $n = 4$ to approximate the integral
$$\int_1^3 \sqrt{1 + 9x^4}\ dx.$$ 

\soln
The $R_4$ Riemann sum approximation is
\begin{eqnarray*}
\int_1^3 \sqrt{1 + 9x^4}\, dx &\doteq& \left(\sqrt{1 + 9(1.5)^4} + \sqrt{1 + 9(2)^4} + 
\sqrt{1 + 9(2.5)^4} + \sqrt{1 + 9(3)^4} \right)(.5) \\
&\doteq& 32.3302.
\end{eqnarray*}

\item[(B)] 

\soln  The midpoint approximation is an \emph{overestimate} because
$\sqrt{1 + 9x^4}$ is increasing on $[1,3]$.  
\end{enumerate}
\vskip 10pt

\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)]  Compute the area of the region $R$. 

\soln
Here is a sketch of the region:

\begin{center}
\includegraphics[width=2.0truein,height=2.0truein]{136FVS.png}
\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)] 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)] 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 easiest way (by far) is to use cylindrical shells:
$$V = \int_0^1 2\pi x (9 - x^2 - 2x)\ dx$$
If you use slices,
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. 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 
$$\frac{dN}{dt} = kN(10 - N).$$
(Note that this can be put into the form of a logistic equation:
$$\frac{dN}{dt} = (10k)N\left(1 - \frac{N}{10}\right);$$
the 10 plays the role of the carrying capacity.)  
\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)]  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. 
\begin{enumerate}
\item[(A)]  Find the sum of the finite geometric series with alternating signs:
$$3 - \frac{3}{2} + \frac{3}{2^2} - \frac{3}{2^3} + \cdots + \frac{3}{2^6}$$
What happens if you continue adding terms of the same form indefinitely?  That is, 
what is the sum 
$$\sum_{k = 0}^\infty \frac{(-1)^k 3}{2^k}?$$

\soln
The geometric series has $c = 3$ and $a = \frac{-1}{2}$, so the sum 
is
$$\frac{3\left(1 - \left(\frac{-1}{2}\right)^7\right)}{1 - \left(\frac{-1}{2}\right)} = \frac{3(1 + \frac{1}{128})}{\frac{3}{2}} = \frac{257}{128}.$$
The infinite sum is 
$$\lim_{N\to\infty} \frac{3\left(1 + \left(\frac{-1}{2}\right)^N\right)}{\frac{3}{2}} = 2.$$
 
\item[(B)]  What would be the monthly payment on a $\$500,000$ 30-year fixed-rate
mortgage at $3.5\%$ annual interest.

\soln
The future value of the $\$500000$ is 
$$FV = \$500000 \left(1 + \frac{.035}{12}\right)^{12\cdot 30} = \$1,426,643.75$$
(rounding to the nearest cent).  On the other hand a stream of monthly payments $P$
extending over $30$ years has future value
$$P\cdot \frac{\left(1 + \frac{.035}{12}\right)^{360} - 1}{1 + \frac{.035}{12} - 1} 
\doteq P \cdot 635.4128591$$
Setting the future values equal, we solve for $P$:
$$P = \frac{\$1,426,643.75}{635.4128591} \doteq \$2245.22$$
(You need to be quite well-off to be able to afford a loan this big!!)

\end{enumerate}

\end{document}
