%% MATH 132-01 Final Exam Solutions, Spring 2005
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\begin{document}

\begin{center}
\textbf{Holy Cross College, Spring Semester, 2005} \\
\textbf{MATH 132, Section 01, Final Exam Solutions} \\
\textbf{May 11} \\
\end{center}

\noindent
1. The following graph shows the function $y = f(x)$.

\centerline{\psfig{figure=1321a.ps,height=2.0in,width=2.4in,angle=-90}}

\noindent
a. [10~points] Let $F(x)$ be the antiderivative of $f(x)$ with $F(0) = 0$.
Compute the entries in the following table of values:

\noindent
{\bf Solution:}
\[
\begin{array}{|c||r|r|r|r|r|}
\hline
x          & 0 & 1 & 2& 3& 4\Strut \\
\hline
F(x)       & \quad 0 & -1 & 0 & .5 & -.5 \Strut \\
\hline
\end{array}
\]
For example, the area between the graph and the $x$-axis from $x = 0$
to $x = 1$ (counted with a negative sign) is $-1$.  So by the Fundamental
Theorem of Calculus:
$$-1 = \int_0^1 f(x)\ dx = F(1) - F(0) = F(1) - 0\Rightarrow F(1) = -1$$
The others are obtained similarly.
\sep
\noindent
b. [10~points] Sketch the graph $y = F(x)$ on the interval $0 \le x \le 4$:

\vskip 10pt
\centerline{\psfig{figure=1321bs.ps,height=2.0in,width=2.0in,angle=-90}}

\noindent 
2. Integrate each of the following.  You must show all work
and cite any table entries you use for full credit.

\vskip 10pt
a. [5~points]  $\ds \int \frac{x^2 + 3x + 5}{x^{3/2}}\ dx$

\vskip 10pt
\noindent
{\bf Solution:}  Divide $x^{3/2}$ into each term on the top, then integrate:
\begin{eqnarray*}
\int \frac{x^2 + 3x + 5}{x^{3/2}}\ dx & = & \int x^{1/2} + 3 x^{-1/2} + 5x^{-3/2}\\
&=& \frac{2}{3} x^{3/2} + 6 x^{1/2} - 10 x^{-1/2} + C \\
\end{eqnarray*}
\sep
\noindent
b. [10~points]  $\ds \int \frac{e^t + 1}{(e^t + t)^2}\ dt$ 

\vskip 10pt
\noindent
{\bf Solution:}  Let $u = e^t + t$, so $du = (e^t + 1)\ dt$.  Then the 
form is
$$\int \frac{du}{u^2} = \frac{-1}{u} + C = \frac{-1}{e^t + t} + C.$$
\sep
\noindent
c. [10~points]  $\ds \int \frac{dx}{(9 - x^2)^{3/2}}$
\vskip 10pt
\noindent
{\bf Solution:}  We use the trigonometric substitution 
$x = 3\sin\theta$, so $dx = 3\cos\theta$ and 
\begin{eqnarray*}
\int \frac{dx}{(9 - x^2)^{3/2}} &=& \int \frac{3\cos\theta\ d\theta}{27\cos^3\theta}\\
&=&\frac{1}{9}\int \frac{d\theta}{\cos^2\theta}\\
&=&\frac{1}{9} \tan\theta + C\\
&=&\frac{1}{9} \frac{\sin\theta}{\cos\theta} + C \\
&=&\frac{1}{9} \frac{x}{\sqrt{9-x^2}} + C\\
\end{eqnarray*}
\sep
\noindent
d. [10~points] $\ds\int_0^1 x^2 e^{2x}\ dx$
\vskip 10pt
\noindent
{\bf Solution:} Use integration by parts twice, or \# 14 in the table
of integrals:
$$\int x^2 e^{2x}\ dx = \left(\frac{x^2}{2} - \frac{x}{2} + \frac{1}{4}\right)e^{2x} + C$$
Then for the definite integral
$$\int x^2 e^{2x}\ dx = \left(\frac{x^2}{2} - \frac{x}{2} + \frac{1}{4}\right)e^{2x}|_0^1
= \frac{1}{4}(e^2 - 1) \doteq 1.597$$

\sep
\noindent
e. [10~points]  $\ds \int \frac{(x - 1)\ dx}{x^3 + 25 x}$
\vskip 10pt
\noindent
{\bf Solution:} Use partial fractions based on the factorization 
$x^3 + 25 x = x(x^2 + 25)$.  We have
$$\frac{x-1}{x(x^2+25)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 25}.$$
Clearing denominators,
$$x - 1 = A(x^2 + 25) + (Bx + C)x = (A+B)x^2 + Cx + 25A$$
Equating coefficients of like powers of $x$:
$(A+B) = 0$, $C = 1$, and $25A = -1$.  So $A = \frac{-1}{25}$ and 
$B = \frac{-1}{25}$.  
Then using \# 25 in the table
$$\int \frac{\frac{-1}{25}}{x} + \frac{\frac{1}{25} x + 1}{x^2 + 25} 
= \frac{-1}{25} \ln|x| + \frac{1}{50} \ln|x^2 + 25| + \frac{1}{5}\arctan\left(\frac{x}{5}\right)
+C$$

\sep
\noindent
3.  All parts of this problem refer to the region in the plane
bounded by $y = 2 + \cos(x)$, $y = 0$, $x = 0$, and $x = \pi$,
shown below:

\centerline{\psfig{figure=1323.ps,height=2.0in,width=2.0in,angle=-90}}

\vskip 10pt
\noindent
a. [10~points] Find the volume of the solid with base $R$ whose cross-sections by
planes perpendicular to the $x$-axis are squares.
\vskip 10pt
\noindent
{\bf Solution:}  We have, using \# 18 in the table
\begin{eqnarray*}
V & = & \int_0^\pi (2 + \cos x)^2\ dx\\
& = & \int_0^\pi 4 + 4\cos x + \cos^2 x\ dx\\
& = & (4x + 4\sin x + \frac{1}{2} x + \frac{1}{2}\sin x\cos x|_0^\pi\\
&=& \frac{9\pi}{2}\\
\end{eqnarray*} 
\sep
\noindent
b. [10~points] True or False:  The solid obtained by rotating $R$ about the $x$-axis
has volume $\pi$ times the volume from part a.  (For full credit,
give the integral that would compute the volume of this solid of
revolution, and answer the question.)
\vskip 10pt
\noindent
{\bf Solution:}  This is TRUE, since the volume of the solid of 
revolution is 
$$\int_0^\pi \pi (2 + \cos x)^2\ dx = \pi \int_0^\pi (2 + \cos x)^2\ dx.$$

\sep
\noindent
c. [15~points] A thin metal plate has the shape of the region $R$ (units in meters) 
and constant mass density $1 kg/m^2$.  Find the $x$-coordinate of its center of mass.
\vskip 10pt
\noindent
{\bf Solution:}  We have, integrating by parts on the top
\begin{eqnarray*}
\overline{x} & = & \frac{\int_0^\pi x(2 + \cos x)\ dx}{\int_0^\pi (2 + \cos x)\ dx}\\
&=& \frac{x^2 + x\sin x + \cos x|_0^\pi}{2x + \sin x|_0^\pi}\\
&=& \frac{\pi^2 - 2}{\pi}\\
\end{eqnarray*}

\sep
\noindent
d. [10~points] Set up, {\it but do not evaluate} the integral giving the 
arclength of the top edge of $R$.
\vskip 10pt
\noindent
{\bf Solution:}  The arclength integral is $L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\ dx$.
Here, since $y = 2 + \cos x$, $\frac{dy}{dx} = -\sin x$ and 
$$L = \int_0^\pi \sqrt{1 + \sin^2 x}\ dx.$$

\sep
\noindent
4. [5~points each] The following graph shows a probability density function (pdf)
for a quantity $x$:

\centerline{\psfig{figure=1325.ps,height=2.0in,width=2.0in,angle=-90}}

\vskip 10pt
\noindent
a.  Which of the following two plots shows the corresponding cumulative
distribution for $x$:

\vskip 10pt
\centerline{\psfig{figure=1325a1.ps,height=2.0in,width=2.0in,angle=-90},
\psfig{figure=1325a2s.ps,height=2.0in,width=2.0in,angle=-90}}
\vskip 10pt
\noindent
{\bf Solution:}  Graph B is the correct cdf for the given density 
function.  We can see this because the given density should be the 
derivative of the cdf.  Plot B has two maxima for the slope,
corresponding to the two ``peaks''.
\sep
\noindent
b.  Using the appropriate graph, estimate the {\it median} of $x$.
\vskip 10pt
\noindent
{\bf Solution:}  The median is the $T$-value where $\int_{-\infty}^T p(x)\ dx = \frac{1}{2}$.
Drawing in a horizontal line in plot B, it appears that the line $y = 1/2$
intersects the graph at about $x = 1$.  That is the  median value.
\sep
\noindent
5. [5~points each] Does each of the following series converge?  For full credit, you must
justify your answer completely by showing how the indicated test leads
to your stated conclusion.
\vskip 10pt

\noindent
a.  $\ds \sum_{n=1}^\infty \frac{(-1)^n}{n^{1/3}}$ -- alternating series test.
\vskip 10pt
\noindent
{\bf Solution:}  $\lim_{n\to\infty} \frac{1}{n^{1/3}} = 0$ and 
for all $n \ge 1$, $\frac{1}{n^{1/3}} > \frac{1}{(n+1)^{1/3}}$.
Hence the alternating series test (see p. 420 in the textbook)
implies that this series {\it converges}.

\sep
\noindent
b.  $\ds \sum_{n=1}^\infty \frac{1}{n^{9/10}}$ -- integral test
\vskip 10pt
\noindent
{\bf Solution:}  The improper integral
$$\int_1^\infty \frac{dx}{x^{9/10}} = \lim_{b\to\infty} 10 x^{1/10}|1^b$$
does not converge, because as $b\to\infty$, so does $b^{1/10}$.  Hence
by the integral test, this series {\it diverges}.

\sep
\noindent
c.  $\ds \sum_{n=1}^\infty \frac{1}{n e^n}$ -- any applicable method
\vskip 10pt
\noindent
{\bf Solution:} Method 1: Since $n \ge 1$, $\frac{1}{ne^n} < \frac{1}{e^n}$.
The series $\ds \sum_{n=1}^\infty \frac{1}{e^n}$ is a geometric series
with ratio $\frac{1}{e} < 1$.  Hence it converges.  By the comparison
test, $\ds \sum_{n=1}^\infty \frac{1}{n e^n}$ {\it converges too}.

\vskip 10pt
\noindent
Method 2:  Apply the ratio test:
$$\lim_{n\to\infty} \left|\frac{1}{(n+1)e^{n+1}}\cdot\frac{ne^n}{1}\right|
=\lim_{n\to\infty} \frac{n}{n+1} \cdot \frac{1}{e} = \frac{1}{e}$$
Since $\frac{1}{e} < 1$, the ratio test implies the series {\it converges}.
\sep
\noindent
6. 
\noindent
[10~points]  Using the ratio test,
determine the radius of convergence of the following power series
$$\sum_{n=1}^\infty \frac{(x+2)^n}{n^2 4^n}$$
\vskip 10pt
\noindent
{\bf Solution:}  
$$\lim_{n\to\infty} \left|\frac{(x+2)^{n+1}}{(n+1)^2 4^{n+1}}\cdot
\frac{n^2 4^n}{(x+2)^n}\right| = \lim_{n\to\infty} \frac{n^2}{(n+1)^2} \frac{1}{4}|x+2|
=\frac{1}{4}|x+2|$$
The radius of convergence is $1/(1/4) = 4$.

\sep
\noindent
7.  All parts of this problem refer to $f(x) = \cos\left(\frac{x}{2}\right)$.

\noindent
a. [15~points]  Using the definition of Taylor polynomials, compute the Taylor polynomial of 
degree $n = 6$ for $f(x)$ at $a = 0$.
\vskip 10pt
\noindent
{\bf Solution:}  The problem said to use the definition of Taylor
polynomials, so we compute $f'(x) = \frac{-1}{2}\sin\left(\frac{x}{2}\right)$,
etc.  The 6th degree polynomial is 
$$p_6(x) = \sum_{i=0}^6 \frac{f^{(i)}(0)}{i!} x^i = 
1 - \frac{1}{8} x^2 + \frac{1}{384} x^4 - \frac{1}{46080} x^6$$

\sep
\noindent
b. [5~points] Use your answer from part a to approximate $\cos(1.2)$.
\vskip 10pt
\noindent
{\bf Solution:}  Since $f(x) = \cos\left(\frac{x}{2}\right)$, to 
approximate $\cos(1.2)$, we substitute $x = 2.4$ into $p_6(x)$ from 
part a.  (Note:  $p_6(x)$ is supposed to approximate 
$\cos\left(\frac{x}{2}\right)$, so to get $1.2$ ``inside'' the cosine,
we want $x = 2.4$.  We get
$$p_6(2.4) = 1 - \frac{1}{8} (2.4)^2 + \frac{1}{384} (2.4)^4 - \frac{1}{46080} (2.4)^6
\doteq 1-.72 _ .0864 - .0041472 \doteq .3622528$$
\sep
\noindent
c. [10~points] Using the error bound for Taylor approximations, estimate how big 
$n$ must be to guarantee 4 decimal place accuracy in approximations to 
$f(x)$ for all $x$ between $0$ and $2$.  
\vskip 10pt
\noindent
{\bf Solution:}  The $(n+1)$st derivative of $f(x)$ is 
$\frac{1}{2^{n+1}}$ times either $\pm\sin x$ or $\pm \cos x$.
Hence we can use $M = \frac{1}{2^{n+1}}$ in the error bound, and
we get for $x$ between $0$ and $2$:
$$|f(x) - p_n(x)| \le \frac{\frac{1}{2^{n+1}}}{(n+1)!} 2^{n+1} = \frac{1}{(n+1)!}.$$
We have $\frac{1}{8!} \doteq .000025$ so $n = 7$ (or $n = 6$, since
$p_6 = p_7$ for $\cos$) is large enough.

\sep
\noindent
8.  When a course ends, it's an unfortunate fact of life that students
start to forget the material they have learned.  One mathematical model for this
process states that {\it the rate at which a student forgets material is
proportional to the difference between the material currently remembered
and some positive constant $a < 1$}.  Let $y$ be the fraction of the
original material remembered $t$ weeks after the course has ended. 
Then  $y(0) = 1$, and the model assumes $y(t) > a$ for all $t$.    
\vskip 10pt
\noindent
a. [5~points] Which of the following differential equations is the 
translation of the model described above:
$$I:  \frac{dy}{dt} = -k (y - a) \qquad II: \frac{dy}{dt} = \frac{-k}{y-a}$$
\vskip 10pt
\noindent
{\bf Solution:}  I is the correct equation.
\sep
\noindent
b. [10~points] Solve the equation you selected in part a by separation
of variables.
\vskip 10pt
\noindent
{\bf Solution:}  Dividing by $y - a$ separates so
\begin{eqnarray*}
\int \frac{dy}{y-a} &=& \int -k\ dt\\
\ln|y - a| &=& -kt + c\\
y & = & a + d e^{-kt}\\
\end{eqnarray*}
where $d = \pm e^c$.  Note that it is given in the problem that
$y(0) = 1$.  This says $1 = y(0) = a + d \cdot 1$, so $d = 1 - a$.
\sep
\noindent
c. [10~points] Suppose $a = .2$, and after 3 weeks the student only 
remembers $80\%$ of the material from the course (that is, $y(3) = .8$). 
How long will it be before the student only remembers one half of the 
material?
\vskip 10pt
\noindent
{\bf Solution:} Using this last comment, $y(t) = .2 + .8 e^{-kt}$.
From the given information $y(3) = .8$, so 
$$.8 = .2 + .8 e^{-3k} \Rightarrow k = \frac{\ln(.6/.8)}{-3} \doteq .0959$$
Then we want to solve for $t$:
$$.5 = .2 + .8 e^{(-.0959)t} \Rightarrow t = \frac{\ln(.3/.8)}{-.0959} \doteq 10.23$$
That is, the student will only remember half of the material 10.23 
weeks after the end of the class.
\end{document}