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\def\header{{\hfill Exam 2 \hfill A. Miller \hfill Fall 2013
                                  \hfill Math 221\hfill}}
\markboth\header\header

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\probno=0

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Circle your answer. \par\bigskip}
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\begin{document}

\bigskip

\begin{center}
Instructions
\end{center}

No notes, no books, no calculators, no cell phones, no pagers,
no electronic devices of any kind.

Show all of your work. Circle your answer.

\par\vskip .25in
\begin{center}
Name\makebox[3in]{\hrulefill} \\
\end{center}

\bigskip 


\bigskip

\begin{minipage}[top]{2.5in}
Circle your section number. 

Hand in to your TA.

\begin{verbatim}
Section               Start
Number                Time
 421 Cheng, Jingrui   07:45
 422 Cheng, Jingrui   08:50
 424 Lynch, John      09:55
 429 Lynch, John      01:20
 426 Powers, Michael  11:00
 427 Powers, Michael  12:05
 430 Soule, Michelle  01:20
 431 Soule, Michelle  02:25
 432 Charles, Zachary 07:45
 433 Charles, Zachary 02:25
\end{verbatim}
\end{minipage}
\begin{minipage}{2in}
\begin{center} \Large
 \begin{tabular}{||c|c|c||} \hline\hline
  Problem & Points & Score \\  \hline\hline
   \input{exam.tbl}
  \hline \hline
 \end{tabular}
\end{center}
\end{minipage}

\immediate\openout\examTBLFile=exam.tbl

\bigskip

Solutions will be posted shortly after the exam:

www.math.wisc.edu/$\sim$miller

\setcounter{page}{0}
\newpage



\prob{8} % fund2.pdf
Prove the Fundamental Theorem Part 2:  If $f$ is a continuous
function on $[a,b]$ and $F'=f$ is any antiderivative of $f$, then
$$\int_a^bf(x)dx=F(b)-F(a)$$

\def\blank{\underline{\phantom{AAAAA}}}
\prob{6} % Extra 28 ---- Extra : 26 27 28 29 30 31
Find, simplify, and place your answer in the blank.

\bigskip
\begin{enum}
\item $\sum_{k=3}^5 k(k-1)\;=\;$\blank

\vfill

\item $\sum_{k=1}^4 \frac6k\;=\;$\blank

\vfill

\item $\sum_{k=0}^3 k\cos(k\frac\pi2)\;=\;$\blank

\vfill

\end{enum}

\answer (a) $38$ (b) $12.5$ (c) $-2$

\prob{8} % 5.12 54 
Show that the equation $\sin x = 1-2x$ has exactly
one solution $x$.  First show that it has at least one
solution and then show that it cannot have more than one.

\answer Let $f(x)=\sin(x)-1+2x$.  Then $f(0)=-1$ and
$f(10)=\sin(10)-1+20\geq 20$.  So by the Intermediate
Value Theorem there is some $x$ between $0$ and $10$ where
$f(x)=0$.  Now $f'(x)=\cos(x)+2\geq 1$ for every $x$ and
so has no zeros.  But Rolle's Theorem implies that if
$f$ has two zeros, that in between them is a zero of
$f'$.  Hence $f$ has at most one zero.  Since
$f(x)=0$ iff $\sin x = 1-2x$ the result follows.

\prob{8} %  5.12 24 same problem
 $$  f(x)=\frac{x}{1+x^2}$$
Find intervals where it is increasing or decreasing.  Find
local maxima and minima. Find inflection points and intervals
where it is concave up or down. Find limits at $\infty$ and
$-\infty$.  Sketch graph.

\answer  Local max at $x=1$. Local min at $x=-1$.  (These
are global max and min.)
There are inflection points at $x=\sqrt{3}$,
$x=-\sqrt{3}$ and $x=0$. It is concave down $(-\infty,-\sqrt{3})$,
up $(-\sqrt{3},0)$, down $(0,\sqrt{3})$, up $(\sqrt{3},\infty)$.
The limit at $\infty$ is $0^+$ and the limit at $-\infty$ is $0^-$.

\prob{8} % extra 20
Suppose
$$f(x)=\left\{
\begin{array}{ll}
ax^3+bx^2 & \mbox{ if } x<1 \\
x^2+1 & \mbox{ if } x \geq 1 \\
\end{array}\right.$$
Find $a$ and $b$ so that $f$ is differentiable everywhere.

\answer $a=-2$ and $b=4$

\prob{8} % 5.14 16 
A $20$ inch piece of wire is cut into two pieces. One piece is
bent into a square and the other is bent into a triangle which is similar
to the right-angle triangle with sides 3, 4, and 5.
What is the minimal total area enclosed by both figures?

\answer Minimal area 10.


\prob{8} % 5.18c Extra 25
Compute the following limits or say the limit is not defined:

\bigskip
(a) $\lim_{x\to 1} \frac{\sin \pi x}{x^2+x-2}$

\vfill

(b) $\lim_{x\to 0} \;\;\;\;\frac1{\sin(x)}-\frac1x$

\vfill


\answer (a) $-\frac{\pi}{3}$  (b) $0$

\prob{9} % 7.05  5,9,24 ---- 7.05  : 1 2 3 4 5 6 7 8 9
Find an antiderivative $F(x)$ for each of the following
  functions $f(x)$. 

\bigskip
(a) $  f(x) = x+2 $  

\vfill

(b) $  f(x) = 1-x+\frac{x^2}{2}-\frac{x^3}{3}+\frac{x^4}{4} $

\vfill
(c) $  f(x) = \sin x - \cos 2x $ 

\vfill



\answer (a) $F(x)=\frac{x^2}2 + 2x$  
\par (b) $F(x)=x-\frac{x^2}2+\frac{x^3}6-\frac{x^4}{12}+ \frac{x^5}{20}$
\par (c) $F(x)=-\cos(x)-\frac12\sin(2x)$


\prob{8} % Extra 32 ---- Extra : 32 33
Recall the $\epsilon-\delta$ definition of the Riemann integral:
  $\int_a^b f(x) dx = I$
  iff for every $\epsilon >0$, there exists a $\delta>0$ such that
  for any partition of $[a,b]$, $a=x_0<x_1<\cdots <x_n=b$
  with $x_k-x_{k-1}=\Delta x_k\leq\delta$
  and sample points $c_k$ with $x_{k-1}\leq c_k\leq x_k$ for $k=1\ldots n$
  we have
  $$\Bigl|f(c_1)\Delta x_1 + f(c_2) \Delta x_2 + \cdots + f(c_n)\Delta
  x_n -I\Bigr|\leq \epsilon$$

Define
$$f(x)=\left\{
\begin{array}{ll}
5       & \mbox { if } x = 1 \\
0       & \mbox { if } x\neq 1 \\
\end{array}\right.$$

\bigskip
(a) Draw the graph of $f$.  What is $I=\int_0^2f(x)dx$?

\vfill
(b)
Using the $\epsilon-\delta$ definition of the integral show
that $f$ is Riemann integrable, i.e., given $\epsilon>0$ what
should $\delta>0$ be? Explain.

\vfill

\answer (a) $I=0$. 
\par (b) $\delta=\frac\epsilon{10}$ works.  At most two of the $c_k$ can
be $1$.  So $|\sum_{k=1}^n f(c_k)\Delta x_k-I|$ is
at most $5\Delta x_k+5\Delta x_{k+1}$ which is at most
$10\delta$.

\prob{8} % 7.10 52 
Find
$$ \int^{-e}_{-e^2} \frac1{3x} \,dx$$

\answer $-\frac13$

\prob{8} %  7.10 59 ---- 7.10 : 54 57 59 62 81 87 90 93
 Compute  
\[
I = \int _0^2 3x^2 \bigl(1+x^3\bigr)^2\, dx
\]
in two different ways:

\bigskip
(a) Expand $(1+x^3)^2$, multiply with $3x^2$, and integrate each term.


\vfill
(b) Use the substitution $u=1+x^3$.

\vfill


\answer $242 \frac23$


\prob{8} 
Suppose 
$$f(x)=\int_{0}^{\sin(x)}e^{t^2}\;dt$$
Find $f'(x)$.

\answer $f'(x)=e^{\sin^2(x)}\cos(x)$.


\def\blank{\underline{\phantom{AAAA}}}
\prob{5} % 5.12 1-3
Matching problem. Fill in the blanks.
\begin{enum}
\item Intermediate Value Theorem \blank
\item Extreme Value Theorem \blank
\item Fermat's Theorem \blank
\item Rolle's Theorem \blank
\item Mean Value Theorem \blank
\end{enum}

\begin{enumerate}
\item For any $a<b$ and $f$ differentiable on $(a,b)$
and continuous on $[a,b]$ there exists $c$ with $a<c<b$ and
$$f'(c)=\frac{f(b)-f(a)}{b-a}$$
\item For any $a<b$ and $f$ differentiable on $(a,b)$,
continuous on $[a,b]$, and $f(a)=f(b)=0$ there exists $c$ with $a<c<b$ and
$$f'(c)=0$$
\item If $f$ is differentiable at $a$ and for some $\epsilon>0$
$f(a)\geq f(x)$ for any $x$ in $(a-\epsilon,a+\epsilon)$, then
$f'(a)=0$.
\item For any $a<b$ and $f$ continuous on $[a,b]$ there exists
$c$ and $d$ in $[a,b]$ such that for any $x$ in $[a.b]$
$$f(c)\leq f(x)\leq f(d)$$
\item For any $a<b$ and $f$ continuous on $[a,b]$ and any $v$ 
between $f(a)$ and $f(b)$,
there exists $c$ in $[a,b]$ with
$$f(c)=v$$
\end{enumerate}




 \printanswers

\finishtbl
\end{document}



