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\pagestyle{myheadings}
\def\header{{\hfill Exam 1 \hfill A. Miller \hfill Fall 2010
                                       \hfill Math 222\hfill}}
\markboth\header\header

\newcount\probno
\newcount\total
\newwrite\examTBLFile

\probno=0

\def\message{$\;\;\;$ Show all of your work.
Circle your answer. \par\bigskip}
\def\nextpage{\vfill \newpage}
%\def\nextpage{\par\bigskip}

\def\prob#1{\advance\probno by 1 
    \nextpage
    \noindent\the\probno . (#1 pts)
    \message
    \advance\total by #1
    \immediate\write\examTBLFile{\the\probno
       \string& #1 \string&
       \string\qquad \string\\ \string\hline}}

\def\finishtbl{
    \immediate\write\examTBLFile{Total \string& \the\total
    \string& \qquad \string\\ \string\hline}
     }

\newwrite\ans
\immediate\openout\ans=answers
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\def\printanswers{\newpage\begin{center} Answers \end{center}
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% \def\answer{\par\bigskip\bigskip\bigskip\noindent \par\bigskip
% Answer: \par\bigskip\bigskip}
% \outer\def\answer{ {\par\bigskip Circle your answer.} \par\medbreak

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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
Number
 321  7:45  Zhao, Jie
 332  9:55  McMahon, Kayla
 322  8:50  Zhao, Jie
 328  1:20  Emrah, Elnur
 331  9:55  Kim, Yoosik
 324 11:00  McMahon, Kayla
 325 12:05  Wang, Kejia
 326 12:05  Emrah, Elnur
 327  1:20  Wang, Kejia
 330  2:25  Kim, Yoosik
\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{9} % 34
Find
$$\int_{-1}^2\;|x^2-x|\;dx$$

\def\simple{ \bigskip Express your answer as a simple fraction.}
\simple

\answer $11\over 6$

\prob{9} % 49 
Find
$$\int{\ln(2x^2)\over x}\;dx$$

\answer $\frac14(\ln(2x^2))^2+C$

\prob{9} % 67
Find
$$\int{1\over 3x^2+6x+6}\;dx$$

\answer ${1\over 3} \arctan(x+1)+C$

\prob{9} % 101 not assigned
Find
$$\int {x^3\over x^2-1}\;dx$$

\answer ${1\over 2}(x^2+\ln|x^2-1|)+C$


\prob{9} % 108
Find
$$\int {1\over x^2(x+1)}\;dx$$

\answer $-\ln|x|-{1\over x}+\ln|x+1|+C$

\prob{9} % 138
Find 
$$\int 2x\ln(x+1)\;dx $$

\answer $x^2\ln(x+1)-{\;\;x^2\over 2}+x-\ln(x+1)+C$

\prob{9} % 152
Find a second order polynomial (i.e., a
quadratic function) $p(x)$ such that
$p(2)=3$, $p^\prime(2)=8$, and $p^{\prime\prime}(2)=-1$.

\answer $p(x)=3+8(x-2)-{1\over 2}(x-2)^2$

\prob{9} % 215
Find the Taylor series for the following function, by substituting
adding, multiplying, applying long division,and/or differentiating
known series, for $\sin(x),\cos(x), e^x, \ln(1+x),$ or ${1\over 1-x}$.

$$f(x)=\left\{
\begin{array}{cl}
{\sin(x)\over x} & \mbox{ if $x\neq 0$} \\ 
1 & \mbox{ if $x=0$} \\  
\end{array}\right.$$

\answer $f(x)=1-{x^2\over 3!}+ {x^4\over 5!}-\cdots$
$=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\;{x^{2n}}$

\prob{7} % 255
(a) Find the second degree Taylor polynomial for the function
$e^t$.  

(b) Use it to give an estimate for
the integral
$$\int_0^1 e^{x^2}\;dx$$

\simple

\answer $\frac{43}{30}$

\prob{6} % 255
Suppose instead we used the 5th degree Taylor polynomial $p(t)$ for $e^t$
to give an estimate for the integral:
$$\int_0^1 e^{x^2}\;dx$$
Give an upper bound for the error:
$$\left|\;\int_0^1 e^{x^2}\;dx-\int_0^1 p(x^2)\;dx\;\right|$$
Note: You need not find $p(t)$ or the integral
$\int_0^1 p(x^2)\;dx$ and will be given no credit for doing so.
You do not need to simplify your answer.
Yes, one million is an upper bound, but you will get no credit
for it.

\answer $\frac{3}{6!\cdot 13}$


\prob{9} % 246
For which real numbers $x$ does the Taylor series of
$f(x)={1\over 2-5x}$ converge to $f(x)$?

\answer $|x|<2/5$.

\def\mykern{\kern -.5in \parbox{.45in}}
\def\tf{\par\medskip\noindent
 {\mykern  {T $\;$ F $\;$}}}
\def\tfx{\par\medskip\noindent
 {\mykern  {\fbox{T}  F $\;$}}}
\def\message{}

\prob{6} Circle T or F for True or False. No penalty for guessing.

\tf Given a function $f$ and
distinct real numbers $a_1<a_2<\cdots<a_n$ there is a unique polynomial
$p(x)$ of degree $\leq n-1$ such that $f(a_k)=p(a_k)$ for
all $k=1,2,\ldots,n$.
\tf Given any function
$f$ continuous on the interval $[a,b]$ and $\epsilon>0$
there is a polynomial $p(x)$ such that $|f(x)-p(x)|<\epsilon$ for all $x$ in
$[a,b]$.
\tf Given any $f$ with $n$-derivatives at
0 there is a unique polynomial $p(x)$ of degree $\leq n$ such
that $f^{(k)}(0)=p^{(k)}(0)$ for all $k=0,\ldots, n$.
\tf $\lim_{n\to\infty} a^n = 0$ if and only if  $-1<a<1$.
\tf $\lim_{n\to\infty} {a^n\over n!} = 0$ for every $a$.
\tf ${\pi\over 4}=1-{1\over 3}+{1\over 5}-{1\over 7}+\cdots$
\tf $\ln(2)=1-{1\over 2}+{1\over 3}-{1\over 4}+\cdots$
\tf ${\pi^2\over 6}=\sum_{n=1}^\infty {1\over n^2}$
\tf If $f(x)=e^{-{1\over x^2}}$ for  $x\neq 0$ and $f(0)=0$,
then $f^{(n)}(0)=0$ for all $n=0,1,2,\ldots$
\tf Since ${1\over 1-x}= 1+x+x^2+x^3+\cdots$, we have
that $${1\over 2}= {1\over 1-(-1)}= 1 +(-1) + 1 + (-1) +\cdots$$
\tf Now that we have studied advanced techniques of integration
we know the correct thing we should have been writing 
all along is:
$$ {d\over dx} x^2 = 2x +C$$
\tf Apply integration by parts to $\int {1\over x}\;dx$ letting
$u={1\over x}$ and $dv=dx$ gives 
$$\int {1\over x}\;dx = ({1\over x})(x) -\int -({1\over x^2})x\;dx$$
simplifying gives us
$$\int {1\over x}\;dx = 1 + \int {1\over x}\;dx$$
and subtracting the integral from both sides gives us $0=1$.

\answer The last three are false.

\printanswers

\finishtbl
\end{document}



