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\documentclass[11pt]{exam}
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\firstpageheader{Math 122\\Calculus II}{Exam 2-A
}{Dr. Karel\\March 26, 2002}
\headrule
\runningheader{Math 122}%
    {Exam 2-A: Page \thepage\ of  \numpages}%
    {March 26, 2002}
    
\begin{document}
    	\boxedpoints
\pointsinmargin %\addpoints
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%This exam has \numpoints{} points, \numquestions{} questions and 
% \numparts{} parts.
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\vspace{0.2in}

\hbox to  \textwidth{{\large\bf Name:}\enspace\hrulefill}
\vspace{0.1in}

\begin{center}
\fbox{\fbox{\parbox{5.5in}{\centering{\large\bf INSTRUCTIONS.}  
Print your name on the line above.\\ Part I is closed book}}}
\end{center}

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%   \begin{center}                                           %
%   \fbox{\parbox{3.5in}{\centering{\large\bf Formulas.}     %
%   \[                                                       %
%   \lim_{t\rightarrow\infty} \arctan(t) = \pi/2             %
%   \]                                                       %
%   \[                                                       %
%   \lim_{t\rightarrow-\infty} \arctan(t) = -\pi/2           %
%   \]                                                       %
%   }}                                                       %
%   \end{center}                                             %
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\begin{questions}

\question[8]  %4  improper ``p-integrals''  %%%% #1

\begin{parts}
\part For which values of $p$ does the following integral converge?
\( \displaystyle\int_0^1 x^{-p} \, dx   \)
\hfill \fbox{\parbox[b][3ex]{.6in}{\hfill}}
\part For which values of $p$ does the following integral converge?
\( \displaystyle\int_1^\infty x^{-p} \, dx   \)
\hfill \fbox{\parbox[b][3ex]{.6in}{\hfill}}
\end{parts}
    
\question[4]  %3  Surface Area             %%%% #2
Set up, but do NOT evaluate, an integral for the area of the surface 
obtained by rotating the  curve \( y=\sin x, \quad %\displaystyle{
\pi/6 \le x \le \pi/2 %\frac{\pi}{6}\le x\le \frac{\pi}{2} } 
\)  about the $x$-axis. \\
\( 
\displaystyle{\int_{\,\fbox{\parbox[t][\width]{.24in}{\hfill}}}^
{\,\fbox{\parbox[b][\width]{.24in}{\hfill}}}
\,\fbox{\parbox[b][5ex]{1.8in}{\hfill}}\,dx} \) 


\question[4]  %2  Arc Length: Fill In       %%%% #3
\begin{parts}
    \part  %[1] 
    The length of a straight segment  ($\Delta x$ across and 
    $\Delta y$ upward) is $\Delta s =$ \makebox{\parbox[b][5ex]{1.3in}{
    \hrulefill }  }.
    \part  %[1]
    Between two points on the graph $y=f(x)$, $\Delta y$ is  
    approximately $\displaystyle{\frac{dy}{dx}}$ times
     \makebox{\parbox[b][5ex]{1in}{\hrulefill}  }.
     \part  %[1]
     The length of the arc connecting the two points in part (b), 
     above, is approximately  
     \( \displaystyle{\sqrt{(\Delta x)^{2}+
     \makebox{\parbox[b][5ex]{1in}{\hrulefill}} }}.\)
     \part  %[1]
     An ``infinitesimal piece of the curve'' has length 
     \( ds=\makebox{\parbox[b][5ex]{1.4in}{\hrulefill}  }. \)
\end{parts}


\question[8]  %6  L'H\^{o}pital's Rule: Fill In   %%%% #4
Describe how to compute 
\(\displaystyle L=\lim_{x\to 0} \frac{\sin x}{x}  \)
by filling in the blanks below.\\
Since  \(\displaystyle \lim_{x\to 0} \sin x =
\parbox[b][5ex]{.5in}{\hrulefill} \)
and   \(\displaystyle \lim_{x\to 0}  x =
\parbox[b][5ex]{.5in}{\hrulefill} \), 
we say that the limit is \\ 
\parbox[b][5ex]{1.6in}{\hrulefill}
of the form \parbox[b][5ex]{.8in}{\hrulefill}.
Therefore, we can use l'H\^{o}pital's Rule, which allows us
to replace the numerator,  $\sin x$, by 
\parbox[b][5ex]{1.6in}{\hrulefill} and the denominator, $x$,
by \parbox[b][5ex]{.8in}{\hrulefill}, and this gives us
\[
\lim_{x\to 0} \frac{\sin x}{x} = \lim_{x\to 0} \frac{\cos x}{1}
= \parbox[b][5ex]{.8in}{\hrulefill}.
\]

\clearpage

    
\question[8]  %1  Arc Length: Set Up         %%%% #5
For the curve $y=x^{3}$  from the point $(x,y)=(0,0)$ to the point 
$(x,y)=(2,8)$ the arc length is given by the definite integral
\( 
\displaystyle{\int_{\,\fbox{\parbox[t][\width]{.15in}{\hfill}}}^
{\,\fbox{\parbox[b][\width]{.15in}{\hfill}}}
\,\fbox{\parbox[b][5ex]{1in}{\hfill}}\,dx} \) \\
Do NOT try to compute the numerical value of the arc length.


\question[8]  %5  Improper Integrals as Limits    %%%% #6
Define each of the following improper integrals as a limit of 
ordinary integrals, but do NOT compute their numerical values.
\medskip
\begin{parts}
    \part
    \(\displaystyle\int_{1}^{\infty} \sin \pi x \, dx \) 
    \parbox[b][5ex]{4in}{\hrulefill}
\medskip
    \part 
    \(\displaystyle\int_{0}^{1} \frac{\, dx}{\sqrt{1-x}} \)
    \parbox[b][5ex]{4in}{\hrulefill}
\medskip
    \part
    \(\displaystyle\int_{0}^{\pi/2}\sec x\, dx \)
    \parbox[b][5ex]{4in}{\hrulefill}
\medskip 
    \part 
    \(\displaystyle\int_{0}^{2} \frac{\, dx}{\sqrt{1-x}} \)
    \parbox[b][5ex]{4in}{\hrulefill}
\end{parts}



\begin{center}
\fbox{\fbox{\parbox{5.5in}{\centering{\large\bf End of  Part I}  
After you have handed in this page and your first bluebook, you
 may start work on Part II, which is open book.}}}
\end{center}


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\clearpage

%\vspace{0.2in}

\hbox to  \textwidth{{\large\bf Name:}\enspace\hrulefill}
\vspace{0.1in}

\begin{center}
\fbox{\fbox{\parbox{5.5in}{\centering{\large\bf PART II INSTRUCTIONS.}  
Print your name on the line above.\\ Part II is open book.}}}
\end{center}

%\vspace{0.4in}    
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\question[9]  %9  L'H\^{o}pital's Rule:  When to Use it.   %%%% #7
For which expressions below should  l'H\^{o}pital's Rule be used
to compute the limit?  Circle the letter for each one that 
should be evaluated by using  l'H\^{o}pital's Rule.
\begin{parts}
\part  %b
 \(\displaystyle\lim_{x\to 0^{+}} \frac{\tan x}{x}.  \)
\part  %a
 \(\displaystyle\lim_{x\to 0} \frac{\sin x}{e^{x}}.  \)
\part  %c
 \(\displaystyle\lim_{x\to 0^{+}} \frac{\ln x}{x}.  \)
\end{parts}

\question[9]  %10  L'H\^{o}pital's Rule: Computations   %%%% #8
Evaluate the following by using l'H\^{o}pital's Rule
one or more times.
\begin{parts}
\part  %b
 \( \displaystyle\lim_{x\to \infty} \frac{\ln x}{x}.  \)
\part  %c
 \( \displaystyle\lim_{x\to 0^{+}} \frac{1-\cos x}{x^{2}}.  \)
\part  %a
 \( \displaystyle{\lim_{x\to \infty} \frac{x^{2}}{e^{x}}.}  \)
\end{parts}

\question[9]  %7  Improper Integrals: Convergence   %%%% #9
For each of the integrals below say whether it is {\em convergent}
or {\em divergent}.  Do NOT compute the value.
\begin{parts}
\part   % part (b)
\(\displaystyle\int_0^{1} \frac{\,dx }{x} \).
%\hfill\fbox{\(={\displaystyle\lim_{t\to 0^{+}}}\int_t^{1}  \frac{\,dx }{x} \, dx 
%=\lim_{t\to 0^{+}}-\ln t \) : DIVERGENT.}
\part   % part (c)
\(\displaystyle\int_{1}^{\infty} \frac{ \, dx}{\sqrt{x}} \).
%  \hfill \fbox{ DIVERGENT:  see \#1 Part (b) above  }
\part   % part (a)
\(\displaystyle\int_0^{3} \frac{ \, dx}{\sqrt{x}} \).
%  \hfill \fbox{ CONVERGENT:  see \#1 Part (b) above  }
\end{parts}

\question[9]  %8 Improper Integrals: Computations    %%%% #10
For each integral below either {\em compute a value for it} or say
that {\em it does not converge.}
\begin{parts}
\part  %c
 \(\displaystyle\int_{0}^{\infty} e^{-x}\,dx  \)
\bigskip
\part  %a
 \(\displaystyle\int_{0}^{\infty} \frac{\, dx}{1+x^{2}}  \)
\bigskip
\part  %b
 \(\displaystyle\int_{0}^{\infty} \frac{2x\, dx}{1+x^{2}}  \)
%\bigskip
\end{parts}






%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\clearpage



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   


\begin{center}
\fbox{\fbox{\parbox{5.5in}{\centering{\large\bf INSTRUCTIONS.}  
Choose the best method to use as the first step for evaluating each 
of the following integrals. Circle the letter and fill in the 
corresponding blanks.}}}
\end{center}


\question[4] %13     %%%% #11
\(  \displaystyle
\int \frac{dx}{1-x^{2}}
\)

\begin{parts}
\part 
Integration by Parts:  \makebox[3in]{$u=\,$ \hrulefill\quad
$dv=\,$\hrulefill}  \medskip
\part 
Substitution:  
\makebox[3in]{ $u=\,$ \hrulefill\quad $du=\,$\hrulefill} \medskip
\part
Inverse Substitution:  
\makebox[3in]{ $x=\,$ \hrulefill\quad $dx=\,$\hrulefill}  \medskip
\part
Partial Fractions:\\ Divide First? \makebox[.5in]{ \hrulefill}\quad \medskip\\ 
Decomposition Scheme:\makebox[2.5in]{\hrulefill} \medskip
\part
Algebraic Simplification: \makebox[3in]{\hrulefill} \medskip
\part
Identities (For $\ln$, $\sin$, etc.): \makebox[3in]{\hrulefill} \medskip
\end{parts}

\question[4] %16       %%%%% #12
\(  \displaystyle  
\int \sqrt{\frac{1+x}{1-x}}\, dx  \text{ with $-1<x<1$ }
\)

\begin{parts}
\part % a
Integration by Parts:  \makebox[3in]{$u=\,$ \hrulefill\quad
$dv=\,$\hrulefill}  \medskip
\part  % b
Substitution:  \makebox[3in]{ $u=\,$ \hrulefill\quad
$du=\,$\hrulefill} \medskip 
\part  % c
Inverse Substitution:  \parbox[b][3ex]{3in}{ $x=\,$ \hrulefill\quad
$dx=\,$\hrulefill}  \medskip 
\part  % d
Partial Fractions:\\ Divide First? \makebox[.5in]{ \hrulefill}\quad \medskip\\ 
Decomposition Scheme:\makebox[2.5in]{\hrulefill} \medskip
\part  % e
Algebraic Simplification:  \parbox[b][3ex]{3in}{\hrulefill} \medskip
\part  % f
Identities (For $\ln$, $\sin$, etc.): \makebox[3in]{\hrulefill} \medskip
\end{parts}




\question[4]  %14      %%%% #13
\(  \displaystyle
\int \frac{(1+e^{x})\, dx}{x+e^{x}}     %\sin^{2}x \cos^{3}x\
\)

\begin{parts}
\part 
Integration by Parts:  \makebox[3in]{$u=\,$ \hrulefill\quad
$dv=\,$\hrulefill}  \medskip
\part 
Substitution:  \makebox[3in]{ $u=\,$ \hrulefill\quad
$du=\,$\hrulefill} \medskip 
\part
Inverse Substitution:  \makebox[3in]{ $x=\,$ \hrulefill\quad $dx=\,$\hrulefill}  \medskip
\part
Partial Fractions:\\ Divide First? \makebox[.5in]{ \hrulefill}\quad \medskip\\ 
Decomposition Scheme:\makebox[2.5in]{\hrulefill} \medskip
%\part
%Algebraic Simplification: \makebox[3in]{\hrulefill} \medskip
\part
Identities (For $\ln$, $\sin$, etc.): \makebox[3in]{\hrulefill} \medskip
\end{parts}


%\clearpage
\clearpage


\question[4]  %12      %%%% #14
\(  \displaystyle 
\int \frac{ \, dx}{\sqrt{9-x^{2}}}
\)  

\begin{parts}
\part 
Integration by Parts:  \makebox[3in]{$u=\,$ \hrulefill\quad
$dv=\,$\hrulefill}  \medskip
\part 
Substitution:  \makebox[3in]{ $u=\,$ \hrulefill\quad $du=\,$\hrulefill} \medskip
\part
Inverse Substitution:  \makebox[3in]{ $x=\,$ \hrulefill\quad $dx=\,$\hrulefill}  \medskip
\part
Partial Fractions:\\ Divide First? \makebox[.5in]{ \hrulefill}\quad \medskip\\ 
Decomposition Scheme:\makebox[2.5in]{\hrulefill} \medskip
\part
Algebraic Simplification: \makebox[3in]{\hrulefill} \medskip
\part
Identities (For $\ln$, $\sin$, etc.): \makebox[3in]{\hrulefill} \medskip
\end{parts}


\question[4] %11    %%%% #15
\(  \displaystyle
\int x e^{x}\,dx
\)  


\begin{parts}
\part 
Integration by Parts:  \makebox[3in]{$u=\,$ \hrulefill\quad
$dv=\,$\hrulefill}  \medskip
%\( \displaystyle\int_1^\infty x^{-p} \, dx   \)
\part 
Substitution:  \makebox[3in]{ $u=\,$ \hrulefill\quad $du=\,$\hrulefill} \medskip
%\( \displaystyle\int_0^1 x^{-p} \, dx   \)
\part
Inverse Substitution:  \makebox[3in]{ $x=\,$ \hrulefill\quad $dx=\,$\hrulefill}  \medskip
\part
Partial Fractions:\\ Divide First? \makebox[.5in]{ \hrulefill}\quad \medskip\\ 
Decomposition Scheme:\makebox[2.5in]{\hrulefill} \medskip
\part
Algebraic Simplification: \makebox[3in]{\hrulefill} \medskip
%\part
%Identities (For $\ln$, $\sin$, etc.): \makebox[3in]{\hrulefill} \medskip
\end{parts}


\question[4] %15       %%%% #16
\(  \displaystyle
\int \frac{x^{4}+1}{(x-1)(x-2)(x+2)}\, dx
\)

\begin{parts}
\part 
Integration by Parts:  \makebox[3in]{$u=\,$ \hrulefill\quad
$dv=\,$\hrulefill}  \medskip
\part 
Substitution:  \makebox[3in]{ $u=\,$ \hrulefill\quad $du=\,$\hrulefill} \medskip
\part
Inverse Substitution:  \makebox[3in]{ $x=\,$ \hrulefill\quad $dx=\,$\hrulefill}  \medskip
\part
Partial Fractions:\\ Divide First? \makebox[.5in]{ \hrulefill}\quad \medskip\\ 
Decomposition Scheme:\makebox[2.5in]{\hrulefill} \medskip
\part
Algebraic Simplification: \makebox[3in]{\hrulefill} \medskip
\part
Identities (For $\ln$, $\sin$, etc.): \makebox[3in]{\hrulefill} \medskip
\end{parts}


\clearpage

\question  %17  Bonus
 {\bfseries Bonus Question.}  [10 points]
Use the Comparison Theorem (stated below) to show that 
\(\displaystyle\int_1^\infty \frac{1+\sin x}{x^{2}} \, dx \)
is divergent by comparing it to 
\(\displaystyle\int_1^\infty \frac{2}{x^{2}} \, dx \).

\vspace{3in}
{\bfseries Comparison Theorem.}
Suppose that $f$, $g$ are continuous functions with  $f(x)\ge g(x)\ge 
0$ for $x\ge a$, where $a$ is a real number.
\begin{parts}
    \part If $\int_{a}^{\infty} f(x) \, dx$ is convergent, then
    $\int_{a}^{\infty} g(x) \, dx$ is also convergent.
    \part If $\int_{a}^{\infty} g(x) \, dx$ is divergent, then
    $\int_{a}^{\infty} f(x) \, dx$ is also divergent.
\end{parts}

\end{questions}
\end{document}


