\def\whichexam{Final Exam}
\def\examdate{May 6, 2004}
\def\whichcourse{Math 126}
\font\impno=cmbx10 at 12pt
\newcount\n\newcount\na\newcount\nb

%% List of problem number at top of each page with last entry = number of problems +1
%% Begin with the problem number at the top of the second page since problem 1 is
%% always at the top of the first page of problems.
\gdef\pagebreaks{\\{3}\\{5}\\{7}\\{9}\\{11}\\{13}\\{15}\\{17}\\{19}%
\\{21}\\{23}\\{25}\\{26}}

%% These lines are bold on the marked answer sheet but not the
%% regular student answer sheet.
\gdef\boldlines{\\{34}}
%%\gdef\boldlines{\\{1}\\{7}\\{12}\\{18}\\{34}}

\newif\ifproblemspacing
%\problemspacingtrue
%\problemspacingfalse


% #1 - path to pdf file
% #2 - scale factor
% #3 #4  - size of box pdf is centered in this box. It can overflow box
\def\LoadPDF#1#2#3#4{
\pdfximage {#1}
\setbox0=\hbox{\pdfrefximage\pdflastximage} % get dimensions of the image
                                            % in order to include the image
                                            % at a specific resolution 
\dimen0=#2\wd0         % calculate the image width at 1200 dpi (0.06 = 72/1200)
\dimen1=#2\ht0
\pdfximage              % include the image at resolution 1200 dpi
    width \dimen0 {#1}  % by setting image width to the calculated value
\divide\dimen0 by 2
\divide\dimen1 by 2
\dimen2=#3 \divide\dimen2 by 2
\dimen3=#4 \divide\dimen3 by 2
\vbox to #4{\vskip\dimen3\hbox to#3{\hskip\dimen2%
\hbox to0pt{\vtop to0pt{\vskip-\dimen1 %
\hbox to 0pt{\hskip -\dimen0\pdfrefximage \pdflastximage\hss}\vss}\hss}%
\hss}\vss}%
}
\def\boxLoadPDF#1#2#3#4{
\pdfximage {#1}
\setbox0=\hbox{\pdfrefximage\pdflastximage} % get dimensions of the image
                                            % in order to include the image
                                            % at a specific resolution 
\dimen0=#2\wd0         % calculate the image width at 1200 dpi (0.06 = 72/1200)
\dimen1=#2\ht0
\pdfximage              % include the image at resolution 1200 dpi
    width \dimen0 {#1}  % by setting image width to the calculated value
\divide\dimen0 by 2
\divide\dimen1 by 2
\dimen2=#3 \divide\dimen2 by 2
\dimen3=#4 \divide\dimen3 by 2
\boxit{\vbox to #4{\vskip\dimen3\hbox to#3{\hskip\dimen2%
\hbox to0pt{\vtop to0pt{\vskip-\dimen1 %
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\hss}\vss}}{1pt}{1pt}%
}

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\ifnum\mc>0 \newpage\null\vskip10pt\fi\fi%
\else\vskip10pt\headline{\hfill}\fi%
\mc=1 \fi}
\newcount\breakpc\breakpc=1

\def\newpage{\vfill\eject}
\def\mathrm#1{{\rm #1}}

\newif\ifprelim
\prelimtrue
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\multichoicetrue
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\def\START#1{\ifnum#1=0{\global\prelimtrue\global\problemspacingfalse%
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\else{\ifnum#1=1{\global\prelimfalse\global\problemspacingtrue%
\global\STARTNumber=1}%
\else{\global\prelimfalse\global\problemspacingfalse%
\global\STARTNumber=2}\fi}%
\fi}

\START 1 %% 0 - initial run for instructors
                  %% 1 - run for the actual exam
                  %% 2 - for the webpage

%%
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\setbox0=\hbox{$99$}%
\dimen0=\wd0 %
\frac{x^{\hbox to0pt{$2$\hss}}}{\hbox to\wd0{\hfil$#1$\hfil}}+%
\frac{y^{\hbox to0pt{$2$\hss}}}{\hbox to\wd0{\hfil$#2$\hfil}}=1}

\def\dd#1#2{{{\textstyle d\> #1}\over{\textstyle d#2}}}
 \def\frac#1#2{{{\textstyle#1}\over{\textstyle#2}}}
 \def\smfrac#1#2{{{#1}\over{#2}}}

\def\Lim{\displaystyle\lim}
\def\Int{\displaystyle\int}
\def\brace#1#2{%
\def\xbx{#1}%
\ifx\xbx\empty\relax(#2)\else%
\csname #1l\endcsname(#2\csname #1r\endcsname)%
\fi}
%%
\magnification1200\footline{\ifnum\pageno=1 \hfil\else\hfil\number\pageno\hfil\fi}

\def\setdata#1{\count0=1 \def\\##1{\advance\count0 by1 }#1 %
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\def\\##1{\count0=##1 }#1 \advance\count0 by-1 %
\edef\numberofproblems{\number\count0 }}%
\setdata\pagebreaks

%%
\long\def\hide#1\endhide{}

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\vtop\bgroup\obeylines\baselineskip20pt\hsize=2.5in\pspacewidth=2.3in
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}
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\parindent=0pt
}
\def\dxd{\displaystyle}
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\hfuzz=5pt
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\edef\sum{\displaystyle\xxsum}
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\def\Pra{\leftremove\list\to\moose%
\expandafter\mooseAntlers\moose@%
\switchbox{\pA}{\Aa}\switchbox{\pB}{\Ab}%
\switchbox{\pC}{\Ac}\switchbox{\pD}{\Ad}\switchbox{\pE}{\Ae}%
\resetboxes%
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\pcc=\pc\advance\pcc by -1 %
\immediate\write16{Problem \number\pcc - Correct Answer \number\pA}%
}
\def\OR{%
\bigskip
\noindent{\bf OR}\bigskip
\expandafter\leftinsertitem\moose\to\list%
}

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\picV=#2 \divide\picV by1000 \multiply\picV by#4 
\picH=#1 \divide\picH by1000 \multiply\picH by#4
\vtop to \picV{
      \hrule width \picH height 0pt depth 0pt
      \vfill
      \special{picture #3 scaled #4}}}

\def\checkPBN#1@{%
\edef\yy{\number#1}\def\\##1{ \edef\yyy{\number##1}\ifx\yyy\yy \gdef\PBN{1}\fi}%
\xxpagebreaks}

\newbox\probbox
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\noindent\vbox{\noindent{\bf\number\pc.}(\number\credit\ pts.)
\global\advance\pc by1 %
\global\advance\breakpc by1 %
#1\endp}\vfil}

\long\def\Ans#1\ui#2\ui#3\ui#4\ui#5\ui#6\ui#7\endp{
\ifcase#1\relax\or\one\or\two\or\three\or\four\or\five\fi
\AN{#2}{#3}{#4}{#5}{#6}\Pra}

%: Multiple choice problem macro
\long\def\Newp #1\ui#2#3\\#4\\#5\\#6\\#7\\#8\endp{%
\newp#1
\Ans#2\ui
#3\ui
#4\ui
#5\ui
#6\ui
#7\ui
\endp}

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\newbox\aa\newbox\ab\newbox\ac
\newbox\ad\newbox\ae

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\hfuzz=20pt

\def\five{\global\Ax=5 \global\Aspace=.2\hsize\relax\global\advance\Aspace by-20pt}
\def\four{\global\Ax=4 \global\Aspace=.25\hsize\relax\global\advance\Aspace by-20pt}
\def\three{\global\Ax=3 \global\Aspace=.3\hsize\relax\global\advance\Aspace by-20pt}
\def\two{\global\Ax=2 \global\Aspace=.5\hsize\relax\global\advance\Aspace by-20pt}
\def\one{\global\Ax=1 \global\Aspace=\hsize\relax\global\advance\Aspace by-20pt}

\long\def\AN#1#2#3#4#5{%
\setbox\Aa=\hbox to\Aspace{\ignorespaces#1\hfil}%
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\setbox\Ad=\hbox to\Aspace{\ignorespaces#4\hfil}\setbox\Ae=\hbox to\Aspace{\ignorespaces#5\hfil}}

\long\def\ans#1�#2�#3�#4�#5�{%
\AN{#1}{#2}{#3}{#4}{#5}\Pra%
}

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\def\pra{\ifcase\Ax \or\praone\or\pratwo\or\prathree\or\prafour \or\prafive\fi}

\def\prafive{\vskip\Ls\noindent(a)\quad\box\Aa(b)\quad\box\Ab%
(c)\quad\box\Ac(d)\quad\box\Ad%
(e)\quad\box\Ae}

\def\prafour{\vskip\Ls\noindent(a)\quad\box\Aa(b)\quad\box\Ab%
(c)\quad\box\Ac(d)\quad\box\Ad\vskip\Ls\noindent(e)\quad\box\Ae}

\def\prathree{\vskip\Ls\noindent(a)\quad\box\Aa(b)\quad\box\Ab%
(c)\quad\box\Ac\vskip\Ls\noindent(d)\quad\box\Ad(e)\quad\box\Ae}

\def\pratwo{\vskip\Ls\noindent(a)\quad\box\Aa(b)\quad\box\Ab%
\vskip\Ls\noindent(c)\quad\box\Ac(d)\quad\box\Ad%
\vskip\Ls\noindent(e)\quad\box\Ae}

\def\praone{\vskip\Ls\noindent(a)\quad\box\Aa\vskip\Ls\noindent(b)\quad\box\Ab%
\vskip\Ls\noindent(c)\quad\box\Ac\vskip\Ls\noindent(d)\quad\box\Ad%
\vskip\Ls\noindent(e)\quad\box\Ae}

\def\resetboxes{\setbox\Aa=\box\aa\setbox\Ab=\box\ab\setbox\Ac=\box\ac%
\setbox\Ad=\box\ad\setbox\Ae=\box\ae}
\def\switchbox#1#2{\ifcase#1 \or\setbox\aa=\box#2\or\setbox\ab=\box#2\or%
\setbox\ac=\box#2\or\setbox\ad=\box#2\or\setbox\ae=\box#2\fi}

\def\xPra#1#2#3#4#5{\switchbox{#1}{\Aa}\switchbox{#2}{\Ab}%
\switchbox{#3}{\Ac}\switchbox{#4}{\Ad}\switchbox{#5}{\Ae}%
\resetboxes\ifcase\Ax \or\praone\or\pratwo\or\prathree\or \or\prafive\fi\vfill}




\def\frac#1#2{{\displaystyle{{\textstyle#1}\over{\textstyle#2}}}}

\long\def\boxit#1#2#3{\vbox{\hrule\hbox{\vrule\kern#2
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\newcount\pc
\def\pr{\vskip10pt\noindent\number\pc.\global\advance\pc by 1}
\def\dd#1#2{{{\textstyle d\> #1}\over{\textstyle d#2}}}

\def\rhs#1#2{\par\noindent\hbox to\hsize{\hfill\hbox to#1{#2\hfill}}}
\def\ul#1{\vrule width#1 height -3pt depth3.1pt}
\def\UL#1#2{\hbox to0pt{\qquad #2\hss}\ul{#1}}
\newif\ifnotmarkanswers\notmarkanswerstrue
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\global\let\xxpagebreaks=\pagebreaks
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\relax\leftremove\pagebreaks\to\nextpb \else\bskip\fi%
\ifnotmarkanswers\relax\let\fst=\rm\else
\ifnum#1=\ansbold\relax\let\fst=\bf\leftremove\boldlines\to\ansbold%
\else\let\fst=\rm\fi%
\fi%
{\fst \tabalign {#1.}&
\global\ansmark=#2 \relax%
\ifnotmarkanswers \global\ansmark=0 \else%
\ifprelim \global\ansmark=1 \fi\fi%
\ifnum\ansmark=1 \relax($\bullet$)\else(a)\fi&
\ifnum\ansmark=2 \relax($\bullet$)\else(b)\fi&
\ifnum\ansmark=3 \relax($\bullet$)\else(c)\fi&
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\hskip\AA\cr}%
\ifprelim
\rightappenditem{12345}\to\list
\else
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\fi%
\global\ansmark=#1 %
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\def\marked{\pageno=1 %
\global\notmarkanswersfalse
\coversheet{ANSWER}}

\def\+{\tabalign}
\long\gdef\answersheet{
\null\vskip10pt
\itemitem{$\bullet$} Be sure that you have all \numberpages\ pages of the test. 
\itemitem{$\bullet$} No calculators are to be used.
\itemitem{$\bullet$} The exam lasts for two hours.
\itemitem{$\bullet$} {\impno When told to begin, remove this answer sheet 
and keep it under the rest of your test.
When told to stop, hand in just this one page.}
\itemitem{$\bullet$}
The Honor Code is in effect for this examination, including keeping your answer
sheet under cover.
\vskip10pt
\def\leaderfill{\leaders\hbox to3pt{\hss.\hss}\hfill}
\def\pskip{\vskip -8pt\hbox to\pspacewidth{\leaderfill}\vskip-8pt}
\def\bskip{\vskip -8pt\hbox to\pspacewidth{\hfill}\vskip-8pt}
\def\HH{\hskip.42in}
\settabs\+9.\qquad &\HH&\HH&\HH&\HH&\HH&\cr
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\par\centerline{\small The dotted lines in the answer box indicate page breaks.}
\hskip-45pt\boxit{%
\noindent\hbox to0pt{%
%: Permutations
\multiplechoicebox
\ll  1 14235
\ll  2 32145
\ll  3 15342
\ll  4 35241
\ll  5 42153
\ll  6 12543
\ll  7 23541
\ll  8 35421
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\ll 10 41253
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\ll 13 23415
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\mcbbreak
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\ll 16 43512
\ll 17 13425
\ll 18 32514
\ll 19 25413
\ll 20 53241
\ll 21 52134
\ll 22 25134
\ll 23 13425
\ll 24 31254
\ll 25 52134
\vskip6pt
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\vskip6pt
\hbox{\hbox to 1in{Previous Total:\hfill} \ul{1in}}
\vskip6pt
\hbox{\hbox to 1in{Course Total:\hfill} \ul{1in}}
\endmcb%
\hss}}{20pt}{5in}%
}

\newcount\anshere\anshere=1
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\headline{\vbox{\hsize=2in%
\centerline{\bf \whichcourse}
\centerline{\bf \whichexam}
\centerline{\bf\examdate}}%
\vbox{\hsize=3in\rhs{3in}{Name:\ \ul{3in}\hss}\vskip10pt%
\rhs{3in}{Instructor:\ \UL{2.75in}{#1}\hss}%
}}
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\leftremove\pagebreaks\to\nextpb
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}
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\def\tds#1#2{\adsdim=90pt\hbox{%
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\hbox to \adsdim{{\bf 2) }#2\hfil}}}
\def\ads#1#2#3{\adsdim=80pt\hbox{%
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\hbox to\adsdim{{\bf 1) }#1\hfil}%
\hbox to \adsdim{{\bf 2) }#2\hfil}%
\hbox to \adsdim{{\bf 3) }#3\hfil}}}


\long\def\Exam{%
\coversheet{ }%
\vfill\eject\headline{\hfil}
\pageno=2 %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%
%%%%%%%     START OF TEST
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%: Question 1
%: Katrina
\Newp 
Let $f(x) = x^7 - x^3 + 2x$.  Find  $(f^{-1})^\prime(-2)$.
\ui5
$1/6$\\
$1$\\
$1/4$\\
$4$\\
$6$\\
\endp

%: Question 2
%: Katrina
\Newp 
Solve the following equation for $x$.
$$7^x 5^{x} = 9 e^2\ .$$
\ui3
$x = \frac{\ln 9 + 2}{\ln 35}$ \\
$x = \frac{2}{\ln 35}$ \\
$x = \frac{\ln 9 + 2}{\ln 12}$ \\
$x = \frac{2\ln7}{\ln 35}$\\
There is no solution.\\
\endp


%: Question 3
%: Katrina
\Newp 
 Find the derivative of
$$y = x \cdot x^{1/x}\ .$$
\ui2
$x^{1/x} + x^{\smfrac{1}{x} - 1}(1 - \ln x) $\\
$x^{1/x} + x^{\smfrac{1}{x} - 2}$\\
$x^{-1} \ln x (1 - \ln x)$\\
$x^{1/x} + x^{\smfrac{1}{x}+1}(\ln (\ln x) - x^{-2})$\\
$x \cdot x^{\smfrac{1}{x} -3} = x^{\smfrac{1}{x} -2}$\\
\endp

%: Question 4
%: Katrina
\Newp 
 Evaluate the following limit.
$$\lim_{x \to \infty} x \cdot x^{1/x}\ .$$
\noindent{\bf Remark:} Note the function is the same in both problems 3 and 4.
\ui5
$\infty$\\
$1$\\
$e^{-1}$\\
$e$\\
$0$\\
\endp

%: Question 5
%: Katrina
\Newp 
Find $f^\prime(x)$ for
$$f(x) =  \ln(2^x + x) + \arctan (e^x)\ $$
\ui2
$\frac{2^x \ln 2+1}{2^x + x} + \frac{e^x}{1+e^{2x}}$\\
$\frac{2^x \ln 2}{2^x + x} + \frac{e^x}{1 + e^{2x}} $\\
$\frac{2^x +1}{2^x + x} + \frac{1}{\sqrt{1 - e^{2x}}}$\\
$\frac{e^x \ln 2+1}{2^x + x} + \frac{e^x}{\sqrt{e^{2x} - 1}}$\\
$\frac{e^x \ln 2}{2^x + x} + \frac{e^x}{\sqrt{1 - 2e^{x}}}$\\
\endp

%: Question 6
%: Katrina
\Newp 
Which line below is the tangent line to the polar curve
$r =  1 + \cos \theta$ at the point where $\theta=\frac{\pi}{2}$?
\ui3
$y = x + 1$\\
$y = \frac{1- \sin \theta}{1+\cos\theta }\ (x+1)$\\
$y = 1$, a horizontal tangent\hss\\
$y = \frac{\pi}{2} x + \frac{\pi}{2}$\\
$x = 0$, a vertical tangent\hss\\
\endp

%: Question 7 
%: Katrina
\Newp 
 Evaluate the following definite integral.
$$\int_{1}^{e} \frac{\ln x}{x^3} dx\ .$$
\ui3
$\frac{1}{4} (1 - 3 e^{-2})$\\
$\frac{1}{2} (e^{-2} - 1)$\\
$\frac{1}{4} (1- e^{-3} + 2 e^{-2})$\\
$\frac{1}{4} (e^{-3} + 2 e^{-2} -1)$\\
$\frac{1}{2} (1 - 2e^{-2})$\\
\endp

%: Question 8
%: Nicholls
\Newp 
Evaluate
$$
\int \frac{x^2}{\sqrt{9 - x^2}} \; dx.
$$
\ui2
$\frac{9}{2} \left[ \arcsin(x/3) - \frac{x \sqrt{9 - x^2}}{9} \right] + C$\\
$9 \arcsin(x/3) + C$\\
$\frac{9}{2} \left[ \arcsin(x/3) - \frac{x^2}{9} \right] + C$\\
$\frac{9}{2} \left[ \arcsin(x/3) - \frac{x}{3}  \right] + C$\\
$\frac{1}{2} x \sqrt{9-x^2} + C$\\
\endp

%: Question 9
%: Nicholls
\Newp 
Evaluate
$$
\int \frac{x+7}{x^2+2x-3} \; dx.
$$
\ui2
$\ln\left| \frac{(x-1)^2}{x+3} \right| + C$\\
$\ln\left( 2 |x-1| - |x+3| \right) + C$\\
$\ln\left| \frac{(x+3)^2}{x-1} \right| + C$\\
$\ln\left| \frac{2 (x+3)}{x-1} \right| + C$\\
$\ln\left| x^2 + 2x - 3 \right| + C$\\
\endp

%: Question 10
%: Nicholls
\Newp 
Find the Midpoint Rule approximation (using four intervals) of
$$
\int_{-1}^3 x^2 \; dx\ .
$$
\ui5
$9$\\
$11$\\
$\frac{28}{3}$\\
$\frac{35}{4}$\\
$\frac{45}{4}$\\
\endp

%: Question 11
%: Nicholls
\Newp 
Find the arclength of $y = 6 - 2 x^{3/2}$, for $0 \leq x \leq 2$?
\ui3 
$\frac{2}{27} [ 19 \sqrt{19} - 1 ]$\\
$\frac{1}{9} [ \sqrt{19} - 1 ]$\\
$\frac{2}{3} [ 19 \sqrt{19} ]$\\
$\frac{2}{27} [ 10 \sqrt{10} - 1 ]$\\
$\frac{20}{3} \sqrt{10}$\\
\endp

%: Question 12
%: Nicholls
\Newp 
Find the center of mass of a plate with shape bounded below by the 
$x$--axis, on the left by $x=-1$, on the right by $x=1$, and above
by the curve
$$
y = \cases{2^x& $-1 \leq x \leq 0$\cr
2^{-x}&$\hphantom{-}0 \leq x \leq 1$\ .\cr}
$$
The area is $\frac{1}{\ln 2}$.
\vskip .6in
\hskip2in\LoadPDF{finalCentroid.pdf}{0.6}{1in}{1in}
\vskip-.3in
\ui3
$\left( 0, \frac{3}{8} \right)$\\
$\left( 0, \frac{1}{4}  \right)$\\
$\left( 0,  \frac{3}{8 \ln 2} \right)$\\
$\left( 0, \frac{3-\ln 2}{5 + 2\ln 2}  \right)$\\
$\left( 0,  \frac{2+ \ln 2}{8}  \right)$\\
\endp

%: Question 13
%: Nicholls
\Newp 
Solve the initial value problem:
$$
y' = (5 - 2x) (1 + y) \qquad y(1) = 0.
$$
\ui3
$y(x) = e^{-4 + 5 x - x^2} - 1$\\
$y(x) = e^{5 x - x^2} - e^4$\\
$y(x) = e^{-4 +5 x - x^2}$\\
$y(x) = e^{x^2 - 5 x} - e^{-2}$\\
$y(x) = e^{-3 + 5 x - x^2} - e$\\
\endp

%: Question 14
%: Nicholls
\Newp
If 50 grams of radioactive material with a half--life of two days 
are present at day zero, how many grams are left at day five?
\ui5
$\frac{50}{\sqrt{32}}$\\
$\frac{50}{\sqrt{8}}$\\
$\frac{50}{10}$\\
$\frac{50}{4^{1/3}}$\\
$\frac{50}{32^{1/3}}$\\
\endp

%: Question 15
%: Taylor (linear ode)
\Newp 
If the function $y(x)$ satisfies the differential equation
$$x y^\prime - 2 y = x - 2$$
subject to the initial value $y(1)=1$, what is $y(3)$?
\ui5
$7$\\
$6$\\
$5$\\
$4$\\
$3$\\
\endp

%: Question 16
%: Taylor (parameterized arc length)
\Newp 
Which integral below computes the length of the parameterized curve \par\noindent
$x(t) = t^3-t$ and $y(t)= t^5+t$ for $0\leq t \leq 1$.
\ui2
$\dxd\int_0^1 \sqrt{2 - 6 t^2 + 19 t^4 + 25 t^8 \ }\ dt$\\
$\dxd\int_0^1 \sqrt{ t^ 3 + t^5\ }\ dt$\\
$\dxd\int_0^1(t^5 + t) (3t^2-1)\ dt$\\
$\dxd\int_0^1 (t^3-t)\sqrt{2 - 6 t^2 + 19 t^4 + 25 t^8 \ }\ dt$\\
$\dxd\int_0^1 (t^5+t)\sqrt{ t^ 3 + t^5\ }\ dt$\\
\endp

%: Question 17
%: Taylor (polar area)
\Newp
Which integral below gives the area inside the polar curve $r = \cos(3\theta)$?
\ui2
$\frac{1}{2} \dxd\int_0^{\pi} \cos ^2(3\theta)\ d\theta$\\
$\frac{1}{2} \dxd\int_0^{\pi} \sqrt{\cos^2(3\theta) + 9\sin^2(3\theta)\ }\ d\theta$\\
$\frac{1}{2} \dxd\int_0^{2\pi} \cos^2(3\theta)\ d\theta$\\
$\frac{1}{2} \dxd\int_{\pi/6}^{\pi/3} \cos^2(3\theta)\ d\theta$\\
$\frac{1}{2} \dxd\int_0^{2\pi} \sqrt{\cos^2(3\theta) + 9\sin^2(3\theta)\ }\ d\theta$\\
\endp

%: Question 18
%: Taylor (limit of terms not 0)
\Newp 
Which statement below is true about the series
$\sum_{n=1}^\infty\ \frac{3^n}{n^3 + 2\cdot 3^n}$
\ui1
$\dxd\lim_{n\to\infty}\ \frac{3^n}{n^3 + 2\cdot 3^n} = 1/2$ so the series diverges.\\
$\dxd\lim_{n\to\infty}\ \frac{3^n}{n^3 + 2\cdot 3^n} = 1/2$ so the series converges.\\
$\dxd\lim_{n\to\infty}\ \frac{3^n}{n^3 + 2\cdot 3^n} = 0$ so the series diverges.\\
$\dxd\lim_{n\to\infty}\ \frac{3^n}{n^3 + 2\cdot 3^n} = 0$ so the series converges.\\
$\dxd\lim_{n\to\infty}\ \frac{3^n}{n^3 + 2\cdot 3^n}$ does not exist so the series converges.\\
\endp

%: Question 19
%: Taylor (sum geometric series)
\Newp 
Sum the series $\sum_{n=1}^\infty\ \frac{7^n}{5^{2n}}$.
\ui5
$\frac{7}{18}$\\
$\frac{3}{22}$\\
$\frac{96}{25}$\\
$\frac{25}{22}$\\
$\frac{5}{7}$\\
\endp

%: Question 20
%: Katrina
\Newp 
The interval of convergence of the series
$$\sum_{n=1}^\infty \frac{(-1)^{n}}{\sqrt{n}} \ (x+3)^n$$
is
\ui3
$(-4,-2]$\\
$(-4,-2)$\\
$(-1,1)$\\
$(2,4)$\\
$[2,4]$\\
\endp

%: Question 21
%: Taylor (absolute convergence)
\Newp 
Which series below absolutely converges?
\ui4
$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^3}$\\
$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{\ln(n+1)}$\\
$\sum_{n=1}^\infty \frac{(-1)^n n!}{n^3}$\\
$\sum_{n=1}^\infty \frac{\sqrt{n^3\;}}{n^2+1}$\\
$\sum_{n=1}^\infty \frac{(-1)^{n-1}\pi^n}{3^n}$\\
\endp

%: Question 22 
%: Taylor (conditional convergence via AST)
\Newp 
Which series below conditionally converges?
{\bf These are the same series as in problem 21}.
\ui4
$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{\ln(n+1)}$\\
$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^3}$\\
$\sum_{n=1}^\infty \frac{(-1)^n n!}{n^3}$\\
$\sum_{n=1}^\infty \frac{\sqrt{n^3\;}}{n^2+1}$\\
$\sum_{n=1}^\infty \frac{(-1)^{n-1}\pi^n}{3^n}$\\
\endp

%: Question 23
%: Nicholls
\Newp 
Find a power series for the function
$$
f(x) = -\frac{3}{(x+1)^2}.
$$
centered at $1$.
\ui2
$\frac{3}{2} \sum_{n=1}^{\infty} (-1)^n n \frac{(x-1)^{n-1}}{2^n}$\\
$\frac{3}{2} \sum_{n=1}^{\infty} n \frac{(x-1)^{n-1}}{2^n}$\\
$\frac{3}{2} \sum_{n=0}^{\infty} (-1)^n \frac{(x-1)^n}{2^n}$\\
$\frac{3}{2} \sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)} \frac{(x-1)^{n+1}}{2^n}$\\
$\frac{3}{2} \sum_{n=1}^{\infty} (-1)^n n \frac{(x-1)^n}{2^n}$\\
\endp

%: Question 24
%: Taylor 
\Newp
Which series below represents $\dxd\frac{\sin x - x}{x^3}$?
\ui3
$\dxd \sum_{n=1}^\infty\ \frac{(-1)^n x^{2n-2}}{(2n+1)!}$\\
$\dxd \sum_{n=1}^\infty\ \frac{(-1)^n x^{2n-2}}{(2n)!}$\\
$\dxd \sum_{n=1}^\infty\ \frac{(-1)^n x^{2n-2}}{n!}$\\
$\dxd \sum_{n=1}^\infty\ \frac{(-1)^n x^{2n-1}}{(2n)!}$\\
$\dxd \sum_{n=1}^\infty\ (-1)^n {{1/2}\choose{n}}x^{2n-2}$\\
\endp
%: Question 25
\Newp
Which series below gives the arc length of the curve $y = \frac{x^3}{3}$ from $x=0$ to $x=1$?
\ui3
$\dxd\sum_{n=0}^\infty\ {{1/2}\choose{n}}\frac{1}{4n+1}$\\
$\dxd\sum_{n=0}^\infty\ {{1/3}\choose{n}}\frac{1}{4n+1}$\\
$\dxd\sum_{n=0}^\infty\ {{1/2}\choose{n}}\frac{1}{4n+2}$\\
$\dxd\sum_{n=0}^\infty\ {{1/3}\choose{n}}\frac{1}{4n+2}$\\
$\dxd\sum_{n=0}^\infty\ {{1/2}\choose{n}}\frac{(-1)^n}{4n+1}$\\
\endp
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