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%TCIDATA{<META NAME="Title" CONTENT="Exercises">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Thu Feb 13 17:07:57 1997}
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\input{tcilatex}
\begin{document}


\section{Exercises\label{Exercises}}

Here are a few problems designed to exercise your human--machine interface.

\begin{enumerate}
\item[1.]  \label{q1}Verify the formula $\frac{d}{dx}\left( x^{8}\right)
=8x^{7}$ by starting with the definition of derivative and choosing submenu
items such as \textsf{Expand} and \textsf{Simplify}.\hfill \hyperref{%
\TCIIcon{BITMAPSETAnswer}{0.2214in}{0.205in}{0in}}{}{}{a1}

\item[2.]  \label{q2}Use Newton's method on the function $f(x)=x^{2}+1$,
starting with $x_{0}=0.5$. What conclusions can you draw?\hfill \hyperref{%
\TCIIcon{BITMAPSETAnswer}{0.2214in}{0.205in}{0in}}{}{}{a2}

\item[3.]  \label{q3}Find the \label{Tangent line problem}equation of one
line that is tangent to the graph of 
\[
f(x)=x(x-1)(x-3)(x-6)
\]
at two different points.\hfill \hyperref{\TCIIcon{BITMAPSETAnswer}{0.2214in}{%
0.205in}{0in}}{}{}{a3}

\item[4.]  \label{q4}For $0<k<1$, the \label{Elliptic integral}elliptic
integral $E=\int_{0}^{\pi /2}\sqrt{1-k\sin ^{2}t}dt$ has no elementary
solution. Use a series expansion of the integrand to estimate $E$.\hfill 
\hyperref{\TCIIcon{BITMAPSETAnswer}{0.2214in}{0.205in}{0in}}{}{}{a4}

\item[5.]  \label{q5}Find all the solutions to $x^{y}=y^{x}$ for unequal
positive integers $x$ and $y$.\hfill \hyperref{\TCIIcon{BITMAPSETAnswer}{%
0.2214in}{0.205in}{0in}}{}{}{a5}

\item[6.]  \label{q6}Blood \label{Blood flow problem}flowing through an
artery flows fastest at the center of the artery, and slowest near the walls
of the artery where friction is a factor. In fact, the velocity is given by
the formula $v(r)=\alpha (R^{2}-r^{2})$, where $\alpha $ is a constant, $R$
is the radius of the artery, and $r$ is the distance from the center.

Set up an integral that gives the total blood flow through an artery. Show
that, if an artery is constricted to one-half of its original radius, the
blood flow (assuming constant blood pressure) is reduced to $\frac{1}{16}$
of its original flow.\hfill \hyperref{\TCIIcon{BITMAPSETAnswer}{0.2214in}{%
0.205in}{0in}}{}{}{a6}

\item[7.]  \label{q7}The mass of an object traveling at a velocity $v$ with
rest mass $m_{0}$ is given by 
\[
m=m_{0}\left( 1-\frac{v^{2}}{c^{2}}\right) ^{-1/2} 
\]
where $c$ is the speed of light. Use a Maclaurin series expansion to show
the increase in mass at low velocities.\hfill \hyperref{\TCIIcon{%
BITMAPSETAnswer}{0.2214in}{0.205in}{0in}}{}{}{a7}

\item[8.]  \label{q8}Evaluate $\int 2^{x}\cos bxdx$ and simplify the
answer.\hfill \hyperref{\TCIIcon{BITMAPSETAnswer}{0.2214in}{0.205in}{0in}}{}{%
}{a8}

\item[9.]  \label{q9}Evaluate $\dint_{-\pi }^{\pi }\dfrac{1+\sin x}{\left(
x-\cos x\right) ^{2}}dx$ .\label{Integral}\hfill \hyperref{\TCIIcon{%
BITMAPSETAnswer}{0.2214in}{0.205in}{0in}}{}{}{a9}

\item[10.]  \label{q10}Evaluate $\lim_{h\rightarrow 0^{+}}\int_{0}^{\infty
}\sin (x^{1+h})\,dx$ .\hfill \hyperref{\TCIIcon{BITMAPSETAnswer}{0.2214in}{%
0.205in}{0in}}{}{}{a10}

\item[11.]  \label{q11}\label{Fundamental theorem}The Fundamental Theorem of
Calculus says that if $f$ is continuous on a closed interval $\left[
a,b\right] $, then

\begin{enumerate}
\item[a.]  If $g$ is defined by $g(x)=\int_{a}^{x}f(t)dt$ for $x\in \left[
a,b\right] $, then $g^{\prime }(x)=f(x)$, and

\item[b.]  If $F$ is any antiderivative of $f$, then.$%
\int_{a}^{b}f(x)dx=F(b)-F(a)$.
\end{enumerate}

Demonstrate that these two conditions hold for each of the three functions $%
f(x)=x^{3}$, $f(x)=xe^{x}$, and $f(x)=\sin ^{2}x\cos x$.\hfill \hyperref{%
\TCIIcon{BITMAPSETAnswer}{0.2214in}{0.205in}{0in}}{}{}{a11}
\end{enumerate}

\section{Solutions\label{Solutions}}

\begin{enumerate}
\item[1.]  \label{a1}By definition, 
\begin{eqnarray*}
\frac{d}{dx}\left( x^{8}\right) &=&\lim_{h\rightarrow 0}\frac{\left(
x+h\right) ^{8}-x^{8}}{h} \\
&=&\lim_{h\rightarrow 0}\frac{8x^{7}h+28x^{6}h^{2}+\cdot \cdot \cdot
+28x^{2}h^{6}+8xh^{7}+h^{8}}{h} \\
&=&\lim_{h\rightarrow 0}\left( 8x^{7}+28x^{6}h+\cdot \cdot \cdot
+28x^{2}h^{5}+8xh^{6}+h^{7}\right) \\
&=&\,8x^{7}
\end{eqnarray*}

\hfill \hfill \hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.2006in}{0.243in}{%
0in}}{}{}{q1}

\item[2.]  \label{a2}Defining $g$ by $g(x)=x-f(x)/f^{\prime }(x)$, from the 
\textsf{Calculus} submenu, choose \textsf{Iterate} to get 
\[
\left[ 
\begin{array}{c}
.5 \\ 
-.75 \\ 
.29167 \\ 
-1.5684 \\ 
-.4654 \\ 
.84164
\end{array}
\right] 
\]
If this result seems to be headed nowhere, it is doing so for good reason.
The function $f$ is always positive, so it has no zeroes. Newton's method is
searching for something that does not exist.\hfill \hfill \hyperref{\TCIIcon{%
BITMAPSETProbSolvHint}{0.2006in}{0.243in}{0in}}{}{}{q2}

\item[3.]  \label{a3}It is sufficient to find three numbers $a$, $b$, and $m$
that satisfy $f^{\prime }(a)=m$, $f^{\prime }(b)=m$, and $\frac{f(b)-f(a)}{%
b-a}=m$. Put these three equations inside a $3\times 1$ matrix, and from the 
\textsf{Solve} submenu, choose \textsf{Exact} to get the solutions 
\[
m=4b^{3}-30b^{2}+54b-18,b=b,a=b 
\]
and 
\begin{eqnarray*}
m &=&-8 \\
a &=&5-\rho \\
b &=&\rho
\end{eqnarray*}
where $\rho $ is a root of $\left( Z^{2}-5Z+1\right) $. The first solution
is not allowed, because the problem requires $a\neq b$.

Leave the insertion point in the equation $Z^{2}-5Z+1=0$, and from the 
\textsf{Solve} submenu, choose \textsf{Exact} to get the solutions 
\begin{eqnarray*}
Z &=&\frac{5}{2}+\frac{1}{2}\sqrt{21} \\
Z &=&\frac{5}{2}-\frac{1}{2}\sqrt{21}
\end{eqnarray*}

Choosing $\rho =\frac{5}{2}+\frac{1}{2}\sqrt{21}$, we have 
\begin{eqnarray*}
a &=&\frac{5}{2}-\frac{1}{2}\sqrt{21} \\
b &=&\,\frac{5}{2}+\frac{1}{2}\sqrt{21}
\end{eqnarray*}
Evaluating and expanding, 
\begin{eqnarray*}
f(a) &=&\,\left( \frac{5}{2}-\frac{1}{2}\sqrt{21}\right) \left( \frac{3}{2}-%
\frac{1}{2}\sqrt{21}\right) \left( -\frac{1}{2}-\frac{1}{2}\sqrt{21}\right)
\left( -\frac{7}{2}-\frac{1}{2}\sqrt{21}\right) \\
&=&\,-21+4\sqrt{21}
\end{eqnarray*}
so that 
\begin{eqnarray*}
y &=&f(a)+m(x-a) \\
&=&-21+4\sqrt{21}-8(x-\frac{5}{2}+\frac{1}{2}\sqrt{21} \\
&=&\,-1-8x
\end{eqnarray*}
Plot the two curves $x\left( x-1\right) \left( x-3\right) \left( x-6\right) $
and $-1-8x$, just for visual verification. Use a viewing window with domain
interval $-1\leq x\leq 6.5$ to generate the following picture.\hfill 
\hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.2006in}{0.243in}{0in}}{}{}{q3}%
\FRAME{dtbpF}{3in}{2.0003in}{0pt}{}{}{}{\special{language "Scientific
Word";type "MAPLEPLOT";width 3in;height 2.0003in;depth 0pt;display
"PICT";function \TEXUX{$x\left( x-1\right) \left( x-3\right) \left(
x-6\right) $};linecolor "red";linestyle 1;linethickness 1;pointstyle
"point";function \TEXUX{$-1-8x$};linecolor "navy";linestyle 1;linethickness
1;pointstyle "point";xmin "-1";xmax "6.5";xviewmin "-1.15";xviewmax
"6.653";yviewmin "-55.31";yviewmax "64.92";rangeset"X";phi 45;theta
45;plottype 4;numpoints 49;axesstyle "normal";xis \TEXUX{x};var1name
\TEXUX{$x$};}}

\item[4.]  \label{a4}The series is given by $\sqrt{1-k\sin ^{2}t}=\,1+\left(
-\frac{1}{2}k\right) t^{2}+\left( \frac{1}{6}k-\frac{1}{8}k^{2}\right)
t^{4}+O\left( t^{6}\right) $. Thus, an estimate for $E$ is given by 
\begin{eqnarray*}
E &\approx &\int_{0}^{\pi /2}\left[ 1+\left( -\frac{1}{2}k\right)
t^{2}+\left( \frac{1}{6}k-\frac{1}{8}k^{2}\right) t^{4}\right] dt \\
&=&\,-\frac{1}{1280}k^{2}\pi ^{5}+\frac{1}{2}\pi +\frac{1}{960}k\pi ^{5}-%
\frac{1}{48}k\pi ^{3}
\end{eqnarray*}
As a check, $k=1$ yields $1.0045$ compared with the exact value 
\[
\int_{0}^{\pi /2}\sqrt{1-\sin ^{2}t}dt=1\, 
\]
and $k=0$ yields $\frac{1}{2}\pi $, which agrees precisely with \hfill 
\hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.2006in}{0.243in}{0in}}{}{}{q4} 
\[
\int_{0}^{\pi /2}dt=\,\frac{1}{2}\pi 
\]

\item[5.]  \label{a5}Compute natural logs on both sides and separate
variables to get 
\[
\frac{\ln x}{x}=\frac{\ln y}{y} 
\]
Plot the graph of $\frac{\ln x}{x}$ on the interval $1\leq x\leq 10$ and
locate the extreme values of $\frac{\ln x}{x}$ by solving 
\[
\frac{d}{dx}\left( \frac{\ln x}{x}\right) =0 
\]
Note that $2$ is the only integer between $1$ and $e$, and verify that $%
2^{4}=4^{2}$ is true.\hfill \hyperref{\TCIIcon{BITMAPSETProbSolvHint}{%
0.2006in}{0.243in}{0in}}{}{}{q5}

\item[6.]  \label{a6}The flow is given by the integral 
\[
\int_{0}^{R}\alpha (R^{2}-r^{2})2\pi rdr=\,\frac{1}{2}\alpha \pi R^{4} 
\]
If $R$ is reduced by one-half, then $R^{4}$ is reduced to $\frac{1}{16}$ of
the original amount.\hfill \hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.2006in%
}{0.243in}{0in}}{}{}{q6}

\item[7.]  \label{a7}The series expansion is given by 
\[
m_{0}\left( 1-\frac{v^{2}}{c^{2}}\right) ^{-1/2}=\,m_{0}+\frac{1}{2}\frac{%
m_{0}}{c^{2}}v^{2}+\frac{3}{8}\frac{m_{0}}{c^{4}}v^{4}+O\left( v^{6}\right) 
\]
If $\frac{v}{c}$ is small, then the model 
\[
m\approx \,m_{0}+\frac{1}{2}\frac{m_{0}}{c^{2}}v^{2} 
\]
is useful for estimating the increased mass.\hfill \hyperref{\TCIIcon{%
BITMAPSETProbSolvHint}{0.2006in}{0.243in}{0in}}{}{}{q7}

\item[8.]  \label{a8}The following sequence requires \textsf{Evaluate}, 
\textsf{Simplify}, \textsf{Combine Trig Functions}, \textsf{Factor}, and 
\textsf{Simplify}. \hfill \hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.2006in%
}{0.243in}{0in}}{}{}{q8} 
\begin{eqnarray*}
\dint 2^{x}\cos bxdx &=&\dfrac{\frac{\ln 2}{b^{2}+\ln ^{2}2}2^{x}+2^{1+x}%
\frac{b}{b^{2}+\ln ^{2}2}\tan \frac{1}{2}bx-\frac{\ln 2}{b^{2}+\ln ^{2}2}%
2^{x}\tan ^{2}\dfrac{1}{2}bx}{1+\tan ^{2}\frac{1}{2}bx} \\
&=&\dfrac{2^{1+x}\ln 2\cos ^{2}\frac{1}{2}bx+2^{1+x}b\sin \frac{1}{2}bx\cos 
\frac{1}{2}bx-\left( \ln 2\right) 2^{x}}{b^{2}+\ln ^{2}2} \\
&=&\dfrac{1}{2\left( b^{2}+\ln ^{2}2\right) }2^{1+x}\ln 2\cos bx+\dfrac{1}{%
2\left( b^{2}+\ln ^{2}2\right) }2^{1+x}\ln 2 \\
&&\quad +\dfrac{1}{2\left( b^{2}+\ln ^{2}2\right) }2^{1+x}b\sin bx-\dfrac{1}{%
b^{2}+\ln ^{2}2}2^{x}\ln 2 \\
&=&\frac{1}{2}\dfrac{2^{1+x}\ln 2\cos bx+2^{1+x}\ln 2+2^{1+x}b\sin
bx-2\left( \ln 2\right) 2^{x}}{b^{2}+\ln ^{2}2} \\
&=&2^{x}\dfrac{\cos bx\ln 2+b\sin bx}{b^{2}+\ln ^{2}2}
\end{eqnarray*}

\item[9.]  \label{a9}The integral 
\[
\int_{-\pi }^{\pi }\frac{1+\sin x}{\left( x-\cos x\right) ^{2}}dx 
\]
is improper, because $x-\cos x=0$ has a root ($\approx .73909$) between $%
-\pi $ and $\pi $ . \textsf{Evaluate} gives 
\begin{eqnarray*}
\int_{-\pi }^{.73909}\frac{1+\sin x}{\left( x-\cos x\right) ^{2}}dx
&=&\,-1.2277\times 10^{5}-\frac{1}{\pi -1} \\
\int_{.73909}^{\pi }\frac{1+\sin x}{\left( x-\cos x\right) ^{2}}dx &=&\,-%
\frac{1}{\pi +1}+1.2277\times 10^{5}
\end{eqnarray*}

Change \textsf{Digits Used in Display} in the \textsf{Settings} dialog box
to $10$. Solving $\cos x=x$ numerically gives $x=.7390851332$. Using this as
a limit, \textsf{Evaluate} gives 
\begin{eqnarray*}
\int_{-\pi }^{.7390851332}\frac{1+\sin x}{\left( x-\cos x\right) ^{2}}dx
&=&\,3.941192327\times 10^{10}-\frac{1}{\pi -1} \\
\int_{.7390851332}^{\pi }\frac{1+\sin x}{\left( x-\cos x\right) ^{2}}dx
&=&\,-\frac{1}{\pi +1}-3.941192327\times 10^{10}
\end{eqnarray*}
providing some evidence that both integrals diverge.\hfill \hyperref{%
\TCIIcon{BITMAPSETProbSolvHint}{0.2006in}{0.243in}{0in}}{}{}{q9}

\item[10.]  \label{a10}\textsl{Scientific Notebook} returns 
\[
\lim_{h\rightarrow 0^{+}}\int_{0}^{\infty }\sin (x^{1+h})\,dx=1 
\]
This result is reasonable, because the integral $f(h)=\int_{0}^{\infty }\sin
(x^{1+h})\,dx$ can be viewed as a convergent alternating series for $h>0,$
and 
\[
g(y)=\int_{0}^{y}\sin x\,dx=1-\cos y 
\]
ranges in value between $0$ and $2,$ with an average value of $1.$\hfill 
\hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.2006in}{0.243in}{0in}}{}{}{q10}

\item[11.]  \label{a11}We need to show that for each of the three functions $%
f(x)=x^{3}$, $f(x)=xe^{x}$, and $f(x)=\sin ^{2}x\cos x$,\hfill \hyperref{%
\TCIIcon{BITMAPSETProbSolvHint}{0.2006in}{0.243in}{0in}}{}{}{q11}

\begin{enumerate}
\item[a.]  If $g$ is defined by $g(x)=\int_{a}^{x}f(t)dt$ for $x\in \left[
a,b\right] $, then $g^{\prime }(x)=f(x)$, and

\item[b.]  If $F$ is any antiderivative of $f$, then.$%
\int_{a}^{b}f(x)dx=F(b)-F(a)$.
\end{enumerate}

For $f(x)=x^{3}$, 
\[
g(x)=\int_{a}^{x}t^{3}dt=\allowbreak \frac{1}{4}x^{4}-\frac{1}{4}a^{4} 
\]
and 
\[
g^{\prime }(x)=\frac{d}{dx}\left( \frac{1}{4}x^{4}-\frac{1}{4}a^{4}\right)
=\allowbreak x^{3} 
\]
The antiderivatives of $f$ are of the form 
\[
F(x)=\int x^{3}dx=\allowbreak \frac{1}{4}x^{4}+C 
\]
for different constants $C$. Now 
\[
F(b)-F(x)=\left[ \allowbreak \frac{1}{4}x^{4}+C\right]
_{x=a}^{x=b}=\allowbreak \frac{1}{4}b^{4}-\frac{1}{4}a^{4} 
\]
which is the same as 
\[
\int_{a}^{b}x^{3}dx=\allowbreak \frac{1}{4}b^{4}-\frac{1}{4}a^{4} 
\]

For $f(x)=xe^{x}$, 
\[
g(x)=\int_{a}^{x}te^{t}dt=\allowbreak xe^{x}-e^{x}-ae^{a}+e^{a} 
\]
and 
\[
g^{\prime }(x)=\frac{d}{dx}\left( xe^{x}-e^{x}-ae^{a}+e^{a}\right)
=\allowbreak xe^{x} 
\]
The antiderivatives of $f$ are of the form 
\[
F(x)=\int xe^{x}dx=\allowbreak xe^{x}-e^{x}+C 
\]
for different constants $C$. Now 
\[
F(b)-F(x)=\left[ xe^{x}-e^{x}+C\right] _{x=a}^{x=b}=\allowbreak
be^{b}-e^{b}-ae^{a}+e^{a} 
\]
which is the same as 
\[
\int_{a}^{b}xe^{x}dx=\allowbreak be^{b}-e^{b}-ae^{a}+e^{a} 
\]

For $f(x)=\sin ^{2}x\cos x$, 
\[
g(x)=\int_{a}^{x}\sin ^{2}t\cos tdt=\allowbreak \frac{1}{3}\sin x-\frac{1}{3}%
\sin x\cos ^{2}x-\frac{1}{3}\sin a+\frac{1}{3}\sin a\cos ^{2}a 
\]
and 
\begin{eqnarray*}
g^{\prime }(x) &=&\frac{d}{dx}\left( \frac{1}{3}\sin x-\frac{1}{3}\sin x\cos
^{2}x-\frac{1}{3}\sin a+\frac{1}{3}\sin a\cos ^{2}a\right) \\
&=&\allowbreak \cos x-\cos ^{3}x=\allowbreak \left( 1-\cos ^{2}x\right)
\left( \cos x\right) =\sin ^{2}x\cos x
\end{eqnarray*}
The antiderivatives of $f$ are of the form 
\[
F(x)=\int \sin ^{2}x\cos xdx=\allowbreak \frac{1}{3}\sin ^{3}x+C 
\]
for different constants $C$. Now 
\[
F(b)-F(x)=\left[ \frac{1}{3}\sin ^{3}x+C\right] _{x=a}^{x=b}=\allowbreak 
\frac{1}{3}\sin ^{3}b-\frac{1}{3}\sin ^{3}a 
\]
while 
\[
\int_{a}^{b}\sin ^{2}x\cos xdx=\allowbreak \frac{1}{3}\sin b-\frac{1}{3}\sin
b\cos ^{2}b-\frac{1}{3}\sin a+\frac{1}{3}\sin a\cos ^{2}a 
\]
Applying \textsf{Combine + Trig Functions} to both of these expression shows
that they are equal. 
\begin{eqnarray*}
\frac{1}{3}\sin b-\allowbreak \frac{1}{3}\sin b\cos ^{2}b-\frac{1}{3}\sin a+%
\frac{1}{3}\sin a\cos ^{2}a &=&\allowbreak \frac{1}{4}\sin b-\frac{1}{12}%
\sin 3b-\frac{1}{4}\sin a+\frac{1}{12}\sin 3a \\
\frac{1}{3}\sin ^{3}b-\frac{1}{3}\sin ^{3}a &=&\allowbreak \frac{1}{4}\sin b-%
\frac{1}{12}\sin 3b-\frac{1}{4}\sin a+\frac{1}{12}\sin 3a
\end{eqnarray*}

Another way to demonstrate this equality is to replace $\cos ^{2}x$ by $%
1-\sin ^{2}x$ and apply \textsf{Expand} to get

\[
\frac{1}{3}\sin b-\allowbreak \frac{1}{3}\left( \sin b\right) \left( 1-\sin
^{2}b\right) -\frac{1}{3}\sin a+\frac{1}{3}\left( \sin a\right) \left(
1-\sin ^{2}a\right) =\allowbreak \frac{1}{3}\sin ^{3}b-\frac{1}{3}\sin ^{3}a 
\]
\medskip
\end{enumerate}

\subsection{\ Related topics}

\begin{itemize}
\item  \hyperref{Evaluating Calculus Expressions}{}{}{DM7-1.tex#Calculus
expressions}

\item  \hyperref{Limits}{}{}{DM7-2.tex#Limit}

\item  \hyperref{Differentiation}{}{}{DM7-3.tex#Derivative}

\item  \hyperref{Indefinite Integration}{}{}{DM7-4.tex#Indefinite integral}

\item  \hyperref{Methods of Integration}{}{}{DM7-5.tex#Methods of integration%
}

\item  \hyperref{Definite Integrals}{}{}{DM7-6.tex#Definite integral}

\item  \hyperref{Sequences and Series}{}{}{DM7-7.tex#Sequences}

\item  \hyperref{Multivariable Calculus}{}{}{DM7-8.tex#Multivariable calculus%
}
\end{itemize}

\rule{0.67in}{0.01in}

\hyperref{\hyperref{\FRAME{itbpF}{0.5275in}{0.3399in}{0.0519in}{}{}{index.wmf%
}{\special{language "Scientific Word";type "GRAPHIC";display
"PICT";valid-file "F";width 0.5275in;height 0.3399in;depth
0.0519in;original-width 36.125pt;original-height 22.5625pt;cropleft
"0";croptop "1";cropright "1";cropbottom "0";filename
'index.wmf';file-properties "XNPEU";}}}{}{}{DMIndex.tex#Math Index}}{}{}{%
DMIndex.tex#Math Index}\quad {\small Index entries: \hyperref{Derivative}{}{%
}{DMIndex.tex#Derivative}, \hyperref{Fundamental theorem of calculus}{}{}{%
DMIndex.tex#Fundamental theorems}, \hyperref{Integral}{}{}{%
DMIndex.tex#Integral}, \hyperref{Limit}{}{}{DMIndex.tex#Limits}, \hyperref{%
Sequences}{}{}{DMIndex.tex#Sequences}, \hyperref{Series}{}{}{%
DMIndex.tex#Series}}

\end{document}
