%Improper Integrals
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%Math263B
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{\Large
Improper Integrals\footnote{Copyright \copyright 2002 Larry Snyder and Todd Young.
All rights reserved. Please address comments to young@math.ohiou.edu.}}
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\begin{enumerate}
\item Enter the following sequence of commands: \\
\verb& syms x &\\
\verb& int(1/sqrt(x^6 + 1), 0, inf) & \dotfill Calculates symbolically\\
\verb& double(ans) & \dotfill Converts to a numeric format\\
\verb& quadl('1./sqrt(x.^6 + 1)', 0, inf) & \dotfill Calculates numerically
\item Use the commands above to evaluate the following integrals (you will encounter
error messages in some of them):
\begin{enumerate}
\item $ {\displaystyle
\int_0^\infty
\frac{1}{x^{2/3}}dx}$
\hfill (Use \verb& 1/x^(2/3)&.)
\item $ {\displaystyle
\int_1^\infty
\frac{1}{x + 1}dx}$
\item $ {\displaystyle
\int_1^\infty
\frac{\ln x}{x^2}}$
\hfill (Use \verb& log & for natural logarithm.)
\item $ {\displaystyle
\int_0^\infty
\sin{^2}(x)dx}$
\hfill(Use \verb$ (sin(x))^2 $)
\end{enumerate}
\item Try to use \textsc{MatLab} to evaluate the following functions using commands in \#1:
\begin{enumerate}
\item $ {\displaystyle
\int_{-1}^1
\frac{1}{x^2}dx}$
\item $ {\displaystyle
\int_0^1
\frac{1}{\sqrt{x}}dx}$
\end{enumerate}
\item What are some problems with calculating improper integrals numerically?
\item Try the following: \\
\verb& int(1/x^5, 1, inf) & \\
\verb& int(sin(x^3)/x^5, 1, inf) &\\
\verb& double(ans) & \\
Comparing the integrands of these two integrals, should the second one converge?
Does the answer for the second integral make sense?
\item Prepare a brief (\verb$< $1 page) written report describing
what happened and answering all the questions. Use complete sentences
and standard mathematical notation. Writing quality will play a part in the grade.
\end{enumerate}
\vfill
\noindent
\textsf{This exercise explores improper integrals both symbolically and numerically.
Evaluating improper integrals symbolically is precarious because
it is hard for the computer to handle the symbol $\infty$
correctly. Evaluating
numerically is also difficult because one cannot actually compute
all the way to $\infty$, one must stop at some finite place.
}
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