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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{Area between curves}
\vspace{0.3cm}

\begin{example}
Find the areas enclosed by the curves \( y=\arctan (x) \) and \( y=\pi x/4 \).
\textit{The constants are supposed to help.} 
\end{example}

\subsubsection*{Solution}

The region is as pictured:

\vspace{0.3cm}
{\par\centering \includegraphics{area-5.ps} \par}
\vspace{0.3cm}

The intersections occur when \( \pi x/4=\arctan (x) \), which happens at \( x=0 \),
\( x=1, \) and \( x=-1 \) only. Since the region is symmetric, the area enclosed
is twice the area in the first quadrant, \[
Area=2\int _{0}^{1}\left( \arctan (x)-\frac{\pi x}{4}\right) dx.\]



\subsubsection*{Ah, but how do you do that integral?}

Here is the answer:

\[
\int \arctan (x)dx=x\cdot \arctan (x)-\frac{1}{2}\ln \left| 1+x^{2}\right| +C.\]


OK, the absolute value signs are not important. Then plug in to the definite
integral, \begin{eqnarray*}
Area & = & 2\int _{0}^{1}\left( \arctan (x)-\frac{\pi x}{4}\right) dx\\
 & = & 2\left. \left( x\cdot \arctan (x)-\frac{1}{2}\ln \left| 1+x^{2}\right| -\frac{\pi x^{2}}{8}\right) \right| _{0}^{1}\\
 & = & 2\left( 1\cdot \arctan (1)-\frac{1}{2}\ln \left| 1+1^{2}\right| -\frac{\pi 1^{2}}{8}\right) -\left( 0\cdot \arctan (0)-\frac{1}{2}\ln \left| 1+0^{2}\right| -\frac{\pi 0^{2}}{8}\right) \\
 & = & 2\left( 1\frac{\pi }{4}-\frac{1}{2}\ln (2)-\frac{\pi }{8}\right) \\
 & = & \frac{\pi }{4}-\ln (2)\\
 & \cong  & 0.092
\end{eqnarray*}


\begin{rem}
By the way, when I first solved this, I dropped that factor of \( 1/2 \) in
front of the natural log, resulting in an answer that was negative. I only noticed
that when I computed the numerical approximation (with a computer, of course).
My reaction was what yours should be, look back in the computation to see why
I got a negative area. These {}``applied{}'' problems provide a reality check;
since areas are always positive (except in Star Trek http://www.fandom.com/star\_trek/),
if you get a negative area, something was wrong. \end{rem}

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