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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{Average value of a function}

\begin{example}
Find the average value of \( \sin (x) \) over  
\end{example}
\begin{enumerate}
\item \( [0,\pi ] \), 
\item \( [0,2\pi ] \) 
\end{enumerate}

\subsubsection*{Solution}

For the first interval, \begin{eqnarray*}
f_{av} &amp; := &amp; \frac{1}{\pi -0}\int _{0}^{\pi }\sin (x)\, dx\\
 &amp; = &amp; \frac{1}{\pi }\left. \left( -\cos (x)\right) \right| ^{\pi }_{0}\\
 &amp; = &amp; \frac{1}{\pi }\left( -\cos (\pi )+\cos (0)\right) \\
 &amp; = &amp; \frac{2}{\pi }.
\end{eqnarray*}
You should first check to see that this is a reasonable answer. It is, since
the graph of \( \sin (x) \) on that interval looks like:

\vspace{0.3cm}
{\par\centering \includegraphics{sine.eps} \par}
\vspace{0.3cm}

Note that an answer greater than \( 1/2 \) should be right (and of course,
it should be bigger than 0 and less than 1!), since most of the time the graph
is above the mark for height 0.5. Always verify your results by checking to
make sure they are reasonable. Areas and the like have to be positive (averages
don't, of course). Averages should certainly be between the maximum and the
minimum of the function on the interval.

For that same function on the interval \( [0,2\pi ] \), things are a little
different:\begin{eqnarray*}
f_{av} &amp; := &amp; \frac{1}{2\pi -0}\int _{0}^{2\pi }\sin (x)\, dx\\
 &amp; = &amp; \frac{1}{2\pi }\left. \left( -\cos (x)\right) \right| ^{2\pi }_{0}\\
 &amp; = &amp; \frac{1}{2\pi }\left( -\cos (2\pi )+\cos (0)\right) \\
 &amp; = &amp; 0.
\end{eqnarray*}


0? Is that a reasonable answer? Actually, of course it is, since the graph:

\vspace{0.3cm}
{\par\centering \includegraphics{sine2.eps} \par}
\vspace{0.3cm}

spends just about as much time above and below the \( x \)-axis. By symmetry,
it spends exactly as much time above the axis as below, so an average of 0 is
to be expected.
\end{document}
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