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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}

The concept of an \textit{average} is easy. To find the average (or mean) score
on an exam, we simply add up all the scores and divide by the number of people
who took the exam. More generally, the \textit{average} of a collection \( \{y_{1},\ldots ,y_{n}\} \)
of \( n \) numbers is \[
y_{av}:=\frac{y_{1}+y_{2}+\ldots +y_{n}}{n}.\]


It is perhaps a little less clear what we would mean by the \textit{average}
value of a function defined over an interval, but the concept is really the
same.  To find the average of \( f(x) \) over the interval \( [a,b] \), you
could simply find a bunch of points in there, add up the values of the function
on all those points, and divide by the total number of points you chose. But,
of course this is only approximately the average, since you just chose a few
points. To make it more accurate, you'd just take more points, and in the end,
you'd have meaningless infinities top and bottom. 

This is the first application of integration where the integral is not quite
so apparently an integral. In this case, if, say you look at values of \( f(x) \)
at \( \{x_{1},\ldots ,x_{n}\} \), \( \{f(x_{1}),\ldots ,f(x_{n})\} \), then
the numerator of the average is \( (f(x_{1})+\ldots +f(x_{n})) \), which is
not an integral, even approximately. But if we multiply each of those terms
by \( dx \), it does look more like an integral formula, \( (f(x_{1})dx+\ldots +f(x_{n})dx) \).
Remember that \( dx \) is just the small distance in the \( x \) direction,
the width of the subinterval we're at. But, of course we have to multiply the
bottom \( (n) \) by \( dx \) as well, which at first doesn't make any sense,
\( n\, dx \) (at least not in the limit as \( dx\rightarrow 0 \) and \( n\rightarrow \infty  \)).
But, if you write \( n=1+\ldots +1 \) (\( n \) times, of course), then \( n\, dx=dx+dx+\ldots +dx \),
which looks, now, like the integral of \( 1 \). 

So, the \textit{average value} of \( f(x) \), for \( x\in [a,b] \), is \[
f_{av}:=\frac{\int _{a}^{b}f(x)\, dx}{\int _{a}^{b}1\, dx}=\frac{1}{b-a}\int _{a}^{b}f(x)\, dx,\]


which is the limit of these sums as \( dx\rightarrow 0 \). Of course, \[
\int _{a}^{b}1\, dx=b-a.\]
As with many of these applications, the examples are not the point. The thing
to concentrate on is the derivation of the formula. However, some examples will
help you with the homework.

\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}

\begin{xca}
Find the average value of \( f(x):=\sin ^{2}(x) \) on the interval \( [0,\pi ] \).
\( f_{av}= \)
\end{xca}
%

\begin{xca}
Egbert is slicing oranges. He slices them by setting them on a table, and giving
them a good thwack with a very sharp knife. Unfortunately, he's doesn't have
very good aim, and his thwack could, with equal likelihood, hit the orange anywhere.
What is the average size of the cut surface of Egbert's oranges? He does always
at least nick the oranges (when he doesn't, he doesn't count that wiff as a
thwack), and the oranges are all perfect spheres of radius 2 inches. \end{xca}

\end{document}
