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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{The Fundamental Theorem of Calculus}

\begin{prop}
\textbf{\emph{{[}MVT, integral form{]}}} If \( f(x) \) is continuous on \( [a,b] \),
then there is a \( c\in [a,b] \) so that \[
f(c)=\frac{\int _{a}^{b}f(x)dx}{b-a}.\]

\end{prop}
\begin{rem}
If you are used to thinking about the definite integral as an antiderivative,
then this is really just the old MVT applied to the antiderivative of \( f \),
but we're not there, just yet. This is a slightly different statement from the
regular MVT, but is even easier to prove. We are assuming that any continuous
function is integrable, which we proved earlier.
\end{rem}
\textbf{\emph{Proof:}}

Since \( f \) is continuous on \( [a,b] \), it has a maximum value \( M \)
and minimum value \( m \) on that interval, \[
m\leq f(x)\leq M,\, \textrm{for all}\, x\in [a,b].\]
But then, by the properties of integration, \[
m(b-a)\leq \int _{a}^{b}f(x)dx\leq M(b-a),\]
 or \[
m\leq \frac{\int _{a}^{b}f(x)dx}{b-a}\leq M.\]
However, any value between the maximum and minimum values of \( f \) on \( [a,b] \)
is a value of \( f \) on \( [a,b] \), by the Intermediate-Value Theorem, so
there is a \( c\in [a,b] \) for which \[
f(c)=\frac{\int _{a}^{b}f(x)dx}{b-a},\]
which is what we needed to show.
\end{document}
