\begin{frame} \frametitle{Average Value of a Function} Let $F$ be an antiderivative of $f$. We know that \begin{talign} f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx \mpause[1]{= \frac{1}{b-a} (F(b) - F(a))} \end{talign} \pause\pause By the Mean Value Theorem there exists $c$ in $[a,b]$ such that \begin{talign} f(c) = \frac{1}{b-a} (F(b) - F(a)) \end{talign} \pause\medskip Thus we obtain: \begin{block}{Mean Value Theorem \emph{for Integrals}} If $f$ is continuous an $[a,b]$, then there exists a number $c$ in $[a,b]$ such that:\vspace{-1ex} \begin{talign} f(c) \;=\; f_{\text{avg}} \;=\; \frac{1}{b-a} \int_a^b f(x) \, dx \end{talign} \pause that is,\vspace{-2.2ex} \begin{talign} f(c) (b-a) \;=\; \int_a^b f(x) \, dx \end{talign} \end{block} \vspace{10cm} \end{frame}