\begin{frame} \frametitle{Average Value of a Function} \begin{block}{Mean Value Theorem \emph{for Integrals}} If $f$ is continuous an $[a,b]$, then there is $c$ 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} \end{block} \pause\medskip \begin{exampleblock}{} Since $f(x) = 1+ x^2$ is continuous on $[-1,2]$, the Mean Value Theorem for Integrals says\ldots \pause \medskip There is a number $c$ in $[-1,2]$ such that \begin{talign} f(c) = \frac{1}{b-a} \int_a^b f(x) \, dx \end{talign} \pause For this $f$, we can find $c$ explicitly. \pause Since $\frac{1}{b-a} \int_a^b f(x) \, dx = 2$ \pause \begin{talign} 2 &= f(c) = 1+ c^2 \mpause[1]{\;\implies\; c^2 = 1} \mpause{\;\implies\; c = \pm 1} \end{talign} \pause\pause\pause Thus there are two numbers $c = -1$ and $c= 1$ that work! \end{exampleblock} \vspace{10cm} \end{frame}