\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}
      f(c) \;=\; f_{\text{avg}} \;=\; \frac{1}{b-a} \int_a^b f(x) \, dx

    Since $f(x) = 1+ x^2$ is continuous on $[-1,2]$,
    the Mean Value Theorem for Integrals says\ldots
    There is a number $c$ in $[-1,2]$ such that
      f(c) = \frac{1}{b-a} \int_a^b f(x) \, dx
    For this $f$, we can find $c$ explicitly. \pause Since $\frac{1}{b-a} \int_a^b f(x) \, dx = 2$
      2 &= f(c) = 1+ c^2 \mpause[1]{\;\implies\; c^2 = 1} \mpause{\;\implies\; c = \pm 1}
    Thus there are two numbers $c = -1$ and $c= 1$ that work!