\frametitle{Average Value of a Function}
  Let $F$ be an antiderivative of $f$. We know that
    f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx
    \mpause[1]{= \frac{1}{b-a} (F(b) - F(a))}
  By the Mean Value Theorem there exists $c$ in $[a,b]$ such that
    f(c) = \frac{1}{b-a} (F(b) - F(a))

  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}
      f(c) \;=\; f_{\text{avg}} \;=\; \frac{1}{b-a} \int_a^b f(x) \, dx
    that is,\vspace{-2.2ex}
      f(c) (b-a) \;=\; \int_a^b f(x) \, dx