\begin{frame}
  \frametitle{Review - Midterm Exam 3}

  \begin{exampleblock}{}
    Find the area caught between $f(x) = x^2-2$
    and the $x$-axis from $x=1$ to $x=2$.
    \pause\medskip
    
    The area corresponds to the following integral\pause
    \begin{talign}
      A = \int_1^2 |f(x)|
    \end{talign}
    \pause
    To evaluate this integral, we need the $x$-intercepts in $[1,2]$:
    \begin{talign}
      f(x) = 0 \;\iff\; \mpause[1]{ x = \pm \sqrt{2} }
    \end{talign}
    \pause\pause
    Only $\sqrt{2}$ is in $[1,2]$. \pause Hence
    $
      A \;=\;\mpause[1]{ \left|\int_1^{\sqrt{2}} f(x)\right| \;+\; \left|\int_{\sqrt{2}}^2 f(x)\right| }
    $
    \pause\pause\medskip
    
    An antiderivative of $f(x)$ is $F(x) = \frac{1}{3}x^3 - 2x$. \pause
    \begin{talign}
      A  \;&=\; \mpause[1]{ |F(\sqrt{2}) - F(1)| + |F(2) - F(\sqrt{2})| }\\
      \mpause{&=\; -(F(\sqrt{2}) - F(1)) + (F(2) - F(\sqrt{2})) }
%       \mpause{&=\; |\frac{1}{3}\sqrt{2}^3 - 2\sqrt{2}| + |(\frac{1}{3}2^3 - 2\cdot 2) - (\frac{1}{3}\sqrt{2}^3 - 2\sqrt{2})|}
      \mpause{\;=\; \frac{8\sqrt{2}}{3} - 3}
    \end{talign}
  \end{exampleblock}
\end{frame}