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\begin{frame}
  \frametitle{Fundamental Theorem of Calculus}
  
  \fundamental
  \medskip

  \begin{exampleblock}{}
    Evaluate\vspace{-1ex}
    \begin{talign}
      \int_{-1}^3 \frac{1}{x^2}dx \quad\mpause[10]{\text{\alert{does not exist}}}
    \end{talign}
    \pause    
    An antiderivative of $f(x) = \frac{1}{x^2}$ is $F(x) = \pause -\frac{1}{x}$. 
    \pause
    Then
    \begin{talign}
      \int_{-1}^3 \frac{1}{x^2}dx = \mpause[1]{-\frac{1}{x} \Big]_{-1}^3} \mpause{= -\frac{1}{3} - (-\frac{1}{-1})} \mpause{= -\frac{4}{3}}
    \end{talign}
    \pause\pause\pause\pause
    Does this make sense? \pause Note that $\frac{1}{x^2}$ is above the $x$-axis!
    \pause\medskip
    
    \alert{The calculation is wrong since $\frac{1}{x^2}$ is not continuous on $[-1,3]$!}
  \end{exampleblock}
  \vspace{10cm} 
\end{frame}