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\begin{frame}
  \frametitle{The Area below a Curve}

  \begin{exampleblock}{}
    Estimate the area below the curve $f(x) = x^2$ from $0$ to $1$.
  \end{exampleblock}
  \pause\vspace{-2ex}

  \begin{talign}
    \lim_{n\to \infty} R_n = \frac{1}{3}
    \mpause[1]{ &&\text{ and similar }&& \lim_{n\to \infty} L_n = \frac{1}{3} }
  \end{talign}
  \pause\pause
  The right- and left-approximations converge to the same value.
  \pause
  \begin{center}
  \scalebox{.65}{
  \begin{tikzpicture}[default]
    \def\mfun{(4*((\x+\mfunshift)/4)^2)}

    \diagram[1]{-.5}{4.5}{-.4}{4.5}{1}
    \diagramannotatez
    \diagramannotatexx{1/.25,2/.5,3/.75,4/1}
    \diagramannotateyy{1/.25,2/.5,3/.75,4/1}
    \def\mfunshift{0}
    \begin{scope}[ultra thick]
      \draw[cred] plot[smooth,domain=0:4,samples=20] (\x,{\mfun});
      \draw[draw=none,fill=cred,opacity=.5] (.5,0) -- plot[smooth,domain=0:4,samples=20] (\x,{\mfun}) -- (4,0) -- (0,0) -- cycle;
    \end{scope}
  \end{tikzpicture}
  }
  \end{center}\vspace{-1ex}
  
  \begin{exampleblock}{}
  We define the area $A$ to be the limit of these approximations\vspace{-.5ex}
  \begin{talign}
    A = \lim_{n\to \infty} R_n = \lim_{n\to \infty} L_n = \frac{1}{3}
  \end{talign}
  \end{exampleblock}
  
  \vspace{10cm}
\end{frame}