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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}