\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}