\begin{frame} \frametitle{The Definite Integral} \begin{center} \scalebox{.7}{ \begin{tikzpicture}[default] \def\mfun{(-.9 + (\x-3+\mfunshift)^2 - .1*(\x-3+\mfunshift)^4)} \diagram[1]{-.5}{6}{-1}{1.7}{1} \diagramannotatez \def\mfunshift{0} \begin{scope}[ultra thick] \draw[fill=cgreen,draw=none,opacity=.5] plot[smooth,domain=.5:2,samples=100] (\x,{\mfun}) -- (.5,0) -- cycle; \draw[fill=cred,draw=none,opacity=.5] plot[smooth,domain=2:4,samples=100] (\x,{\mfun}) -- cycle; \draw[fill=cgreen,draw=none,opacity=.5] plot[smooth,domain=4:5.5,samples=100] (\x,{\mfun}) -- (5.5,0) -- cycle; \draw[cred] plot[smooth,domain=.5:5.5,samples=100] (\x,{\mfun}); \node[anchor=north] at (.5,0) {$a$}; \node[anchor=north] at (5.5,0) {$b$}; \node[scale=1.8] at (.9,.5) {+}; \node[scale=1.8] at (5.15,.5) {+}; \node[scale=3] at (3,-.5) {-}; \end{scope} \end{tikzpicture} } \end{center} \begin{exampleblock}{} \begin{minipage}{.7\textwidth} Evaluate the integral \begin{talign} \int_0^1 \sqrt{1-x^2}dx \end{talign} by interpreting it in terms of the area. \end{minipage} \pause \begin{minipage}{.29\textwidth} \begin{center} \scalebox{.9}{ \begin{tikzpicture}[default] \def\diaborderx{.25cm} \def\diabordery{.25cm} \diagram[1]{-.5}{1.5}{-.5}{1.5}{1} \diagramannotatez \diagramannotatex{1} \diagramannotatey{1} \begin{scope}[cgreen,ultra thick] \draw[fill=cgreen,draw=none,opacity=.5] plot[smooth,domain=0:1,samples=50] function{(1-x**2)**(.5)} -- (0,0) -- cycle; \draw[cgreen,ultra thick] plot[smooth,domain=0:1,samples=50] function{(1-x**2)**(.5)}; \end{scope} \end{tikzpicture} } \end{center} \end{minipage} \pause\medskip Thus the area is $1/4$ of the area of a circle with radius $1$: \begin{talign} \int_0^1 \sqrt{1-x^2}dx = \frac{\pi}{4} \end{talign} \end{exampleblock} \end{frame}