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\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}{.6\textwidth}
Evaluate the integral
\begin{talign}
\int_0^3 (x-1)dx
\end{talign}
by interpreting it in terms of the area.
\end{minipage}
\pause
\begin{minipage}{.39\textwidth}
\begin{center}
\scalebox{.9}{
\begin{tikzpicture}[default]
\def\diaborderx{.25cm}
\def\diabordery{.25cm}
\diagram[1]{-.5}{3.5}{-1}{2}{1}
\diagramannotatez
\diagramannotatex{1}
\diagramannotatey{1}

\begin{scope}[cgreen,ultra thick]
\draw[cred,ultra thick] plot[smooth,domain=0:3,samples=50] function{(x-1)};
\mpause[1]{
\draw[fill=cred,draw=none,opacity=.5] plot[smooth,domain=0:1,samples=50] function{(x-1)} -- (0,0) -- cycle;
\draw[fill=cgreen,draw=none,opacity=.5] plot[smooth,domain=1:3,samples=50] function{(x-1)} -- (3,0) -- cycle;
}
\end{scope}
\end{tikzpicture}
}
\end{center}
\end{minipage}
\pause\pause

Thus the integral is:
\begin{talign}
\int_0^3 (x-1)dx  = {\color{cgreen}\frac{1}{2}(2\cdot 2)} - {\color{cred}\frac{1}{2}(1\cdot 1)} = 1.5
\end{talign}
\end{exampleblock}
\end{frame}