\begin{frame} \frametitle{The Definite Integral} \begin{block}{} The definite integral can be interpreted as the \emph{net area}, that is: \begin{talign} \int_a^b f(x) dx = A_1 - A_2 \end{talign} where \begin{itemize} \item $A_1$ is the area of above the $x$-axis, below the curve, \item $A_2$ is the area of below the $x$-axis, above the curve. \end{itemize} \end{block} \medskip \begin{center} \scalebox{.9}{ \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} \vspace{10cm} \end{frame}