\begin{frame} \frametitle{Area Between Curves} \begin{exampleblock}{} Find the area between $f(x) = x^2 - 6x + 8$ from $1$ to $6$. \medskip\pause An antiderivative of $f$ is $F(x) = \pause \frac{1}{3}x^3 - 3x^2 + 8x$. \pause\smallskip We need to find the $x$-intercepts in $[1,6]$: \begin{talign} f(x) = \mpause[1]{(x - 2)(x-4)}\mpause{ = 0 \quad\iff\quad x = 2 \;\text{ or }\; x = 4} \end{talign} \pause\pause\pause\vspace{-2.5ex} \only<-11>{\vspace{-1ex} \begin{center} \scalebox{.9}{ \begin{tikzpicture}[default,yscale=.7] \def\mfun{(\x^2 - 6*\x + 8)} \diagram[1]{-.5}{7}{-1}{3}{1} \diagramannotatez \def\mfunshift{0} \begin{scope}[ultra thick] \draw[fill=cgreen,draw=none,opacity=.5] plot[smooth,domain=1:2,samples=20] (\x,{\mfun}) -- (1,0) -- cycle; \draw[fill=cgreen,draw=none,opacity=.5] plot[smooth,domain=2:4,samples=20] (\x,{\mfun}) -- cycle; \draw[fill=cgreen,draw=none,opacity=.5] plot[smooth,domain=4:5,samples=20] (\x,{\mfun}) -- (6,3) -- (6,0) -- cycle; \draw[cred] plot[smooth,domain=1:5,samples=20] (\x,{\mfun}); \node[anchor=north] at (1,0) {$1$}; \node[anchor=north] at (6,0) {$6$}; \draw[gray] (2,.2) -- node[at end,below,black] {$2$} (2,-.2); \draw[gray] (4,.2) -- node[at end,below,black] {$4$} (4,-.2); \end{scope} \end{tikzpicture} } \end{center}\vspace{-1ex} } \pause Then the area between the curve and the $x$-axis from $1$ to $6$ is \begin{talign} A \quad&=\quad \mpause[1]{\left| \int_1^2 f(x)dx \right|} \mpause{\;+\; \left|\int_2^4 f(x)dx\right|} \mpause{\;+\; \left|\int_4^6 f(x)dx\right|} \\\mpause{} &\mpause{= \quad \left|F(2) - F(1)\right| \;+\; \left|F(4)-F(2)\right| \;+\; \left|F(6) - F(4)\right|} \\ &\mpause{= \quad \left|\frac{20}{3} - \frac{16}{3}\right| \;+\; \left|\frac{16}{3} - \frac{20}{3}\right| \;+\; \left|\frac{36}{3} - \frac{16}{3}\right|} \\ &\mpause{= \quad \left|\frac{4}{3}\right| \;+\; \left|-\frac{4}{3}\right| \;+\; \left|\frac{20}{3}\right|} \\ &\mpause{= \quad \frac{28}{3}} \end{talign}\vspace{-2ex} \end{exampleblock} \end{frame}