\begin{frame} \frametitle{Indefinite Integrals: Applications} \begin{exampleblock}{} We consider an object moving in a straight line: \begin{itemize} \pause \item $v(t)$ is the velocity \end{itemize} \pause Then \begin{talign} \int_{t_1}^{t_2} \alert{|}v(t)\alert{|} dt \end{talign} is the \emph{total distance} the object traveled during the time interval. \end{exampleblock} \pause \begin{center} \scalebox{.9}{ \begin{tikzpicture}[default] \def\mfun{(-.9 + (\x-3+\mfunshift)^2 - .1*(\x-3+\mfunshift)^4)} { \def\diax{t} \def\diay{v(t)} \diagram[1]{-.5}{6}{-.75}{1.5}{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) {$t_1$}; \node[anchor=north] at (5.5,0) {$t_2$}; \node[scale=1] at (.9,.5) {$A_1$}; \node[scale=1] at (5.15,.5) {$A_3$}; \node[scale=1] at (3,-.5) {$A_2$}; \end{scope} \node at (2.75,-1.75) {displacement = $\int_{t_1}^{t_2} v(t) dt = A_1 - A_2 + A_3$}; \node at (2.75,-2.5) {total distance = $\int_{t_1}^{t_2} \alert{|}v(t)\alert{|} dt = A_1 \alert{+} A_2 + A_3$}; \end{tikzpicture} } \end{center} \vspace{10cm} \end{frame}