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\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}
\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}