\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}