\begin{frame}
  \frametitle{Indefinite Integrals: Applications}

  \begin{exampleblock}{}
    We consider an object moving in a straight line with velocity
    \begin{talign}
      v(t) = t^2 - t - 6 \text{m/s}
    \end{talign}
    Find the displacement and distance traveled during $1 \le t \le 4$.
    \pause\medskip
    
    \only<-7>{
    The displacement is
    \begin{talign}
      \int_{1}^{4} v(t) dt &= 
      \mpause[1]{(\frac{1}{3}v^3 - \frac{1}{2}t^2 - 6t)\big]_1^4} \\
      \mpause{&= (\frac{1}{3}4^3 - \frac{1}{2}4^2 - 6\cdot 4) - (\frac{1}{3}1^3 - \frac{1}{2}1^2 - 6\cdot 1)} \\
      \mpause{&= (\frac{64}{3} - \frac{16}{2} - 24) - (\frac{1}{3} - \frac{1}{2} - 6)} 
      \mpause{= -\frac{9}{2}} 
    \end{talign}
    \pause\pause\pause\pause\pause
    The displacement is $-4.5$m.
    }
    \only<8->{
    \pause[8]
    The total distance traveled is \quad
    $
      \int_{1}^{4} |v(t)| dt 
    $
    \pause\medskip
    
    We need to find the $x$-intercepts of $v(t)$ in $[1,4]$:\pause
    \begin{talign}
      v(t) = (t+2)(t-3) = 0 \quad\iff\quad t = -2 \;\text{ or }\; t = 3
    \end{talign}
    \pause
    Thus we have:
    \begin{talign}
      \int_{1}^{4} |v(t)| dt  &= \left| \int_{1}^{3} v(t)  \right| \;+\;  \left| \int_{3}^{4} v(t)  \right| \\
      \mpause[1]{
        &= \left| (\frac{1}{3}v^3 - \frac{1}{2}t^2 - 6t)\big]_1^3 \right| +
        \left| (\frac{1}{3}v^3 - \frac{1}{2}t^2 - 6t)\big]_3^4 \right| 
      }\\
      \mpause{
        &= \left| -22/3 \right| +
        \left| 17/6 \right|
      }
      \mpause{
        = 61/6
      }
    \end{talign}\vspace{-.5ex}
    \pause\pause\pause\pause
    The total distance traveled is $61/6$\;m.
    }
  \end{exampleblock}
  \vspace{10cm}
\end{frame}