\begin{frame}
  \frametitle{Antiderivatives / Integrals}

  \begin{exampleblock}{}
    A particle moves in a straight line and has 
    \begin{itemize}
      \item acceleration $a(t) = 6t + 4$
      \item initial velocity is $v(0) = -6$cm/s
      \item initial displacement is $s(0) = 9$cm
    \end{itemize}
    Find the position function $s(t)$.
    \pause\medskip
    
    The velocity is an antiderivative of the acceleration:
    \begin{talign}
      v(t) = \mpause[1]{3t^2} \mpause{+ 4t}\mpause{ + C}
    \end{talign}
    \pause\pause\pause\pause
    As $v(0) = -6$cm/s, it follows that $C = \pause -6$.
    \pause\medskip
    
    The position function is an antiderivative of the velocity:
    \begin{talign}
      s(t) = \mpause[1]{t^3} \mpause{+ 2t^2}\mpause{ - 6t}\mpause{+ D}
    \end{talign}
    \pause\pause\pause\pause\pause
    As $s(0) = 9$cm/s, it follows that $D = \pause 9$.
    \pause\medskip
    
    Thus the position function is:
    \begin{talign}
      s(t) = t^3 + 2t^2 - 6t + 9 \quad \text{in cm}
    \end{talign}
  \end{exampleblock}
\end{frame}