\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}