\begin{frame} \frametitle{Review - Midterm Exam 3} \begin{exampleblock}{} Let $v(t)$ in m/s be the velocity of a particle moving along a line. \pause\medskip \emph{What does $\int_0^x v(t)dt$ tell us?} \\\pause\smallskip Net change of the position, thus the position after $x$ seconds. \pause\medskip \emph{If $v(t) = t^2 - 3t + 2$, find the particles position after $1$s.} \pause\smallskip An antiderivative of $v$ is $V(t) = \frac{1}{3}t^3 - \frac{3}{2}t^2 + 2t$. \pause Thus \begin{talign} p(1) = \int_0^1 v(t)dt = \mpause[1]{V(1) - V(0)} \mpause{= \frac{1}{3} - \frac{3}{2} + 2}\mpause{ = \frac{5}{6} \text{ m}} \end{talign} \pause\pause\pause\pause is the position of the particle after $1$s. \pause\medskip \emph{What is the average velocity during the first second?} \pause\smallskip The average velocity is \begin{talign} \frac{\Delta p}{\Delta t} \mpause[1]{= \frac{p(1)-p(0)}{1}} \mpause{= \frac{5}{6} \text{m/s}} \end{talign} \end{exampleblock} \end{frame}