\begin{frame} \frametitle{Review - Midterm Exam 3} \begin{exampleblock}{} Find the area caught between $f(x) = x^2-2$ and the $x$-axis from $x=1$ to $x=2$. \pause\medskip The area corresponds to the following integral\pause \begin{talign} A = \int_1^2 |f(x)| \end{talign} \pause To evaluate this integral, we need the $x$-intercepts in $[1,2]$: \begin{talign} f(x) = 0 \;\iff\; \mpause[1]{ x = \pm \sqrt{2} } \end{talign} \pause\pause Only $\sqrt{2}$ is in $[1,2]$. \pause Hence $ A \;=\;\mpause[1]{ \left|\int_1^{\sqrt{2}} f(x)\right| \;+\; \left|\int_{\sqrt{2}}^2 f(x)\right| } $ \pause\pause\medskip An antiderivative of $f(x)$ is $F(x) = \frac{1}{3}x^3 - 2x$. \pause \begin{talign} A \;&=\; \mpause[1]{ |F(\sqrt{2}) - F(1)| + |F(2) - F(\sqrt{2})| }\\ \mpause{&=\; -(F(\sqrt{2}) - F(1)) + (F(2) - F(\sqrt{2})) } % \mpause{&=\; |\frac{1}{3}\sqrt{2}^3 - 2\sqrt{2}| + |(\frac{1}{3}2^3 - 2\cdot 2) - (\frac{1}{3}\sqrt{2}^3 - 2\sqrt{2})|} \mpause{\;=\; \frac{8\sqrt{2}}{3} - 3} \end{talign} \end{exampleblock} \end{frame}