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\begin{frame}
\frametitle{Review - Midterm Exam 3}

\begin{exampleblock}{}
Find the area of the region bounded by
the curves \vspace{-.5ex}
\begin{talign}
y = \sin x && y = \cos x &&
x = 0 && x = \pi/2
\end{talign}
\pause
Area is equal to the area between $\sin x - \cos x$ and the $x$-axis:\vspace{-.5ex}
\begin{talign}
A = \int_0^{\pi/2} |\sin x - \cos x|dx
\end{talign}
\pause
We have to find the $x$-intercepts in the interval $[0,\pi/2]$:\vspace{-.5ex}
\begin{talign}
\sin x - \cos = 0 \mpause[1]{\;\iff\; \sin x = \cos x} \mpause{\;\iff\; x =} \mpause{ \pi/4}
\end{talign}
\pause\pause\pause\pause
Antiderivative of $f(x) = \sin x - \cos x$ is $F(x) = \pause -\cos x - \sin x$:\vspace{-.5ex}
\pause
\only<-15>{
\begin{talign}
\left| \int_0^{\pi/4} f(x)dx \right|
+ \left| \int_{\pi/4}^{\pi/2} f(x)dx \right|
}
\mpause{
= | F(x) \big]_0^{\pi/4} | + | F(x) \big]_{\pi/4}^{\pi/2} |
} \\[-.5ex]
\mpause{
&= | (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) - (-1 - 0)|
+ | (-0 - 1) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}})|
}\\
\mpause{
&= | -\frac{2}{\sqrt{2}} +1| + | -1 + \frac{2}{\sqrt{2}}|
}
\mpause{
= \sqrt{2} -1 + -1 + \sqrt{2}
}
\mpause{
}
\end{talign}
}
\only<16>{
\begin{center}
\scalebox{.8}{
\begin{tikzpicture}[default,xscale=2]
\def\mfuna{(cos (180/pi *\x))}
\def\mfunb{(sin (180/pi *\x))}

\diagram[1]{-.5}{3}{-1}{1.2}{1}
\diagramannotatez
\def\mfunshift{0}
\begin{scope}[ultra thick]
\draw[cred] plot[smooth,domain=-.5:3,samples=100] (\x,{\mfuna}) node[above] {$f(x)$};
\draw[cblue] plot[smooth,domain=-.5:3,samples=100] (\x,{\mfunb}) node[above,yshift=1.5mm] {$g(x)$};
\draw[fill=cgreen,draw=none,opacity=.5] plot[smooth,domain=0:pi/4,samples=100] (\x,{\mfuna}) -- plot[smooth,domain=pi/4:0,samples=100] (\x,{\mfunb}) -- cycle;
\draw[fill=cgreen,draw=none,opacity=.5] plot[smooth,domain=pi/4:pi/2,samples=100] (\x,{\mfuna}) -- plot[smooth,domain=pi/2:pi/4,samples=100] (\x,{\mfunb}) -- cycle;
\end{scope}
\end{tikzpicture}
}
\end{center}\vspace{-2.5ex}
\begin{talign}
\alert{A = 2\sqrt{2} -2}
\end{talign}
}
\end{exampleblock}
\end{frame}