\documentclass{article}
\input{/home/william/preamble.tex}
\usepackage{graphicx}
\begin{document}
	{\noindent 21-120 Differential and Integral Calculus \hfill Summer Session I 2010\\}
	{\centering Quiz 16\medskip\\}
	\hrule
	{
		\begin{prob*} \mbox{}
			Find the area bounded by the curves
			\begin{gather*}
				y = x^2 \\
				y = 4x-x^2
			\end{gather*}
		\end{prob*}
		\begin{soln*}\mbox{}
			Here is the graph of this boundary:
			\begin{figure}[h]
				\centering
				\includegraphics[bb=0 0 814 470,scale=.4]{quiz16.png}
			\end{figure}

			We need to know where they intersect. So, we set them equal, and solve.
			\begin{align*}
				& x^2 = 4x-x^2 \\
				\implies & 2x^2 -4x = 0 \\
				\implies & 2x(x-2) = 0 \\
				\implies & x=0\text{ or } x = 2
			\end{align*}
			Thus, the points of intersection are $(0,0)$ and $(2,4)$. We make slices perpendicular to the $x$-axis (for no other reason other than to avoid having to find the inverse of $y=4x-x^2$) and note that $y=4x-x^2$ is the top function. Thus, to find the area we integrate
			\begin{align*}
				\int\limits_0^2 \left((4x-x^2) - (x^2) \right)dx 	&= \int\limits_0^2 \left(4x-2x^2 \right)dx \\
											&= [2x^2 - \frac{2}{3}x^3]_0^2\\
											&= 2(2)^2 - \frac{2}{3}(8) \\
											&= 8(1 - \frac{2}{3}) \\
											&=\frac{8}{3}
			\end{align*}
		\end{soln*}
	}
\end{document}