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\begin{center}
{\Large Warm-Up Problems and Lecture Problems

Solutions

February 17, 2003}
\end{center}

\begin{enumerate}

\item  \label{one}
	Find the points of intersection for the following sets of graphs.
	\begin{enumerate}
	\item  $y=12-x^2$ and $y=x^2$
		\\\sol Solve the equation $12-x^2=x^2$ and get $x=\pm \sqrt{6}$.
		So the points of intersection are $(-\sqrt{6}, 6)$ and $(\sqrt{6},6)$.
	\item  $y=x^2$ and $y=2x+8$
		\\\sol Solve the equation $x^2=2x+8$ and get the points
		$(-2,4)$ and $(4,16)$.
	\item  $x=y^2$ and $x=y+6$
		\\\sol Solve the equation $y^2=y+6$ and get the points:
		$(4,-2)$ and $(9,3)$.
	\item  $x=y^2-2y-2$ and $x=4+y-2y^2$
		\\\sol Solve the equation $y^2-2y-2=4+y-2y^2$ and get:
		$(1,-1)$ and $(-2,2)$.
	\end{enumerate}

\item  \label{two}
	Using your calculator, estimate the points of intersection of the following sets of graphs:
	\begin{enumerate}
	\item  $y=x^3$ and $y=32-x^4$
		\\\sol You can graph these or use a polynomial solver on your calculator.
		I got the following points:
		$(-2.67388, -19.11726)$ and $(2.16284, 10.11749)$.
	\item  $y=\frac{7x}{(x^2+1)^{3/2}}$ and $y=\frac{x^2}{3}$
		\\\sol By graphing you can find the points of intersection:
		$(0,0)$ and $(1.96331, 1.28487)$.
	\end{enumerate}

\newpage
\lecture

\item  Find the area between the curves (hint: see your work in problem~\ref{one})
	\begin{enumerate}
	\item  $y=12-x^2$ and $y=x^2$
		\\\sol
		\[
		\int_{-\sqrt{6}}^{\sqrt{6}} (12-x^2) - (x^2) \; dx = 16\sqrt{6}
		\]
	\item  $y=x^2$ and $y=2x+8$
		\\\sol
		\[
		\int_{-2}^{4} (2x+8)-(x^2) \; dx = 36
		\]
	\item  $x=y^2$ and $x=y+6$
		\\\sol
		\[
		\int_{-2}^3 (y+6) - (y^2) \; dy = \frac{125}{6}
		\]
	\item  $x=y^2-2y-2$ and $x=4+y-2y^2$
		\\\sol
		\[
		\int_{-1}^{2} (4+y-2y^2) - (y^2-2y-2) \; dy = \frac{27}{2}
		\]
	\end{enumerate}

\item  Find the area between the curves (hint: see your work in problem~\ref{two})
	\begin{enumerate}
	\item  $y=x^3$ and $y=32-x^4$
		\\\sol
		\[
		\int_{-2.67388}^{2.16284} (32-x^4) - (x^3) \; dx = 125.28175
		\]
	\item  $y=\frac{7x}{(x^2+1)^{3/2}}$ and $y=\frac{x^2}{3}$
		\\\sol
		\[
		\int_{0}^{1.96331} \frac{7x}{(x^2+1)^{3/2}} - \frac{x^2}{3} \; dx
		= 2.98210
		\]
	\end{enumerate}







\end{enumerate}



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