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\title[Problem Set 7]{Problem Set 7.  Due Thursday, 14 September}

\begin{document}

\maketitle



\noindent {\bf Reading.} {\em Quick Calculus}, pp. 185--198.

\noindent {\bf Supplementary reading.} Simmons, sections 5.4, 7.3,
7.4, 10.4, and 10.6.

\begin{enumerate}

\item (4pts) Graph the following regions.  Rotate them around the $x$-axis and
compute their volumes.

\begin{enumerate}
\item The region below $f(x)=\sqrt{x}$, above $y=0$ and to the left of 
$x=4$.
\item The region below $f(x)=4x-x^2$ above $y=0$.
\item The region below $f(x)=4x-x^2$ and above $g(x)=3(x-2)^2$.
(Hint: $f(x)=g(x)$ when $x=1,3$.)
\end{enumerate}

\item (4pts) Graph the following regions.  Rotate them around the $y$-axis and
compute their volumes.

\begin{enumerate}
\item The region below $f(x)=\sqrt{x}$, above $y=0$ and to the left of 
$x=4$.
\item The region above $f(x)=x^3$, below $y=8$ and to the right of $x=0$.
\item The region below $f(x)=\sin(x)$, above $y=0$ from $x=0$ to $x=\pi$.
\end{enumerate}

\item (10pts) Compute the following integrals.  For some, you may need
to apply more than one technique to compute the final integral.

\begin{enumerate}
\item $\int \frac{x^2}{\sqrt{1-x^2}}\ dx$
\item $\int \frac{x}{\sqrt{1-x^2}}\ dx$
\item $\int\frac{25}{(x-4)(2x+1)}\ dx$
\item $\int\frac{6x^2-4}{x^2(x-2)}\ dx$
\item $\int\frac{4e^x}{e^{2x}-4}\ dx$
\end{enumerate}

\item (2pts) Let's compute the integral $\int\sec(\theta)\ d\theta$.  We can
transform this
$$
\int\sec(\theta)\ d\theta = \int\frac{d\theta}{\cos(\theta)}
    = \int\frac{\cos(\theta)\ d\theta}{\cos(\theta)}
    = \int\frac{\cos(\theta)}{1-sin^2(\theta)}\ d\theta.
$$
Now make a substitution, and then use partial fractions to complete
the integral.





\end{enumerate}





\end{document}