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\title[Problem Set 6]{Problem Set 6.  Due Wednesday, 13 September}

\begin{document}

\maketitle



\noindent {\bf Reading.} {\em Quick Calculus}, pp. 151--157; 171--190.

\noindent {\bf Supplementary reading.} Simmons, Chapter 6.

\begin{enumerate}

\item Compute the following integrals by substitution.

\begin{enumerate}
\item $\int (1+\frac{1}{x})^2\frac{1}{x^2}\ dx$
\item $\int \tan(x)\ dx=\int \frac{\sin(x)}{\cos(x)}\ dx$
\item $\int \cos(x)\cos(\sin(x))\ dx$
\end{enumerate}

\medskip


\item Compute the following two trigonometric integrals.

\begin{enumerate}
\item  Remember that $\sin^2(x)=\frac{1}{2}-\frac{1}{2}\cos(2x)$.  Use
this to compute
$$
\int \sin^2(x)\ dx.
$$
\item Now use the fact that $\cos^2(x)=1-\sin^2(x)$ to compute 
$$
\int \cos^2(x)\ dx.
$$
\end{enumerate}

\item Use right-hand Riemann sums (in this case, the same as upper
Riemann sums) to show that the area under the graph of $y=x^3$ from
$x=0$ to $x=b$ is $\frac{b^4}{4}$.

\item Each of the following functions has one arch above the
$x$-axis.  Find the area of the region under that arch.

\begin{enumerate}
\item $f(x)=9-x^2$
\item $f(x)=x^3-9x$
\item $f(x)=4x-x^3$
\end{enumerate}

\item Evaluate the following definite integrals using the Fundamental
Theorem of Calculus.  (This is their {\em algebraic area}.)

\begin{enumerate}
\item $\int_{0}^{2\pi} \sin(x)\ dx$
\item $\int_{-3}^{2} x^4+2x^3-5x^2-6x\ dx$
\item $\int_{0}^{\frac{3\pi}{2}} \cos(x)\ dx$
\end{enumerate}


\item Compute the geometric area of the following functions on the
corresponding intervals.  These are the same functions and intervals
as in the previous problem.  Note the difference between geometric and 
algebraic area!

\begin{enumerate}
\item $f(x)=\sin(x)$ on $[0,2\pi]$
\item $f(x)=x^4+2x^3-5x^2-6x=x(x-2)(x+1)(x+3)$ on $[-3,2]$
\item $f(x)=\cos(x)$ on $[0,\frac{3\pi}{2}]$
\end{enumerate}





\end{enumerate}





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