Exam# 1        Answers are in red

Directions: Show all of your work. Start each problem on a fresh sheet of paper. Do not write on the backside of the pages.

1. Complete the following sentences.
   1. $\int f(x)\hspace{0.5em}\mathrm{dx}=F(x)+C$ means $\frac{d\hspace{0.5em}}{\mathrm{dx}}F(x)=f(x)$.
   2. If $\frac{d\hspace{0.5em}}{\mathrm{dx}}F(x)=\frac{d\hspace{0.5em}}{\mathrm{dx}}G(x)$ for all $x$ such that $a\le x\le b$, then there is a constant $C$ such that $F(x)=G(x)+C$ for all $x$ such that $a\le x\le b$.
2. Assuming that
   

   ${\displaystyle \int x\sin (x)\mathrm{dx}=-x\cos (x)+\int \cos (x)\mathrm{dx}}$

   
   [then]
   

   ${\displaystyle {\int }_{-\pi }^{\pi }x\sin (x)\mathrm{dx}}$

   
   equals $2\pi$
3. The curves $y=x^2$ and $y=3x+4$ intersect at the points $(-1,1)$ and $(4,16)$. Write down the integral or integrals that give the area of the region enclosed by the curves $y=x^2$ and $y=3x+4$.
   Since the curve $y=3x+4$ is above the curve $y=x^2$ for $-1\le x\le 4$
   

   ${\displaystyle \text{area}={\int }_{-1}^4(3x+4-x^2)\mathrm{dx}}$

   
   
4. Write down an integral for the volume of a pyramid with height  $10$ whose base is a $3$ by  $3$ square. Carefully explain how you arrived at your integral. Well chosen drawings might be helpful.
   The cross-sections parallel to the base are squares whose sides change linearly from $s=3$ when $y=0$ to $s=0$ when $y=10$. Thus, $s=3-3y/10=3(10-y)/10$. Therefore,
   

   ${\displaystyle \text{volume}=\frac{3^2}{{10}^2}{\int }_0^{10}(10-y{)}^2\mathrm{dy}}$

   
   
5. Use calculus, not calculators, to work out the following integrals. (It's best to check your answers.)
   1. $\int \ln (x)\hspace{0.5em}\mathrm{dx}$ $=x\ln x-x+C$
   2. $\int 2x\cos (x^2)\hspace{0.5em}\mathrm{dx}$ $=\sin (x^2)+C$
   3. $\int x{e}^{2x}\hspace{0.5em}\mathrm{dx}$ $=\frac{x}{2}{e}^{2x}-\frac{1}{4}{e}^{2x}+C$
   4. $\int \frac{1}{x^2-4}\hspace{0.5em}\mathrm{dx}$ $=\frac{1}{4}\ln |x-2|-\frac{1}{4}\ln |x+2|+C$

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On 19 Feb 2006, 22:06.