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\begin{document}

\begin{center}
%{\bf Each question is worth 6 points}
\end{center}


\centerline{\bf \large Math 108 Exam 1 Solutions -- Fall
2002}

\medno
\begin{enumerate}

\item %1
Integrating $\ds \frac{dy}{dt}= \frac{4}{t^3}$  we
obtain $\ds y(t)= 4 \cdot \frac{-1}{2} \cdot t^{-2} +
c$. Since $1 =y(1)=-2 \cdot \frac{1}{1^2} + c $, or
$ 1 = -2 + c$, we have $c = 3$.
Hence, we have $y(t) = -2 \cdot t^{-2} + 3$.


\item %2
 The average value of the given function  $f(x)$
over $[0, 2]$ is
 
$ \frac{1}{2-0} \int_0^2 \sqrt{2}x^{3/2} +2 e^{x/2} dx
=\frac{1}{2} \left( \frac{2\sqrt{2}}{5} x^{5/2} + 4
e^{x/2} \right)_0^2
=\frac{1}{2} \left( \frac{16}{5} + 4e^{1}-  0 -
4e^0 \right) =-\frac{2}{5} + 2e$

\item %3   
 $\ds \int^1_0\sqrt{x+2} \ dx \approx 
\frac{(1-0)}{5}(\sqrt{0 +2} + \sqrt{.2 + 2} + \sqrt{.4
+2} + \sqrt{.6 + 2} + \sqrt{.8 + 2})$

\hskip1.05in$=\frac{1}{5}(\sqrt{2.0} + \sqrt{2.2} +
\sqrt{2.4} + \sqrt{2.6} +\sqrt{2.8})$

\item %4
The function $f(x) = x^2 -2$ crosses the
x-axis when $x^2 -2=0$ or $x^2=2$ or $x=\pm \sqrt2$. 
$f(x)
\leq 0$ on the interval $[0,\sqrt2]$ and $f(x) \geq
0$ on the interval $[\sqrt2,2]$.  

Hence the total area
$=\ds \int_0^{\sqrt2} 0- (x^2 -2)dx + \int_{\sqrt2}^2 
x^2 -2 dx =\left[\frac{-x^3}{3}+2x\right]_0^{\sqrt2}+
\left[\frac{x^3}{3}-2x\right]_{\sqrt2}^2$

$=\frac{-2\sqrt2}{3}+2\sqrt2
-0 +\frac{8}{3}-4 - (\frac{2\sqrt2}{3}-2\sqrt2)
=\frac{8}{3}\sqrt2-\frac{4}{3}$

\item %5
$\ds  \int x^2 (x^3-1)^{5/3} \, dx = \int u^{5/3} \
\frac{du}{3}$ \hskip1.7in $u=x^3-1 \ \ \ \ \ \
du=3x^2dx$

\hskip 1.35in $=\frac{3}{8}u^{8/3} + C =
\frac{3}{8}(x^3 -1)^{8/3} + C$





\begin{minipage}[b]{4in}
\item %6
 $\ds \int_4^6  \frac{2}{3x^2+3x-6} \, dx= 
\frac{2}{3}\int_4^6 \frac{1}{(x-1)(x+2)} \, dx$


\hskip1.45in $=\ds \frac{2}{3}\int_4^6
\frac{\frac{1}{3}}{x-1}+\frac{\frac{-1}{3}}{x+2}\, dx$


\hskip1.45in
$=\frac{2}{3}\left[\frac{1}{3}\ln{|x-1|}
-\frac{1}{3}\ln{|x+2|} \right]_4^6$

\hskip1.45in $=\frac{2}{9}\ln{|\frac{x-1}{x+2}|}
_4^6$


\hskip1.45in
$=\frac{2}{9}\ln{\frac{5}{8}}-\ln{\frac{1}{2}}
=\frac{2}{9}\ln{\frac{5}{4}}$
\end{minipage} \ \hskip.15in \
\begin{minipage}[b]{2.5in}
  \begin{center}
\begin{tabular}
	{| l |} \hline
$\ds \frac{1}{(x-1)(x+2)}=\frac{A}{x-1} +\frac{B}{x+2}$
\\

$1=A (x+2) + B (x-1)$ \\

$1=A(1+2) + B  (1-1)= 3A$ \\

$1=A (-2+2) + B (-2-1)=-3B$


 \\ \hline 
\end{tabular}
\end{center}
\vskip .3 in
\end{minipage}



\item %7
$\ds  \int 2t \ \ln{t} \ dt = \int \ln{t} \cdot 2t \
dt$ 
\hskip1.7in $u=\ln{t} \ \ \ \ \ \ dv=2t \ dt$

\hskip 1in $=t^2 \ln{t}  -\int t^2\frac{1}{t}dt$
\hskip1.4in $du=\frac{1}{t} dt \ \ \ \ \ \ v=t^2 $

\hskip 1in$=t^2 \ln{t}   -\frac{t^2}{2} + C$
\vskip .02in

\begin{minipage}[b]{2.5in}
\item %8 
 Consumer Surplus=

 $\frac{1}{2} \cdot (8-4) \cdot 30=60$
\vskip .8in.
 
\end{minipage} \ \hskip.15in \
\begin{minipage}[b]{2.5in}
  $\epsfxsize=1.5in\epsfbox{ex1pict1temp.eps}$
\end{minipage}

\item %9
Change in cost from 80 to 100 = $C(100)-C(80)
=\int_{80}^{100} 45t^{-1} dt
=45\ln{t}|_{80}^{100}$

\hskip 2.2in $= 45(\ln{100}-\ln{80})=
45\ln{\frac{5}{4}}$
 

\item %10
$D(q)=S(q)$ means $-q^2 - 50 q + 800=q^2 + 10 q$ so
$2q^2 +60q -800=0$ or $2(q-10)(q+40)=0$. Hence $q_e=10$.

\item %11
$q_e=10$ so $p_e=S(10)=10^2+10 \cdot 10 =200$. Thus

$PS=\int_0^10 -q^2 - 50 q + 800 dq - 10 \cdot 200 
= -\frac{q^3}{3} -25q^2 +800q|_0^{10}-2000$

\hskip .3in $=-\frac{1000}{3}-2500+8000 -0 -2000 =
\frac{9500}{3}$


\item %12
The present value for a one-time investment with
continuous interest is

$PV=FV \ e^{-rT}=1000000e^{-.15 \cdot 40}
=1000000e^{-6}$


\item  %13
Solve for present value of a constant investment:

$PV=\frac{S}{r}(1-e^{-rt})
=\frac{5000}{.1}(1-e^{-.1 \cdot 30})
=50000(1-e^{-3})$


\item %14
$\ds  \int_1^2 2x \ e^{-2x} dx = \int_1^2 2x \cdot
e^{-2x} dx$ 
\hskip2.1in $u=2x \ \ \ \ \ \ dv=e^{-2x} dx$

\hskip 1in $=[2x \ \frac{-1}{2}e^{-2x}]_1^2
- \int_1^2 \frac{-1}{2}e^{-2x} 2dx$
\hskip1.15in $du= 2 dx \ \ \ \ \ \ v=\frac{-1}{2}e^{-2x}
$

\hskip 1in$=[-xe^{-2x}]_1^2
+ \frac{-1}{2}e^{-2x}|_1^2$

\hskip 1in$=-2e^{-4}+e^{-2} + \frac{-1}{2}e^{-4}
+\frac{1}{2}e^{-2}$

\hskip 1in$=-\frac{5}{2}e^{-4}+\frac{3}{2}e^{-2}$

\item%15
(a)  $\ds \int_1^2 e^{2r} \, dr \approx 
\frac{(2-1)}{4}(e^{2 \cdot 1.125} + e^{2 \cdot 1.375} +
e^{2 \cdot 1.625} + e^{2 \cdot 1.875})$

\hskip.95in$=\frac{1}{4}(e^{2.25} + e^{2.75} + e^{3.25}
+ e^{3.75})$

\hskip.95in$= 23.36044741$

(b) $\ds \int_1^2 e^{2r} \, dr = \frac{1}{2}e^{2r}|_1^2
=\frac{1}{2}(e^4-e^2)$

\hskip1in$=23.60454697$

(c) Actual Error $= |\frac{1}{4}(e^{2.25} + e^{2.75} +
e^{3.25} + e^{3.75}) - \frac{1}{2}(e^4-e^2)|$

\hskip1.23in$=|23.36044741-23.60454697|$

\hskip1.23in $=.2440995600$


\end{enumerate}
\end{document}
