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\begin{center}
\large \bf Math 108\\
Exam 2 Solutions
\end{center}




\vspace{2mm}
\noindent
{\bf 1.a. }  Intersection of Demand and Supply Curves 
$\displaystyle{\frac{25}{q+1} = q + 1}$ therefore $ (q + 1)^2 =
25, q + 1 = 5, q_e = 4$ and $p_e? = 5$.

\vspace{2mm}
\noindent
{\bf 1.b. } 
\begin{eqnarray*}
CS =  \int^4_0 \frac{25}{q+1} dq -(4)(5)  &=& 25ln(q+1)\big
|^4_0 - 20\\
&=& 25 ln 5 - 20
\end{eqnarray*}


\vspace{2mm}
\noindent
{\bf1.c.} $\displaystyle{ PS = (4)(5) - \int^4_0 (q+1) = 20 -
\frac{(q+1)^2}{2} \big |^4_0 = 20 - \left ( \frac{25}{2} - 0\right
) = \frac{15}{2}}$

\vspace{2mm}
\noindent
{\bf 2.a.}  $\displaystyle{\int \frac{dy}{4-y} = \int dt + }$
const.  = $ t + $ const.
\[
\bigg \| \begin{array}{ll}
n &= 4-y\\
dn &= - dy
\end{array}
\]
\[
- \int \frac{dn}{n} = - ln |n| + \mbox{ const.} = - ln |4-y| +
\mbox{const.}
\]
\begin{eqnarray*}
ln|4-y| &=& -t + c\\
|4 - y| &=& e^{-t+c} = Ke^{-t}\\
4 - y &=& (\pm K)e^{-t}\\
4 - y &=& k e^{-t}.
\end{eqnarray*}

\vspace{2mm}
\noindent
{\bf 2.b .}  Equilibrium Solution $y = 4$


\vspace{2mm}
\noindent
{\bf 2.c. }  
\begin{eqnarray*}
y > 4 &\Longrightarrow& 4-y < 0 \Longrightarrow
\frac{dy}{dt} < 0  \Longrightarrow y \mbox{ decreasing}\\
y < 4 &\Longrightarrow& 4 - y > 0 \Longrightarrow
\frac{dy}{dt} > 0 \Longrightarrow y \mbox{ increasing}
\end{eqnarray*}
\[
\mbox{ Therefore } y = 4 \mbox{ is a stable solution.}
\]

\vspace{2mm}
\noindent
{\bf  3.} 
\[
\frac{1}{x - x^2} = \frac{1}{x(1-x)} = \frac{A}{x} +
\frac{B}{1-x} = \frac{A-Ax + Bx}{x(1-x)} = \frac{A +
(B-A)}{n(1-n)}.
\]
Therefore $A + (B-A)x = 1 $ forall $ x $

\smallskip
\noindent
Therefore $ B - A = 0 $ and $ A = 1$.  Therefore $B = 1.$


\smallskip
\noindent
To??
\begin{eqnarray*}
\int \frac{1}{x(1-x^2)} dx &=& \int \frac{1}{x} \overset{dx}{+}
\int \frac{1}{1-x}ax ??\\
&=& ln|x| - ln|1-n|  + C\\
&=& ln \frac{|n|}{(n-1)} + C.
\end{eqnarray*}



\vspace{2mm}
\noindent
{\bf 4.b.}   
\[
\frac{dy}{y^2} = e^t dt
\]
\[
- \frac{1}{y}  = e^t + c
\]
\[
y(0) = - \frac{1}{8} \ \ \frac{-1}{(\frac{-1}{8})} = 1 + c \
\mbox{ therefore } c = 7
\]
\[
\mbox{Therefore } y = \frac{-1}{e^t + 7}
\]

\vspace{2mm}
\noindent
{\bf  5. c.}   
\begin{eqnarray*}
\frac{-2(x-60)^4}{4} + 4,000 x \bigg |^{70}_{60} &=& 
\frac{-(x-60)^4}{2} + 4000x \bigg |^{70}_{60}  \\
&=& -5000 + 280,000 - (0 + 240,000)\\
&=& 35000
\end{eqnarray*}

\vspace{2mm}
\noindent
{\bf  6.e.} 
\begin{eqnarray*}
\int^\infty_0 \frac{1}{(x+1)^5} dx &=& \lim_{b \to \infty}
\int^b_0 \frac{1}{(x+1)^5} dx\\
\int^b_0 \frac{1}{(x+1)^5} dx  &=& \frac{(x+1)^{-4}}{-4} \bigg
|^b_0 = \frac{-1}{4} \left \{ \frac{1}{(b+1)^4} - \frac{1}{14}
\right \}\\
\lim_{b \to \infty} \int^b_0 \frac{1}{(x+1)^5} dn &=&
\frac{1}{4}
\end{eqnarray*}

\vspace{2mm}
\noindent
{\bf  7.d. } 

\[
\frac{1}{3}\ \  \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \bigg |^4_1
\hspace{5mm}  AV = \frac{\int^4_1 \sqrt{x} \ dx}{3}
\]
\[
\frac{2x^{\frac{3}{2}}}{9} \bigg |^4_1 = \frac{2}{9}
\left \{ (4)^{\frac{3}{2}} - 1^{\frac{3}{2}}\right \} = \frac{2}{9}
(8-1) = \frac{14}{9}
\]


\vspace{2mm}
\noindent
{\bf  8.b.} 
\begin{eqnarray*}
\int^\infty_0 14000 e^{-.07t} dt &=& \lim_{b \to \infty}
\int^b_0 14000 \cdot e^{-(0.7)t} dt\\
&=& \lim_{b \to \infty} (14000) \frac{e^{-(0.7)t}}{(-0.7)}
\bigg |^b_0\\
&=& \frac{14000}{-0.7} \{0 - 1\}  = \frac{14000}{7} \cdot
100\\
&=& 200,000
\end{eqnarray*}

\vspace{2mm}
\noindent
{\bf  9.1.}   We have $f(x) = \sqrt{x^2+1}, x_1 = 0, x_2 = 1, x_3
= 2, x_4 = 3, x_5 = 4$ and $bx = 1$.  Therefore
\begin{eqnarray*}
\int^4_0 \sqrt{x^2 + 1} \ dx &\simeq& \left [ \sqrt{1} + 2
\sqrt{1^2 + 1} + 2 \sqrt{2^2+1} + 2\sqrt{3^2+1} + 2??
\sqrt{4^2+1} \right ] \frac{1}{2}\\
&=& \frac{1}{2} \left ( 1 + 2\sqrt{2} + 2 \sqrt{5} + 2\sqrt{10}
+ \sqrt{15} \right ) \simeq 9.374.????? 9.373??
\end{eqnarray*}


\vspace{2mm}
\noindent
{\bf  10.}   We have $\displaystyle{\frac{dM}{dt} = 0.1M -
50,000.}$

\vspace{2mm}
\noindent
{\bf  11.d. }   For $y  = x^3 + k$ to be a solution we must have 
$x(x^3 + k)^1 - 3(x^3 + k) = 15, \mbox{ or } x \cdot 3x^2 -
3x^3 - 3k = 15,$ or $ -3k = 15$  or \fbox{$k = -5$}.

\vspace{2mm}
\noindent
{\bf  12.}  $\displaystyle{\frac{dW}{dt} = 1,000,000 e^{0.1t}}$
\begin{eqnarray*}
\mbox{Change in } W &=& \int^{10}_0 1,000,000 e^{0.1t} df\\
&=& 1,000,000 \frac{e^{(0.1)t}}{(0.1)} \bigg |^{10}_0\\
&=& 10,000,000 (e-1).\\
&-& 17,000.000.
\end{eqnarray*}


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