\magnification=\magstep1
\centerline{Mathematics 126--Calculus for Social Science 2}
\centerline{Solutions -- Final Examination}
\centerline{\it December 10, 2001}
\bigskip
\noindent
I.  
\item{A)}
$$\int_1^2 x^6 - x^{1/2} + x^{-1}\ dx 
= \left({x^7\over 7} - {2\over 3}x^{3/2} + \ln|x|\right)|_1^2 \doteq 17.62$$
\item{B)}  (This one had the typo.)  The correct integral is
$$\int_0^1 (x^3 + 1)e^{x^4+4x}\ dx$$
Let $u = x^4 + 4x$, so $du = 4x^3 + 4 = 4(x^3 + 1)\ dx$ and 
$(x^3+1)\ dx = {1\over 4} du$.  Substituting the whole integral to 
the new variable $u$, we have
$$\int_{u=0}^{u=5} {1\over 4} e^u\ du = {1\over 4}e^u|_0^5 = {1\over 4}(e^5-1)\doteq
36.85$$
\item{C)} Integrate by parts with $u = (\ln(x))^2$, $dv = xdx$.  Then
$du = 2\ln(x){1\over x}$ and $v = {x^2\over 2}$.  So after simplifying
$\int v\ du$, 
$$\int x(\ln(x))^2\ dx = {x^2\over 2}(\ln(x))^2 - \int x\ln(x)\ dx$$
To finish, we must integrate by parts again, with $u = \ln(x)$, 
$dv = xdx$.  This time $du = {1\over x}$, and $v = {x^2\over 2}$, so
$$\eqalign{\int x(\ln(x))^2\ dx &= {x^2\over 2}(\ln(x))^2 - \int x\ln(x)\ dx\cr
&={x^2\over 2}(\ln(x))^2 - {x^2\over 2}\ln(x) + \int {1\over 2} x\ dx\cr
&={x^2\over 2}(\ln(x))^2 - {x^2\over 2}\ln(x) + {x^2\over 4} + C\cr}$$
\item{D)}  Let $u = e^x$.  Then $du = e^x dx$, so $dx = {du\over e^x} = {du\over u}$.
Substituting, the integral has the form of \# 6 in the table with $a = 4$, $b = 7$:
$$\int {dx\over \sqrt{4 + 7e^x}} = \int {du\over u\sqrt{4+7u}} = 
{1\over 2}\ln\left|{\sqrt{4+7u}-2\over \sqrt{4+7u}+2}\right| + C
={1\over 2}\ln\left|{\sqrt{4+7e^x}-2\over \sqrt{4+7e^x}+2}\right| + C$$
\vfill\eject
\noindent
II.
\item{A)} The value of the midpoint Riemann sum with $n = 4$ is
$$\eqalign{
{(.25)\over 2\sqrt{2\pi}} &\left( e^{-(5.125-6)^2/8} +
 e^{-(5.375-6)^2/8}+e^{-(5.625-6)^2/8} +e^{-(5.875-6)^2/8}\right)\cr
&\doteq{(.25)\over 2\sqrt{2\pi}}\left(e^{-.0957}+e^{-.0488}+e^{-.0176}+e^{-.0020}\right)\cr
&\doteq .1916\cr}$$
\item{B)}  The Simpson's Rule approximation is
$$\eqalign{
{(.25)\over 3\cdot 2\sqrt{2\pi}} &\left( e^{-(5-6)^2/8}+ 4e^{-(5.25-6)^2/8} +
 2e^{-(5.5-6)^2/8}+4e^{-(5.75-6)^2/8} +e^{-(6-6)^2/8}\right)\cr
&\doteq .1915\cr}$$
\item{C)} {\it Note: only one person in the class saw how to do this part, 
so I adjusted the exam scores by adding 5 points to everyone's score (including
the person who did this correctly).}  The idea is that the given function
$$f(x) = {1\over 2\sqrt{2\pi}} e^{(x-6)^2/8}$$
is the probability density function for a normal random variable $x$ with
mean $\mu = 6$, and $\sigma = 2$.  The given integral computes the 
probability $P(5\le x\le 6) = P\left({5 - 6\over 2} \le z \le {6-6\over 2}\right)$,
where $z$ is the standard normal variable with $\mu = 0$, and $\sigma = 1$.
From the table
$$P(-.5\le z \le 0) = P(z \le 0) - P(z \le -.5) = .5000-.3085 = .1915$$
\bigskip
\noindent
III.  The phrase ``indefinitely into the future'' indicates that we want the
value of the improper integral
$$\int_0^\infty (30+2t)e^{(-.08)t}\ dt$$
Using parts with $u = 30+2t$ and $dv = e^{(-.08)t}\ dt$, we have
$du = 2dt$, and $v = {1\over (-.08)} e^{(-.08)t}$, so
$$\eqalign{\int (30+2t) e^{(-.08)t}\ dt &= {(30+2t)\over -.08}e^{(-.08)t} + \int {2\over .08}e^{(-.08)t}\ dt\cr
&={(30+2t)\over -.08}e^{(-.08)t} - {2\over (.08)^2}e^{(-.08)t}\cr}$$
Then to evaluate the improper integral, we want:
$$\eqalign{&\lim_{b\to \infty} {(30+2t)\over -.08}e^{(-.08)t} - {2\over (.08)^2}e^{(-.08)t}|_0^b\cr
&=\lim_{b\to\infty}\left({(30+2b)\over -.08}e^{(-.08)b} - {2\over (.08)^2}e^{(-.08)b} + 
{30\over .08} + {2\over (0.08)^2}\right)\cr
&= 0 + 375 + 312.5\cr
&= 687.5\cr}$$
(thousand dollars).
\bigskip
\noindent
IV.
\item{A)} It's not necessary to solve the differential equation to do this part.
Just check that the relation is satisfied.  For $y = 30 - 10e^{-4t}$,
${dy\over dt} = 40e^{-4t}$.  The right side is $-4(-10)e^{-4t} = 40 e^{-4t}$ also.
Then $y(0) = 30 - 20\cdot 1 = 10$.
\item{B)} This is a separable differential equation for $x$:
${p\over x}{dx\over dp} = -p(.4+2p)$
so 
$$\eqalign{\int {dx \over x} &= \int -.4 -2p\ dp\cr
                 \ln|x| = -.4p - p^2 + c\cr
                     x = be^{-4p-p^2}\cr}$$
The initial condition was supposed to be $x(0) = 1$ (another typo, D'OH!), 
which says $b = 1$.
I did not take off credit if the constant was not determined.
\bigskip
\noindent
V.
\item{A)}  The idea on this part is that a probability density function
must satisfy $\int_{-\infty}^\infty f(x)\ dx = 1$.  This lets us determine
a value for $c$ as follows:
$$\eqalign{\int_0^{100} c(100x^2 - x^3)\ dx &= 1\cr
           c\cdot \left( {100x^3\over 3} - {x^4\over 4}\right)|_0^{100} &= 1\cr
           c{10^8\over 12} &= 1\cr 
           c &= 12\times 10^{-8}\cr}$$
\item{B)} The expected value is 
$$\eqalign{E(x) &= \int_0^{100} (12\times 10^{-8})x(100x^2 - x^3)\ dx\cr
&=(12\times 10^{-8})\left({100 x^4\over 4} - {x^5\over 5}\right)|_0^{100}\cr
&=(12\times 10^{-8})\cdot (10^{10}/20) = 60\cr}$$
\item{C)} The standard deviation is square root of 
$$\eqalign{
Var(x) &= \int_0^{100} (12\times 10^{-8}) x^2 (100x^2 - x^3)\ dx - 60^2\cr  
&= (12\times 10^{-8})\left({100 x^5\over 5} - {x^6\over 6}\right)|_0^{100} - 3600\cr
&= 4000 - 3600 = 400}$$
So $\sigma = \sqrt{Var(x)} = 20$.
\bigskip
\noindent
VI.  To maximize the volume, we clearly want the sum of length and
girth to be as large as possible, so $z + 2x + 2y = 108$.  Use this
to solve for $z$ in terms of $x,y$:  $z = 108-2x-2y$.  Then the volume
is $V = xyz = xy(108-2x - 2y)$.  We have
$$\eqalign{
V_x &= 108y - 4xy - 2y^2 = y(108 - 4x - 2y)\cr
V_y &= 108x - 2x^2 - 4xy = x(108 - 2x - 4y)\cr}$$
The critical points satisfy either $y = 0$ or $4x + 2y = 108$, and
either $x = 0$ or $2x + 4y = 108$.  The critical points with $x = 0$
and/or $y = 0$ will clearly not maximize the volume (because they make 
$V = 0$).  So 
we want the critical point with $4x + 2y = 2x + 4y = 108$. Solving
simultaneously gives $x = y = 18$, and $z = 108 - 36 - 36 = 36$.
It can be seen that this is a maximum for the volume by the 
second derivative test.  
\bigskip
\noindent
VII.  Using Lagrange Multipliers, let $F(x,y,\lambda) = (x^2+y^2-xy) + 
\lambda(x + 2y - 14)$.  Then our system of equations is
$$\eqalign{
F_x &= 2x -y + \lambda =0\cr
F_y &= 2y - x + 2\lambda = 0\cr
F_\lambda&= x+2y-14=0\cr}$$
From the first two equations
$$ -4x + 2y = 2\lambda = x - 2y$$
so $5x = 4y$, or $y = {5\over 4}x$.  Substitute this in the 
constraint equation:  $x + {10\over 4}x = 14$, so $x = 4$, $y = 5$.
Since this is the only critical point satisfying the constraint, it
must be the minimum we want ($f(x,y) = x^2 + y^2 - xy = (x-y/2)^2 + 3y^2/4$)
has no local maximum on the constraint line).
\bye
