%
\long\def\newp #1.{
\par\noindent\vtop\bgroup\noindent{\bf #1.}
}
\long\def\problembreak{\par\egroup\vfill\eject\vtop\bgroup}
\def\endp{\par\noindent\vrule width\hsize depth 0pt height .1pt\egroup\par\bigskip}
\def\frac#1#2{{{#1}\over{#2}}}
\let\ds=\displaystyle
\font\sm=cmr7
\magnification1200
\centerline{Exam I}
\centerline{February 15, 2001}
\def\Int{\displaystyle\int}
\bigskip
\newp 11.
\item{(a)}
First we need to determine the corner points.
They are $(\frac14,4)$, $(1,4)$, $(4,1)$ and $(4,\frac14)$.
It is also good to note at this point that the integral will need to be split
into two pieces.
We also note that we may take the density to be $1$ in this problem.
Hence if we integrate along the $x${--}axis we get for the area ($=$ mass)
$$\eqalign{
{\rm Area} =& \int_{\frac14}^1\biggl(4-\frac 1x\biggr)\>dx+
\int_1^4\biggl(\frac 4x-\frac 1x\biggr)\>dx\cr
=&\bigl[ 4x-\ln x\bigr]^1_{\frac14}+\bigl[3\ln x\bigr]^4_1\cr
=&\bigl(4-\ln 1\bigr)-\bigl(1-\ln \frac14\bigr)+3\ln 4-3\ln 1\cr
=&3+\ln\frac14+3\ln4=3-\ln 4+3\ln 4=3+2\ln 4\cr
}$$
If you preferred to integrate up the $y${--}axis you get an identical formula except
that everywhere you see an $x$ above, put a $y$.
\medskip
\item{(b)}You have four choices.
Do you want to compute the moment about the $x${--}axis
or the $y${--}axis and do you want to integrate along the $x${--}axis or the $y${--}axis?
To compute the moment about the $y${--}axis and integrate along the $x${--}axis,
proceed as follows.
$$\eqalign{
{\rm Moment}_y=&
\int_{\frac14}^1
\vtop{\hsize=.2in\baselineskip=4pt
\centerline{$\underbrace{\vrule width0pt height 0pt depth 10pt\ x\ }$}
\centerline{\sm dist.} 
\centerline{\sm from}
\centerline{\sm $\scriptstyle y${--}axis}
}
\vtop{\hsize=1in
\centerline{$\underbrace{\biggl(4-\frac1x\biggr)\>dx}$}
\centerline{\sm mass of rect.}
}
 +
\int_{1}^4 x\biggl(\frac4x-\frac1x\biggr)\>dx
\cr
=&\bigl[2x^2-x\bigr]^1_{\frac14}+\bigl[ 3x\bigr]^4_1=2-1-\frac2{16}+\frac14+12-3=\frac{81}{8}\cr
}$$
\problembreak
\item{}If you choose instead to calculate the moment about the $x${--}axis while still integrating along the
$x${--}axis you would proceed as follows.
$$\eqalign{
{\rm Moment}_x=&
\int_{\frac14}^1\ \frac12\ 
\vtop{\hsize=.4in\baselineskip=4pt
\centerline{$\underbrace{\biggl(4+\frac1x\biggr)}$}
\centerline{\sm dist. from} 
\centerline{\sm center of mass}
\centerline{\sm to $\scriptstyle x${--}axis}
}
\vtop{\hsize=1in
\centerline{$\underbrace{\biggl(4-\frac1x\biggr)\>dx}$}
\centerline{\sm mass of rect.}
}
 +
\int_{1}^4\frac12 \biggl(\frac4x+\frac1x\biggr)\ \biggl(\frac4x-\frac1x\biggr)\>dx
\cr
=&\int_{\frac14}^1\ 8-\frac{1}{2x^2}\>dx+\int_1^4\ \frac{15}{2x^2}\> dx\cr
=&\biggl[8x+\frac{1}{2x}\biggr]^1_{\frac14}+\biggl[ \frac{-15}{2x}\biggr]^4_1
=\biggl(8+\frac12\biggr)-\biggl(2+2\biggr)+\biggl(-\frac{15}8-\Bigl(\frac{-15}2\Bigr)\biggr)
\cr
=&4+\frac48+\frac{-15}{8}+\frac{60}{8}=\frac{32}{8}+\frac48+\frac{-15}{8}+\frac{60}{8}
=\frac{96-15}{8}=\frac{81}{8}\cr
}$$
If you want the integrals for integrating along the $y${--}axis, just switch $x$ and $y$ in the formulae
above.
\bigskip
\item{(c)}
By symmetry, the moment about the $x${--}axis is the same as the moment about the $y${--}axis
and the center of mass has equal $x$ and $y$ coordinates, each one being
$$\frac{\frac{81}{8}}{3+2\ln 4}$$
\endp
\newp 12.
Complete the square: $x^2+2x+2=(x+1)^2+1$.
This suggests the trig.\ substitution
$\tan\theta=x+1$: $dx=\sec^2\theta\>d\theta$ and 
$x^2+2x+2=(x+1)^2+1=\tan^2\theta+1=\sec^2\theta$.
When $x=-1$, $x+1=0=\tan\theta$, so $\theta=0$;
when $x=0$, $x+1=1=\tan\theta$, so $\theta=\frac\pi4$.
Hence
$$\Int_{-1}^{0} \frac{1}{x^2+2x+2}\ dx =\int_0^{\frac\pi4}\frac{\sec^2\theta}{\sec^2\theta}\>d\theta=
\int_0^{\frac\pi4}\>d\theta=\theta\bigg\vert_0^{\frac\pi4}=\frac\pi4-0=\frac\pi4\ .$$

\endp
\newp 13.
Separating variables gives
$\ds\frac{dy}{\sqrt{1-y^2\,}}=x$:
integrating gives
$$\arcsin(y)=\frac{x^2}{2} +C.$$

Now since $y(0)=0$ we see that the constant $C=0$.

Hence the unique function is
$$y=\sin(\frac{x^2}{2})$$
and $y(1)=\sin(\frac{1}{2})$.
\endp
\newp 14.
To find the general solution to the differential equation
$$y^\prime\ =\ \frac{2y}{x}+x\hskip10pt (x>0)$$
first put it into standard form:
$$y^\prime + \frac{-2}{x}\cdot y=x$$
so $P(x)=\frac{-2}{x}$ and $Q(x)=x$.
Then find the integrating factor $v=e^{\int P(x)\>dx}$.\par\noindent
$\ds\int \frac{-2}x=-2\ln x+C$ so we may take
$v(x)=e^{-2\ln x}$ which is $v(x)=x^{-2}$.
Then we do the second integral
$\ds\int v(x)\>Q(x)\>dx=\int x^{-2}\cdot  x\>dx=\int x{-1}\>dx=\ln x+C$.
Since we are told $x>0$, there is no need for absolute values in the $\ln x$.
Finally,
$$y(x)=x^{-2}\bigl(\ln x+C\bigr)$$
is the general solution.
\endp

\end
