
\def\newp#1.{\par\noindent{\bf #1.}}
\def\endp{\par\noindent\vrule width\hsize depth 0pt height .1pt}
\def\frac#1#2{{{\textstyle#1}\over{\textstyle#2}}}
\def\leaderfill{\leaders\hbox to .5em{\hss.\hss}\hfil}

\def\ans{\par\noindent\hbox to \hsize{\leaderfill}
\par\medskip}

\magnification1200
\def\frac#1#2{{{#1}\over{#2}}}

\newp 13.
Find the center of mass (centroid) of the region bounded by the
curves
$y=\cos x$, $y=0$, $x=-\frac{\pi}2$ and $x=\frac{\pi}2$.
\par\noindent You may use symmetry as part of the 
justification for your answer.
\ans
First we find the area 
$A=\displaystyle\int_{-\frac\pi2}^{\frac\pi2}\cos(x)\ dx$.
The alternative of
$A=\displaystyle\int_0^12\arccos(y)\ dy$ does not look as easy.
Then
$M_x=\displaystyle\int_{-\frac\pi2}^{\frac\pi2}
\frac{\cos(x)}{2}\ \cos(x)\ dx$
and
$M_y=\displaystyle\int_{-\frac\pi2}^{\frac\pi2}
x\ \cos(x)\ dx$.
By symmetry, $M_y=0$: the integral can also be done by parts.
$A=\sin(x)\big\vert_{-\frac\pi2}^{\frac\pi2}=1-(-1)=2$.
$M_x=\displaystyle\int_{-\frac\pi2}^{\frac\pi2}
\frac{\cos^2(x)}{2}\ dx=\int_{-\frac\pi2}^{\frac\pi2}
\frac{1+\cos(2x)}{4}\ dx=\frac{x}{4}+\frac{\sin(2x)}{8}
\Big\vert_{-\frac\pi2}^{\frac\pi2}=
\biggl(\frac{\pi}{8}+0\biggr)-
\biggl(\frac{-\pi}{8}+0\biggr)=\frac{\pi}{4}
$.

Hence the coordinates of the center of mass are at
$\displaystyle\biggl(\frac{M_y}{A},\frac{M_x}{A}\biggr)=
\biggl(0,\frac{\pi}{8}\biggr)$.
\endp

\newp 14.
Solve the initial value problem
$$\cases{\displaystyle xy^\prime+xy+y=e^{-x}\cr\noalign{\vskip 4pt}
y(1)=\displaystyle\frac{2}{e}\cr}$$

\ans
This is a linear differential equation. First we bring the
equation to the standard form by dividing by $x$. We get:

$$\frac{dy}{dx} +
\Bigl(1+\frac{1}{x}\Bigr)y=e^{-x}/x$$
We identify $P(x)=(1+\frac{1}{x})$ and 
$Q(x)=\frac{e^{-x}}{x}$.

We define

$$I(X)=e^{\int P(x)dx}=e^{\int1+\frac1x dx}
=e^{x+lnx}=e^x\>x\ .$$

The solution for the differential equation is:


$$y=\frac{1}{xe^x}\int e^{x}x \ e^{-x}/x dx=\frac{1}{xe^x}\int
dx= \frac{1}{xe^x}(1+C)\ .$$



Evaluating at $x=1$, we get $\frac{2}{e}=y(1)=\frac{1+C}{e}$,

which implies $C=1$,l so the answer is
$$y=\frac{2}{x\> e^x}$$
\endp
 
\newp 15.
Determine whether or not the improper integral 
$\displaystyle\int_1^{+\infty}\frac{2\cos(x^2)+100}{x^{5/4}}\ dx$
is convergent. To receive credit for this problem you must 
justify your answer.
\endp
\ans
We have no hope of evaluating the indefinite integral
$\displaystyle\int\frac{2\cos(x^2)+100}{x^{5/4}}\ dx$
directly so we turn to a comparison theorem.
Note first that
$-1\leq \cos(x^2)\leq1$ so
$$\displaystyle0<\frac{98}{x^{5/4}}\leq
\frac{2\cos(x^2)+100}{x^{5/4}}\leq
\frac{102}{x^{5/4}}$$
for $x\geq 1$.
Hence we can compare our integral with either 
$\displaystyle\int_1^{+\infty}\frac{98}{x^{5/4}}\ dx$
or
$\displaystyle\int_1^{+\infty}\frac{102}{x^{5/4}}\ dx$.
But
$\displaystyle\int_1^{+\infty}\frac{98}{x^{5/4}}\ dx=
98\ \int_1^{+\infty}\frac{1}{x^{5/4}}\ dx$ and
$\displaystyle\int_1^{+\infty}\frac{102}{x^{5/4}}\ dx=
102\ \int_1^{+\infty}\frac{1}{x^{5/4}}\ dx$
so it suffices to evaluate
$\displaystyle\int_1^{+\infty}\frac{1}{x^{5/4}}\ dx$.
Since $\frac54>1$ this integral converges, hence does the original one.

In more detail,
$\displaystyle\int_1^{+\infty}\frac{1}{x^{5/4}}\ dx=
\lim_{t\to\infty}\displaystyle\int_1^{t}\frac{1}{x^{5/4}}\ dx=
\lim_{t\to\infty}\frac{x^{-5/4+1}}{-5/4+1}\Bigg\vert_1^t=
\lim_{t\to\infty}\frac{-4}{x^{1/4}}\Bigg\vert_1^t=
4-\lim_{t\to\infty}\ \frac{1}{t^{1/4}}=4
$
so
$\displaystyle\int_1^{+\infty}\frac{1}{x^{5/4}}\ dx$ converges.

\end

