%
\def\whichcourse{Math 126}
\def\whichexam{Exam II}
\def\examdate{March 20, 2001}

\font\sm=cmr7
\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\vfill}
\def\frac#1#2{{{#1}\over{#2}}}
\let\ds=\displaystyle

\long\def\boxit#1#2#3{\vbox{\hrule\hbox{\vrule\kern#2
\vbox{\hsize=#3\kern#2#1\kern#2}\kern#2\vrule}\hrule}}

\magnification1200
\centerline{\whichcourse}
\centerline{\whichexam}
\centerline{\examdate}
\bigskip
\newp 9.
First by long division the integrand is equal to
$$2x+\frac{1}{x(x-1)}.$$

Expand the second term using partial fractions: 
$$\frac{1}{x(x-1)}=\frac{A}{x-1} +\frac{B}{x}$$
or
$$1=Ax +B(x-1)\ .$$
Solving the equations gives
$$\frac{1}{x(x-1)}=\frac{1}{x-1} -\frac{1}{x}.$$

We can then use standard integration formulas to get the
general antiderivative
$$x^2+\ln |x-1| -\ln |x| +C.$$

\endp\vfill\null
\newp 10.
Use the substitution $x=\frac34\sin\theta$: then $dx=\frac34\cos\theta\>d\theta$ and
$$\eqalign{\int\frac{x^2}{\sqrt{9-16x^2\>}}=&
\biggl(\frac34\biggr)^2\int\frac{\sin^2\theta\>\cos\theta\ d\theta}
{\sqrt{9-9\sin^2\theta\>}}\cr
=&\frac9{64}\int\sin^2\theta\ d\theta\cr
=&\frac9{128}\bigl(\theta-\sin\theta\>\cos\theta\bigr)+C\hskip10pt (\hbox{from memory or see below})\cr
=&\frac9{128}\Bigl(\arcsin\frac{4x}{3} - \frac{4x}{3}\sqrt{1-\biggl(\frac{4x}{3}\biggl)^2\> }\>\Bigr)+C\cr
=&\frac9{128}\Bigl(\arcsin\frac{4x}{3} - \frac{4x}{9}\sqrt{9-16x^2\> }\>\Bigr)+C\cr
}$$
The integral $\int\sin^2\theta\>d\theta$ can be computed several different ways if you didn't
memorize the answer.
\item{1.} Use the trig.\ identities $\sin^2\theta=\frac12\bigl(1-\cos(2\theta)\bigr)$ and
$\sin(2\theta)=2\sin\theta\>\cos\theta$:
$\displaystyle\int\sin^2\theta\>d\theta=\int\frac12\bigl(1-\cos(2\theta)\bigr)\> d\theta=
\frac12\bigl(\theta-\frac12\sin(2\theta)\bigr)+C=\frac12\bigl(\theta-\sin\theta\>\cos\theta\bigr)+C$
\item{2.} Use integration by parts: $u=\sin\theta$, $dv=\sin\theta d\theta$;\par
\item{}$\displaystyle\int\sin^2\theta\>d\theta=-\sin\theta\>\cos\theta+\int\cos^2\theta\>d\theta=
-\sin\theta\>\cos\theta+\int1-\sin^2\theta\>d\theta=
-\sin\theta\>\cos\theta+\theta-\int\sin^2\theta\>d\theta$ so 
$2\displaystyle\int\sin^2\theta\>d\theta=\theta-\sin\theta\>\cos\theta+C$.
\endp
\newp 11.
$$\eqalign{
\int e^{\sqrt x}\>dx
\vtop{\hsize .7in\centerline{$=$}
\boxit{\vtop{\hsize .6in
\centerline{\sm Substitution}
\centerline{$\scriptstyle u=\sqrt x$}
\centerline{$\scriptstyle du=\frac1{2\sqrt x}dx$}
\centerline{$\scriptstyle 2\sqrt x\>du=dx$}
\centerline{$\scriptstyle 2u\>du=dx$}}}{2pt}{2pt}
}&
\int 2u\>e^u\>du
\vtop{\hsize .5in\centerline{$=$}
\boxit{\vtop{\hsize .5in
\centerline{\sm Parts}
\centerline{$\scriptstyle w=2u$}
\centerline{$\scriptstyle dv=e^u\>du$}
\centerline{$\scriptstyle dw= 2 du$}
\centerline{$\scriptstyle v=e^u$}}
}{2pt}{2pt}\vskip10pt}
2ue^u-\int2e^u\>du=2ue^u-2e^u+C\cr
\vtop{\hsize .7in\centerline{$=$}}&2\sqrt x e^{\sqrt x}-2e^{\sqrt x}+C\cr
}$$

\vskip10pt
This problem CAN be done via integration by parts, but it is so weird that no one actually did
it this way.
Still
$$\eqalign{
\int e^{\sqrt x}\>dx=&\int \hskip-20pt
\vtop{\hsize=1.4in\centerline{$
\underbrace{\sqrt x\vrule depth 10pt width 0pt height 0pt}_{u}\ 
\underbrace{\frac1{\sqrt x}e^{\sqrt x}\>dx}_{dv}$}
\centerline{$du=\frac{1}{2\sqrt x}\ ;\ v=2e^{\sqrt x}$}
\vskip18pt}
=\ 2\sqrt x\>e^{\sqrt x}-\int\frac1{\sqrt x}e^{\sqrt x}\>dx\cr
=&\hskip10pt2\sqrt x\>e^{\sqrt x}-2e^{\sqrt x}\> +C\cr
}$$
\endp
\newp 12.
Since the answer is not obvious, rewrite:
$$\lim_{x \rightarrow 0}(1+2x)^{1/x}=e^{^{\ds\lim_{x \rightarrow 0}\ln(1+2x)^{1/x}}}$$
and we concentrate on
$\ds\lim_{x \rightarrow 0}\ln(1+2x)^{1/x}=\lim_{x \rightarrow 0}\frac{\ln(1+2x)}{x}$.

Since $\ln(1+2\cdot 0)=\ln(1)=0$ this last limit is of the form $\frac00$ so we may apply
l'H\^opital's rule.
This requires us to compute
$$\lim_{x \rightarrow 0}\frac{\frac{d \ln(1+2x)}{dx}}{\frac{dx}{dx}}=
\lim_{x \rightarrow 0}\frac{\frac{2}{1+2x}}{1}=2\ .$$
Hence 
$$\lim_{x \rightarrow 0}(1+2x)^{1/x}=e^2\ .$$
\bigskip
Another way to go, which no one completed successfully, is to {\sl remember\/}\par
\noindent$\ds\lim_{t\to\infty}\biggl(1+\frac at\biggr)^{t}=e^a$ and proceed as follows.
First compute $\ds\lim_{x \rightarrow 0^+}(1+2x)^{1/x}$ by letting $x=\frac1t$
and
$\ds\lim_{x \rightarrow 0^+}(1+2x)^{1/x}=
\lim_{t\to\infty}\biggl(1+\frac 2t\biggr)^{t}=e^2$.
After this, you know $\ds\lim_{x \rightarrow 0}(1+2x)^{1/x}=e^2$ OR it does not exist depending on
what happens with $\ds\lim_{x \rightarrow 0^-}(1+2x)^{1/x}=
\lim_{t\to-\infty}\biggl(1+\frac 2t\biggr)^{t}$.
If we let $T=-t$, then
$\ds\lim_{t\to-\infty}\biggl(1+\frac 2t\biggr)^{t}=\lim_{T\to\infty}\biggl(1+\frac {-2}T\biggr)^{-T}=
\frac{1}{\ds\lim_{T\to\infty}\biggl(1+\frac {-2}T\biggr)^{T}}=\frac{1}{e^{-2}}=e^2$ so
$\ds\lim_{x \rightarrow 0^-}(1+2x)^{1/x}=e^2$ and $\ds\lim_{x \rightarrow 0}(1+2x)^{1/x}=e^2$.
\endp

\end
