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\begin{center}

Math 1210 Fall 2004 K. M. Golden

\bigskip

{\bf EXAM III Solutions}
\end{center}


\begin{enumerate}
\item[(20 points) ~~1.]  Calculate the following.  Be sure to
show all of your work.

\begin{enumerate}
\item[(a)] $\ds \lim_{x \rightarrow 0}
                 \frac{\sin{x}-x}{x^3}$

\bigskip
This is Indeterminate $\frac{0}{0}$ so use L'Hopital:
 $$\ds \lim_{x \rightarrow 0}
                 \frac{\sin{x}-x}{x^3}
                 =
      ds \lim_{x \rightarrow 0}
                 \frac{\cos{x}-1}{3x^2}
                 $$
                 Intdeterminate $\frac{0}{0}$ so use L'Hopital.
 $$\ds \lim_{x \rightarrow 0}
                 \frac{\sin{x}-x}{x^3}
                 =
      ds \lim_{x \rightarrow 0}
                 \frac{-\sin{x}}{6x}
                 $$
                                  Intdeterminate $\frac{0}{0}$ so use
L'Hopital.
 $$\ds \lim_{x \rightarrow 0}
                 \frac{\sin{x}-x}{x^3}
                 =
      ds \lim_{x \rightarrow 0}
                 \frac{-\cos{x}}{6} = - \frac{1}{6}.
                 $$

\item[(b)] $\ds \lim_{x \rightarrow 1}
                 \frac{x^3-1}{x-1}$
\bigskip

Factor $\frac{x^3-1}{x-1}=(x^2+x+1)$ for $x \neq 1$ so
             $$\ds \lim_{x \rightarrow 1}
                 \frac{x^3-1}{x-1}
                 =
     \ds \lim_{x \rightarrow 1}
                 x^2+x+1 =3.
                 $$
or, this is Indeterminate $\frac{0}{0}$ so use L'Hopital:
$$\ds \lim_{x \rightarrow 1}
                 \frac{x^3-1}{x-1}
                 =
     \ds \lim_{x \rightarrow 1}
                 \frac{3x^2}{1}  =3
                 $$

                 or this limit is equivalent to the definition of
the derivative of $f(x)=x^3$ at $x=1$ (so technically you
shouldn't be using L'Hopital's rule which uses its derivative
to find its derivative!

\bigskip

\item[(c)] $\ds \lim_{x \rightarrow 0}
                 \frac{\ds \int_0^x{\sin{(t^2)}\;dt}}{x^3}$

This is Indeterminate $\frac{0}{0}$ so use L'Hopital and the Fundamental
Theorem:

$$\ds \lim_{x \rightarrow 0}
                 \frac{\ds \int_0^x{\sin{(t^2)}\;dt}}{x^3}
             =
                 \ds \lim_{x \rightarrow 0}
                 \frac{\ds \sin {(x^2)} } {3x^2}$$

This is Indeterminate $\frac{0}{0}$ so use L'Hopital:

$$\ds \lim_{x \rightarrow 0}
                 \frac{\ds \int_0^x{\sin{(t^2)}\;dt}}{x^3}
             =
                 \ds \lim_{x \rightarrow 0}
                 \frac{\ds 2x \cos {(x^2)} } {6x}$$

This is Indeterminate $\frac{0}{0}$ so use L'Hopital:

$$\ds \lim_{x \rightarrow 0}
                 \frac{\ds \int_0^x{\sin{(t^2)}\;dt}}{x^3}
             =
                 \ds \lim_{x \rightarrow 0}
                 \frac{\ds 2 \cos {(x^2)}-2x\sin{(x^2)} } {6} =
\frac{2}{6}=\frac{1}{3}$$


or use $\sin x \sim x$ so $\sin{(t^2)}\sim t^2$ and $\int_0^x \sin{(t^2)}
dt \sim \frac{t^3}{3}$
gives the same result.


\end{enumerate}
\eject

\item[(20 points) ~~2.] A population of frogs in a swamp is found
                        to grow at a rate proportional to the
              square root of the population size. The initial
population $P$ is 400 frogs, and 5 years later there are 900
of them.
\begin{enumerate}
\item[(a)] Write the differential equation for the frog population
$P(t)$ with the two corresponding conditions.

\bigskip
$$P'(t)=kP^{1/2}   \quad \quad P(0)=400 \quad \quad P(5)=900 . $$

\item[(b)] Solve this differential equation, that is, find the
        particular solution which incorporates both conditions.

$$\int \frac{dP}{P^{1/2}} = \int k dt $$
so
$$ 2 P^{1/2} = kt+C$$
or
$$P(t)= \frac{(kt+C)^2}{4}.$$

Plugging in $P=400$ and $t=0$ gives $C=2(400)^{1/2} =40$
and plugging in $P=900$ and $t=5$ gives
$k=\frac{2(900)^{1/2}-40}{5} = 4$.

\bigskip

\item[(c)] How long does it take for the frog population to
      quadruple (reach 1600) from its intitial value of 400?


Setting $P=1600$ and solving for $t$,

$$t=\frac{ 2(1600)^{1/2}-40}{4}=10$$
so it takes $10$ years.


\end{enumerate}

%\vspace{30ex}

\eject

\item[(10 points) ~~3.] Determine whether or not
        $\ds f(x)=\frac{1}{\sqrt{\sin{x}}}\;\;$
         is Riemann integrable on the interval (0,1],
         and {\bf fully} explain your result (you may
        want to do an
         explicit calculation to explain your result).


\bigskip
Using $\sin x \sim x$ as $x \to 0$
we get
$\frac{1}{\sqrt{\sin{x}}}\;\; \sim x^{-1/2}$
or, more precisely, using
$\sin x > \frac{x}{2}$ for $x \in (0,1]$
we get
$\frac{1}{\sqrt{\sin{x}}}\;\; < 2 x^{-1/2}$
for $x \in (0,1]$.

\bigskip
Because of this
$$  \int_a^1 \frac{1}{\sqrt{\sin{x}}}\; dx  < 2 \int_a^1 x^{-1/2} \; dx\;
$$
for any $a \in (0,1]$.

\bigskip
Then because

$$\lim_{a \to 0^+}  \int_a^1 x^{-1/2} \; dx\;
=
\lim_{a \to 0^+}  |_a^1  2x^{1/2} = 2$$
is finite, so is

$$\lim_{a \to 0^+} \int_a^1 \frac{1}{\sqrt{\sin{x}}}\; dx
\;$$
so
by comparison,
$\int_0^1  \ds \frac{1}{\sqrt{\sin{x}}}\;\;$
is a convergent improper integral, i.e.
$\ds f(x)=\frac{1}{\sqrt{\sin{x}}}\;\;$
is Riemann integrable on the interval $(0,1]$.





\eject

\item[(10 points) ~~4.]
Suppose the velocity $v(t)$ of a mass at the end of a spring
is given by $v(t)=\cos{t}$. If the position $x(t)$ of the mass and
its velocity are related by $\ds v(t)=\frac{dx}{dt}$, find
$x(t)$, given that its initial position is $x(0)=1$.


$$x(t)-x(0)= \int_0^t \frac{dx}{dt} dt = \int_0^t v(t) dt = \int_0^t \cos
t dt$$

so integrating and solving for $x(t)$ using $x(0)=1$,

$$x(t)=1+ \sin t .$$





\eject

\item[(20 points)~~ 5.]  Using the fact that
     $\ds \sum^n_{i = 1} \; i^2 = \frac{n(n + 1)(2n+1)}{6}\;$, calculate
     $\ds\int^2_0
      x^2\;dx$ directly from the {\bf definition} of
      the integral, i.e., using Riemann sums. {\it Check} your result
      using the {\bf fundamental theorem of calculus}.

\bigskip
Divide the interval $[0,2]$ into $n$ equal intervals of length $\Delta x
=\frac{2}{n}$
with endpoints $x_i = i \Delta x$, $i=1,\ldots,n$,
and define the integral as the limit of the Riemann sums corresponding to
those
intervals, $\sum_{i=1}^n  f(x_i) \Delta x$:

$$ \ds\int^2_0 x^2\;dx  =
\lim_{n \to \infty}
\sum_{i=1}^n (\frac{2i}{n})^2 \frac{2}{n} .$$
Pulling out three factors of $2$
out of the sum and the limit, and using the given explicit form
of the sum,


$$\ds\int^2_0 x^2\;dx  =
= 8\lim_{n \to \infty}
\sum_{i=1}^n i^2 \frac{1}{n^3} =
8\lim_{n \to \infty}
\frac{ n(n+1)(2n+1) }{6} \frac{1}{n^3} $$
and simplifying,

$$\ds\int^2_0 x^2\;dx  =
8\lim_{n \to \infty}
\frac{1}{3} + \frac{1}{2n} + \frac{1}{6n^2}= \frac{8}{3}.$$


Checking our result,
$$ \ds\int^2_0 x^2\;dx  = F(2)-F(0)$$
where $F'(x)=x^2$, i.e., $F$ is an antiderivative of $f(x)=x^2$. We know
that if
$F(x)=\frac{x^3}{3}+C$ for any $C$, then $F'(x)=x^2$, and
$F(2)-F(0)=\frac{8}{3}$.


\eject
\item[(20 points)~~ 6.]  Calculate the following integrals:

\bigskip
In the definite integral problems, to find $\int_a^b f(x)dx$, find an
antiderivative
$F(x)$ such that $F'(x)=f(x)$ and then $\int_a^b f(x)dx= |_a^b F =
F(b)-F(a)$.
All antiderivatives of the same function differ by a constant, which
cancels out in the subtraction, so typically we use a `simplest' $F$
for definite integrals, e.g., for $f(x)=x^n$, $F(x)=\frac{x^{n+1}}{n+1}$,
that is, there is no point in writing $+C$ or $+7$ when it will just
disappear after we subtract.
\bigskip

\begin{enumerate}
\item[(a)] $\ds \int^3_1 \, (4x^3-2x^2+x-5) \; dx$


$$\ds \int^3_1 \, (4x^3-2x^2+x-5) \; dx
=
|^3_1 \, x^4-2\frac{x^3}{3}+\frac{x^2}{2}-5x$$
or
$$\ds \int^3_1 \, (4x^3-2x^2+x-5) \; dx
=
(81-18+\frac{9}{2}-15)-(1-\frac{2}{3}+\frac{1}{2}-5)=56\frac{1}{3}=169/3.$$


\item[(b)]  $\ds \int \tan^2{x}\; \sec^2{x}\; dx$

\bigskip
This is an indefinite integral, so just
use substitution to find the general antiderivative, and
in this case you must include the $+C$. Let $u=\tan x$
and $du=\sec^2 x dx$. Then
$$\ds \int \tan^2{x}\; \sec^2{x}\; dx =
\ds \int u^2 \; du\; = \frac{u^3}{3} +C = \frac{\tan x ^3}{3} +C .$$
You can always check an antiderivative for both indefinite and
definite integrals by taking its derivative.

\bigskip

\item[(c)]  $\ds \int^1_{0} \; \frac{u(u+1)^2}{\sqrt{u}} \; du$
Multiplying the integrand out to a more convenient form,

$$ \ds \int^1_{0} \; u^{1/2}+2u^{3/2}+u^{5/2}  \; du
=
\; |^1_{0} \; \frac{u^{3/2}}{3/2} +2\frac{u^{5/2}}{5/2}
+\frac{u^{7/2}}{7/2} .
$$
Plugging in and subtracting (the $u=0$ terms all vanish)
$$\ds \int^1_{0} \; \frac{u(u+1)^2}{\sqrt{u}} \; du
=
=\frac{184}{105}.$$

\item[(d)]  $\ds \int^{\pi}_{0} \; \sin^2{x} \; dx$

Either use the $\sin$-$\cos$ reverse symmetry on $[0,\pi]$
so that
$$\ds \int^{\pi}_{0} \; \sin^2{x} \; dx
=
\frac{1}{2} \ds \int^{\pi}_{0} \; (\sin^2{x} + \cos^2 x )\; dx =
\frac{1}{2} \ds \int^{\pi}_{0} \; 1\; dx = \frac{\pi}{2} $$

or use $\sin^2 x = \frac{1}{2} - \frac{1}{2} \cos 2x$ so
$$\ds \int^{\pi}_{0} \; \sin^2{x} \; dx
=
\ds \int^{\pi}_{0} \; (\frac{1}{2} - \frac{1}{2} \cos 2x )\; dx .$$

After substituting $u=2x$ in the second term and antidifferentiating,
$$\ds \int^{\pi}_{0} \; \sin^2{x} \; dx
=
|_0^\pi \frac{1}{2}  x - \frac{1}{4} \sin 2x
$$
and since
$\sin 0 = \sin \pi =0$,
$$\ds \int^{\pi}_{0} \; \sin^2{x} \; dx
=
\frac{1}{2}(\pi - 0) = \frac{\pi}{2} . $$




\end{enumerate}
\eject


\end{enumerate}

\end{document}


