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\centerline{Mathematics 132 -- Calculus for Physical and Life Sciences 2}
\centerline{Exam 2 -- Review Sheet}
\centerline{March 17, 2008}


\noindent
{\it Sample Exam Questions -- Solutions}
\vskip 10pt
\noindent
This list is much longer than the actual exam will be (to give you some
idea of the range of different questions that might be asked).  
Unless otherwise directed, you may use any entry of the Table of 
Integrals from the text that applies.
\vskip 10pt
\noindent
I.  
\item[(A)] Do you need partial fractions to 
compute $$\int {t^2 + 1\over t^3 + 3t + 3}\ dt?$$
Explain, and give a simpler method.
\vspace{.1in}
No. Just do a $u$-substitution. $u=t^3+3t+3, \ du= (3t^2+3) dt= 3(t^2+1) dt$. Then $\displaystyle \int {t^2 + 1\over t^3 + 3t + 3}\ dt = \frac{1}{3} \int \frac{1}{u}\, du = \frac{1}{3} \ln|u| +C=\frac{1}{3} \ln|t^3-3t+3| + C.$

\vspace{.1in}

\item[(B)]  Apply partial fraction decomposition to 
compute $$\int {1\over x(x - 1)(x + 2)}\ dx$$

\vspace{.1in}
From the form of the factorization of the denominator we have the following
partial fractions:
$$\displaystyle \frac{1}{x(x-1)(x+2)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+2}.$$
Then $A(x-1)(x+2)+Bx(x+2)+Cx(x-1)=1$. If $x=0$, we get $-2A=1$. Thus, $\displaystyle A=-\frac{1}{2}$. If $x=1$, we get $3B=1$. Thus $\displaystyle B=\frac{1}{3}$. If $x=-2$, we get $6C=1$. Thus $\displaystyle C=\frac{1}{6}$. Therefore 

$\displaystyle \int {1\over x(x - 1)(x + 2)}\ dx =-\frac{1}{2} \int \frac{1}{x}\, dx + \frac{1}{3} \int \frac{1}{x-1}\, dx + \frac{1}{6} \int \frac{1}{x+2}\, dx = $ 

\vspace{.1in}

 $-\frac{1}{2} \ln|x| + \frac{1}{3} \ln|x-1| + \frac{1}{6} \ln|x+2| +C.$

\vspace{.1in}

\noindent (G)  Which trigonometric substitution would you apply
to compute $\int {1\over u \sqrt{a^2 - u^2}}\ du$?
What trigonometric integral do you get after making the substitution?
Complete the derivation of the integral.
\bigskip\vspace{.1in} 

Use the trigonometric substitution $u=a \sin \theta, \ du = a \cos \theta \, d\theta$. Then 

$\displaystyle \int {1\over u \sqrt{a^2 - u^2}}\ du = \int \frac{1}{a \sin \theta \sqrt{a^2 - a^2 \sin^2 \theta} } a \cos \theta \, d\theta = \int \frac{\cos \theta}{\sin\theta a \sqrt{\cos^2 \theta }}\, d \theta =$ \vspace{.1in}

$ \displaystyle \frac{1}{a} \int \frac{\cos \theta}{\sin\theta \cos \theta} \, d \theta = \frac{1}{a} \int \frac{1}{\sin \theta}\, d \theta= \frac{1}{a} \int \csc \theta \, d \theta$. Now we use the table of integrals (formula $15$) to obtain that the integral equals $\displaystyle \frac{1}{a} \ln|\csc \theta - \cot \theta| + C$. If $u=a \sin \theta$, then $\csc \theta=a/u$ and $\displaystyle \cot \theta = \frac{\sqrt{a^2-u^2}}{u}$. Therefore $$\int {1\over u \sqrt{a^2 - u^2}}\ du= \frac{1}{a} \ln|\frac{a}{u}-\frac{\sqrt{a^2-u^2}}{u}|+C.$$

\vskip 10pt
\noindent
I.
\begin{enumerate} 
\item[(A)] Use the midpoint rule, the trapezoidal rule, and Simpson's rule 
with $n = 4$ to estimate the value of the integral $\int_0^2 \sqrt{1 + x^4}\ dx$, 
rounding your answers to 6 decimal places.
\vskip 10pt
\noindent
\emph{Solution:}  With $n = 4$, $\Delta x = \frac{2 - 0}{4} = .5$, and the 
intermediate points are $x_0 = 0$, $x_1 = .5$, $x_2 = 1$, $x_3 = 1.5$, and $x_4 = 2$.
The midpoints are $m_1 = .25$, $m_2 = .75$, $m_3 = 1.25$, and $m_4 = 1.75$, so the
midpoint rule approximation is
\begin{eqnarray*}
\int_0^2 \sqrt{1 + x^4}\ dx &\doteq& \sqrt{1 + (.25)^4} (.5) + \sqrt{1 + (.75)^4} (.5) \\
&& \quad + \sqrt{1 + (1.25)^4} (.5) + \sqrt{1 + (1.75)^4} (.5)\\
&\doteq& (1.001951 + 1.147347 + 1.855103 + 3.221631)(0.5)\\
&\doteq& 3.613016.
\end{eqnarray*}
The trapezoidal rule approximation is 
\begin{eqnarray*}
\int_0^2 \sqrt{1 + x^4}\ dx &\doteq& \frac{(.5)}{2}\left(\sqrt{1+(0)^4} + 2\sqrt{1+(.5)^4} 
+ 2\sqrt{1+(1.0)^4}\right.\\
&& \quad \left. + 2\sqrt{1 + (1.5)^4} + \sqrt{1 + (2)^4}\right)\\
&\doteq& (.25)(1 + 2(1.030776) + 2(1.414213) + 2(2.462214) + 4.123106)\\
&\doteq& 3.734379.
\end{eqnarray*}
The Simpson's rule approximation is 
\begin{eqnarray*}
\int_0^2 \sqrt{1 + x^4}\ dx &\doteq& \frac{(.5)}{3}\left(\sqrt{1+(0)^4} + 4\sqrt{1+(.5)^4} +
2\sqrt{1 + (1.0)^4}\right.\\
&& \quad \left. + 4\sqrt{1 + (1.75)^4} + \sqrt{1 + (2)^4}\right)\\
&\doteq& (.166667)(1 + 4(1.030776) + 2(1.414213) + 4(2.462214) + 4.123106)\\
&\doteq& 3.653470.
\end{eqnarray*}
\item[(B)] Which of your answers in part A are overestimates and which
are underestimates.  Explain how you can tell.
\vskip 10pt
\emph{Solution:}  The function $y = \sqrt{1 + x^4}$ has $y' = \frac{2x^3}{\sqrt{1+x^4}}$
and $y'' = \frac{2x^2(3 + x^4)}{(1 + x^4)^{3/2}}$.  Since both of these are positive
on $[0,2]$, the function is increasing and concave up on this interval.  This
implies that the midpoint rule is an underestimate and the trapezoidal rule is 
an overestimate.  For Simpson's rule, we do not have enough information to 
determine whether the approximation is an overestimate or an underestimate.  
\item[(C)] Which of the estimates in part A would you expect to have the
smallest error.  Explain.  (Review Lab 1 to see the idea here.) 
\vskip 10pt
\emph{Solution:}  We expect Simpson's rule to give the most accurate approximation
because this Simpson's rule estimate is a weighted average of the midpoint and
trapezoidal rule estimates with $n = 2$ (not $n = 4$):
\[
S_4 = \frac{1}{3}T_2 + \frac{2}{3}M_2.
\]
The weighted average is designed to cancel out the errors.  Recall that (at least
for large enough $n$), the trapezoidal rule error has absolute value close
to twice the absolute midpoint rule error, but with the opposite sign.
\end{enumerate}
\vskip 10pt
\noindent
II.  For each of the following integrals, say why the integral is improper,
determine if the integral converges, and if so, find its value.
\vskip 10pt
\begin{enumerate}
\item[A)]  $\int_1^\infty \frac{1}{{\root 5 \of x}}\ dx$
\vskip 10pt
\emph{Solution:}
This is an improper integral because of the infinite interval.
It converges if
\[
\lim_{b\to\infty} \int_1^b x^{-1/5}\ dx
\]
is finite.  But
\begin{eqnarray*}
\lim_{b\to\infty} \int_1^b x^{-1/5}\ dx &=& \lim_{b\to\infty} \left.\frac{5}{4}x^{4/5}\right|_1^b\\
&=& \lim_{b\to\infty} \frac{5}{4}(b^{4/5} - 1)\\
&=& +\infty.
\end{eqnarray*}
The limit does not exist as real number, so the integral diverges.  
\item[B)]  $\int_0^2 \frac{dx}{x^2 - 7x + 6}$

\emph{Solution:}
This is an improper integral because the integrand 
\[
\frac{1}{x^2 - 7x + 6} = \frac{1}{(x-1)(x-6)}
\]
has an infinite discontinuity at $x = 1$ in the interval $[0,2]$.  
It converges if \emph{both limits in} 
\[
\lim_{b\to 1^-} \int_0^b \frac{dx}{x^2 - 7x + 6} + \lim_{a\to 1^+} \int_a^2 \frac{dx}{x^2 - 7x + 6}
\]
exist (are finite).  But the first integral gives (by the partial fraction method):
\begin{eqnarray*}
\lim_{b\to 1^-} \int_0^b \frac{dx}{x^2 - 7x + 6} &=& \lim_{b\to 1^-} \int_0^b \frac{-1/5}{x-1} + \frac{1/5}{x-6}\, dx \\
&=& \lim_{b\to 1^-}\left(\left.\frac{-1}{5}\ln|x-1| + \frac{1}{5}\ln|x-6|\right|_0^b\right)\\
&=& \lim_{b\to 1^-}\left(\frac{-1}{5}\ln|b-1| + \frac{1}{5}\ln|b-6| - \frac{1}{5}\ln(6)\right).
\end{eqnarray*}
The limit does not exist as real number, so the integral diverges.  
\item[C)]  $\int_0^\infty x e^{-3x}\ dx$
\vskip 10pt
\emph{Solution:}
This is an improper integral because of the infinite interval.
It converges if
\[
\lim_{b\to\infty} \int_0^b x e^{-3x}\ dx
\]
is finite.  Integrating by parts, 
\begin{eqnarray*}
\lim_{b\to\infty} \int_0^b xe^{-3x}\ dx &=& \lim_{b\to\infty} \left(\left. \frac{-x}{3}e^{-3x} - \frac{1}{9} e^{-3x} \right|_0^b\right)\\
&=& \lim_{b\to\infty} \frac{-b}{3} e^{-3b} - \frac{1}{9}e^{-3b} + \frac{1}{9} \\
&=& \frac{1}{9}
\end{eqnarray*}
(using L'Hopital's Rule to evaluate $\displaystyle{\lim_{b\to\infty}\frac{-b}{3e^{3b}} = 0}$).  
The integral converges to $\frac{1}{9}$.    
\item[D)]  For which values of $a$ is $\int_0^\infty e^{ax}\sin(x)\ dx$
convergent?  Evaluate the integral for those $a$. 
 
\emph{Solution:}
This is an improper integral because of the infinite interval.
It converges if
\[
\lim_{b\to\infty} \int_0^b e^{ax}\sin(x)\ dx
\]
is finite.  This integral can be evaluated either by integrating by parts
twice and solving for the integral, or else consulting \# 98 in the table:
\[
\int e^{ax} \sin(x)\ dx = \frac{e^{ax}}{a^2 + 1}(a\sin(x) - \cos(x)) + C.
\]
Therefore, 
\[
\int_0^b e^{ax}\sin(x)\ dx = \frac{e^{ab}}{a^2 + 1}(a\sin(b) - \cos(b)) + \frac{1}{a^2 + 1}.
\]
This has a finite limit as $b \to +\infty$ if and only if $a < 0$.  For those $a$,
the first term goes to zero as $b \to +\infty$, and the value of the integral is
$\frac{1}{a^2 + 1}$.  
\end{enumerate}
\vskip 10pt
\noindent
III.
\begin{enumerate}  
\item[(A)]  Let $R$ be the region in the plane bounded by $y = 3 - x^2$ and the $x$-axis.
\begin{enumerate}
\item[(1)]  Sketch the region $R$.  
\vskip 10pt 
\emph{Solution:}  The region is bounded above by the parabola $y = 3 - x^2$, and 
below by the $x$-axis.  It extends from $x = -\sqrt{3}$ to $x = \sqrt{3}$:
\begin{center}
\includegraphics[width=2.8truein,height=2.8truein]{132Ex2RevIIIA1.ps}
\end{center} 
\item[(2)]  Find the area of $R$.
\vskip 10pt
\emph{Solution:}  The area is 
\[
A = \int_{-\sqrt{3}}^{\sqrt{3}} 3 - x^2\ dx =2\int_0^{\sqrt{3}} 3 - x^2\ dx = 2\left(\left. 3x - \frac{x^3}{3}\right|_0^{\sqrt{3}}\right) = 4\sqrt{3}.
\]
\item[(3)]  Find the volume of the solid generated by rotating $R$ about the
$x$-axis.
\vskip 10pt
\emph{Solution:}  Rotating $R$ around the $x$-axis, the cross-sections are disks
with radius $y = 3 - x^2$, so
\begin{eqnarray*}
V &=& \int_{-\sqrt{3}}^{\sqrt{3}} \pi (3 - x^2)^2 \ dx\\
&=& 2\pi\int_0^{\sqrt{3}} 9 - 6x^2 + x^4 \ dx\\
&=& 2 \pi\left(\left. 9x - 2x^3 + \frac{x^5}{5} \right|_0^{\sqrt{3}}\right)\\
&=& \frac{48\pi\sqrt{3}}{5}.
\end{eqnarray*}
\end{enumerate}
\item[(B)]  Let $R$ be the region in the plane bounded by $y = 3x$ and $y = x^2$.  
\begin{enumerate}
\item[(1)]  Sketch the region $R$.
\vskip 10pt 
\emph{Solution:}  The region is bounded above by the line $y = 3x$ and below
by the parabola $y = x^2$.  It extends from $x = 0$ to $x = 3$:
\begin{center}
\includegraphics[width=2.8truein,height=2.8truein]{132Ex2RevIIIB1.ps}
\end{center}   
\item[(2)]  Find the area of $R$.
\vskip 10pt
\emph{Solution:} The area is 
\[
A = \int_0^3 3x - x^2\ dx = \left(\left.\frac{3x^2}{2} - \frac{x^3}{3}\right|_0^3\right)
= \frac{9}{2}.
\]  
\item[(3)]  Find the volume of the solid generated by rotating $R$ about the
$x$-axis.
\emph{Solution:}  The cross-sections by planes perpendicular to the $x$-axis
are washers with inner radius $r_{in} = x^2$ and outer radius $r_{out} = 3x$.
The volume is 
\begin{eqnarray*} 
V &=& = \int_0^3 \pi (3x)^2 - \pi (x^2)^2\ dx\\
&=& \pi \int_0^3 9x^2 - x^4\ dx\\
&=& \pi \left(\left. 3x^3 - \frac{x^5}{5}\right|_0^3\right)\\
&=& \frac{162\pi}{5}.
\end{eqnarray*}
\item[(4)]  Find the volume of the solid generated by rotating $R$ about the
$y$-axis.
\vskip 10pt
\emph{Solution:}  For rotating about the $y$-axis, we slice horizontally 
and set up the integral in terms of $y$.  The horizontal cross-sections are 
washers too, with inner radius $r_{in} = y/3$ (from the line),
and outer radius $r_{out} = \sqrt{y}$ (from the parabola).  The solid
extends from $y = 0$ to $y = 9$ along the $y$-axis.  This gives
\begin{eqnarray*} 
V &=& = \int_0^9 \pi (\sqrt{y})^2 - \pi (y/3)^2\ dy\\
&=& \pi \int_0^9 y - \frac{y^2}{9}\ dy\\
&=& \pi \left(\left. \frac{y^2}{2} - \frac{y^3}{27}\right|_0^9\right)\\
&=& \frac{27\pi}{2}.
\end{eqnarray*} 
\end{enumerate}
\item[(C)]  Let $R$ be the region in the plane bounded by $y = \cos(\pi x)$, $y = 1/2$,
$x = -1/3$ and $x = 1/3$.
\begin{enumerate}
\item[(1)]  Sketch the region $R$.  
\vskip 10pt 
\emph{Solution:}  The region is bounded above by an arc of the cosine graph
and below by the line $y = 1/2$:
\begin{center}
\includegraphics[width=2.8truein,height=2.8truein]{132Ex2RevIIIC1.ps}
\end{center} 
\item[(2)]  Find the area of $R$.
\vskip 10pt
\emph{Solution:}
The area is (using symmetry):
\begin{eqnarray*}
A &=& \int_{-1/3}^{1/3} \cos(\pi x) - 1/2\ dx\\
&=& 2\int_0^{1/3} \cos(\pi x) - 1/2\ dx\\
&=& 2\left(\left. \frac{1}{\pi} \sin(\pi x) - \frac{x}{2}\right|_0^{1/3}\right)\\
&=& \frac{\sqrt{3}}{\pi} - \frac{1}{3}.
\end{eqnarray*}
\item[(3)]  Find the volume of the solid generated by rotating $R$ about the
$x$-axis.
\vskip 10pt
\emph{Solution:}  The cross-sections by planes perpendicular to the $x$-axis
are washers with inner radius $r_{in} = 1/2$ and outer radius $r_{out} = \cos(\pi x)$. 
By symmetry (and integrating with the 1/2-angle formula for $\cos^2(\pi x)$), 
our volume will be 
\begin{eqnarray*}
V &=& \int_{-1/3}^{1/3} \pi \cos(\pi x)^2 - \pi \left(\frac{1}{2}\right)^2\ dx\\
&=& 2\pi \int_0^{1/3} \frac{1}{2}(1 + \cos(2\pi x)) - \frac{1}{4}\ dx\\
&=& 2\pi \left(\left. \frac{x}{4} + \frac{1}{4\pi}\sin(2\pi x)\right|_0^{1/3}\right)\\
&=&  \frac{\sqrt{3}}{4} + \frac{\pi}{6}.
\end{eqnarray*}
\end{enumerate}
\end{enumerate}
\vskip 10pt
\noindent
IV.  The height of a monument is 20m.  The horizontal cross-section of
the monument at $x$ meters from the top is an isosceles right triangle
with legs $x/4$ meters.  Find the volume of the monument.  
\vskip 10pt
\emph{Solution:}  The area of the cross-section $x$ meters from the top
is $A(x) = \frac{1}{2}\left(\frac{x}{4}\right)^2 = \frac{x^2}{32}$.
So the volume is 
\[
V = \int_0^{20} \frac{x^2}{32}\ dx = \left(\left. \frac{x^3}{96}\right|_0^{20}\right) = \frac{250}{3}.
\]
cubic meters.
\vskip 10pt
\noindent
V. 
\begin{enumerate}
\item[(A)] Set up and evaluate the integral to compute the arclength of the 
curve $x = 3t^2$, $y = 2t^3$, $0 \le t \le 2$.
\vskip 10pt
\emph{Solution:}  By the arclength integral formula, 
\begin{eqnarray*}
L &=& \int_0^2 \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\ dt\\
&=& \int_0^2 \sqrt{(6t)^2 + (6t^2)^2}\ dt\\
&=& \int_0^2 6t \sqrt{1 + t^2}\ dt.
\end{eqnarray*}
Now make a $u$-substitution $u = 1 + t^2$, so $du = 2t\ dt$.  The integral converts
to 
\begin{eqnarray*}
&=& \int_{u=1}^{u=5} 3\sqrt{u}\ du\\
&=& \left(\left. 3\cdot \frac{2}{3} u^{3/2}\right|_1^5 \right)\\
&=& 2(5\sqrt{5} - 1).
\end{eqnarray*}
\item[(B)]  Set up and evaluate the integral to compute the arclength of the 
curve $y = \frac{1}{6}(x^2 + 4)^{3/2}$, $0 \le x \le 3$.  (Hint: the arclength
integral simplifies to a manageable form if you are careful with the algebra.)
\vskip 10pt
\emph{Solution:}  By the arclength integral formula,
we must integrate 
\begin{eqnarray*}
\sqrt{1 + \left(\frac{dy}{dx}\right)^2} 
&=& \sqrt{1 + \left(\frac{1}{4}(x^2 + 4)^{1/2}(2x)\right)^2}\\
&=& \sqrt{1 + \left(\frac{x}{2}(x^2 + 4)^{1/2}\right)^2}\\
&=& \sqrt{1 + x^2 + \frac{x^4}{4}}\\
&=& \sqrt{\left(1 + \frac{x^2}{2}\right)^2}\\
&=& 1 + \frac{x^2}{2}.
\end{eqnarray*}
(In other words, $1 + \left(\frac{dy}{dx}\right)^2$ is itself a perfect square
in this case!)  Hence
\[
L = \int_0^3 1 + \frac{x^2}{2}\ dx = \left(\left. x + \frac{x^3}{6}\right|_0^3\right)
= \frac{15}{2}.
\]
\end{enumerate}
\vskip 10pt
\noindent
VI.  
\begin{enumerate}
\item[(A)] Find the average value of $f(x) = \sqrt{1 - x^2}$ on the interval
$[0,1/2]$.  (Use trigonometric substitution, not the table.)
\vskip 10pt
\emph{Solution:}  The average value is 
\begin{eqnarray*}
f_{ave} &=& \frac{1}{1/2 - 0} \int_0^{1/2} \sqrt{1 - x^2}\ dx\\
&=& 2 \int_0^{\pi/6} \sqrt{1-\sin^2\theta}\cos\theta\,d\theta  \text{  (letting } x = \sin\theta, dx = \cos\theta\,d\theta\text{)}\\ 
&=& 2 \int_0^{\pi/6} \cos^2\theta\,d\theta\\
&=& 2 \int_0^{\pi/6} \frac{1}{2}(1 + \cos(2\theta))\, d\theta \text{  (half-angle formula)}\\
&=& \left(\left. \theta + \frac{1}{2}\sin(2\theta)\right|_0^{\pi/6}\right)\\
&=& \frac{\pi}{6} + \frac{\sqrt{3}}{4}.
\end{eqnarray*}
\item[(B)] Find the average value of $f(x) = x \sqrt{1 + x^4}$ on the interval $[0,2]$. 
  \vskip 10pt
\emph{Solution:}  The average value is 
\begin{eqnarray*}
f_{ave} &=& \frac{1}{2 - 0} \int_0^2 x\sqrt{1 + x^4}\ dx\\
&=& \frac{1}{4}\int_{u=0}^{u=4} \sqrt{1+u^2}\, du  \text{  (letting } u = x^2, du = 2x\, dx\text{)}\\ 
&=& \left(\left. \frac{u}{8}\sqrt{1+u^2} + \frac{1}{8}\ln(u + \sqrt{1 + u^2})\right|_0^4\right) \text{  (by \# 21 in table)}\\
&=& \frac{\sqrt{17}}{2} + \frac{1}{8}\ln(4 + \sqrt{17}).
\end{eqnarray*}
\end{enumerate}
\end{document}  
