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\leftline{Mathematics 134 -- Intensive Calculus for Science 2}
\leftline{Solutions for Exam 2 -- March 24, 2006}
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\noindent
I.  Compute each of the integrals below, using some combination
of integration by substitution, parts, and partial fractions.  
{\it You must show all work for full credit.  Applying an entry 
from the integral table is OK if necessary, but that will result 
in a deduction of 3 points per part.}
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\item{A)}  Use integration by parts with $u = x$, $dv = \cos(4x)\ dx$.
Then $du = dx$, $v = {1\over 4} \sin(4x)$ and 
$$\int x \cos(4x)\ dx = {x\over 4}\sin(4x) - \int {1\over 4} \sin(4x)\ dx
= {x\over 4}\sin(4x) + {1\over 16} \cos(4x) + C$$
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\item{B)} By partial fractions:  The degree of the top is already
strictly smaller than the degree of the bottom so we do not
need to divide.  Then we factor $x^2 - 6x + 8 = (x - 2)(x - 4)$
and the partial fractions are:
$${x+3\over x^2 - 6x + 8} = {A \over x - 2} + {B\over x - 4}.$$
We clear denominators to get
$$x + 3 = A(x - 4) + B(x - 2).$$
Setting $x = 2$ gives $5 = -2A$, so $A = -5/2$.  Then setting
$x = 4$ gives $7 = 2B$, so $B = 7/2$.  
$$\int {-5/2 \over x - 2} + {7/2 \over x - 4}\ dx = 
{-5\over 2}\ln|x -2| + {7\over 2}\ln|x - 4| + C$$
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\noindent
II.  Compute each of the following integrals using an entry or 
entries from the table of integrals.  Say which entry or entries you 
are using.
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\item{A)} Use \#17 with $n = 4$, and then again with $n = 2$:
$$\int \sin^4(x)\ dx = {-1\over 4} \sin^3(x)\cos(x) - 
{3\over 8}\cos(x)\sin(x)+{3x\over 8}+C$$
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\item{B)} Use \#14 with $p(x) = x^2 + 12$ and $a = 3$:
$$\int (x^2 + 12) e^{3x}\ dx = 
{1\over 3}(x^2 + 12)e^{3x} - {2x\over 9}e^{3x} + {2\over 27}e^{3x} + C $$
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\noindent
III.  (20)  Let $R$ be the region bounded by $y = x^2 + 2$, $y = x$,
$x = 0$ and $x = 1$.   Find the volume of the solid obtained if $R$
is rotated about the line $y = 5$.
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{\it Solution:}  The parabola $y = x^2 + 2$ lies above the line
$y = x$ for all $x$.  But there is a gap between
the region $R$ and the line $y = 5$ (the axis of rotation).  
So the cross-sections of the solid
perpendicular to the $x$-axis are {\it washers} with inner radius
$5 - (x^2 + 2) = 3 - x^2$ and outer radius $5 - x$.  The volume is
the integral of the area of the cross-section:
$$\eqalign{V &= \int_0^1 \pi (5 - x)^2 - \pi (3 - x^2)^2\ dx\cr
             &= \pi\int_0^1 16 - 10x + 7x^2 - x^4\ dx\cr
             &= \pi\left(16x - 5x^2 + {7\over 3}x^3 - {x^5\over 5}\right|_0^1 \cr
             &= {197\pi\over 15}\cr}$$
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\noindent
IV.  A long thin straight wire extends from $x = 0$ to $x = 1$ cm.  
The density of the wire at $x$ is $\delta(x) = 2 + \cos(\pi x)$ grams per cm.
\item{A)} (10) Set up, {\it but do not evaluate}
the formula computing the {\it center of mass} of the wire.
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{\it Solution:}  Note the problem says ``long straight'' wire -- think
of it as a segment of the $x$-axis.  The center of mass is at
$$\overline{x} = {\int_0^1 x(2 + \cos(\pi x))\ dx \over \int_0^1 2 + \cos(\pi x)\ dx}$$
(Note: the $2 + \cos(\pi x)$ is the density of the metal in the 
wire, not the {\it shape} of the wire!)
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\item{B)} (10) If the midpoint Riemann sum with $n = 10$ is
used to approximate the integral for the total mass, will the result be
an overestimate or an underestimate?  Explain.
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{\it Solution:}  The mass integral is 
$$M = \int_0^1 \cos(\pi x) + 2\ dx.$$
The function $\cos(\pi x) + 2$ has a graph that is 
concave down from $x = 0$ to $x = 1/2$, then concave 
up from $x=1/2$ to $x = 1$.  By the symmetry of the cosine
graph, we can see that the overestimate areas on the first half
are exactly balanced by the underestimate areas on the second
half.  So the correct answer is: {\it neither} -- the midpoint
Riemann sum gives the {\it exact value}(!)
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\noindent
{\it Scoring note:}  This was a tricky question and no one in
the class actually got it correct (for the correct reason).  
So what I did to score this problem was:  
\item{$\bullet$}  If you said the answer depended on 
the concavity of the graph $y = \cos(\pi x) + 2$ and got
the estimates going the right way (that is, overestimate 
where concave down, underestimate where concave up), then 
I gave full credit (10 points)
\item{$\bullet$} If you said the answer depended on 
the concavity of the graph $y = \cos(\pi x) + 2$ and got
the estimates going the wrong way, then 
I gave part credit (8 points)
\item{$\bullet$} If you didn't make the connection 
with concavity, instead of giving no credit, I {\it dropped 
the question entirely}, and computed your grade using the other 
90 possible points.  
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(This seemed the fairest way to do it and it actually helped everyone
in all three categories.)
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V.  (10)  The graph $y = f(x)$ below shows either a probability density (pdf)
or a cumulative distribution (cdf).  Say which it is, and find the value
of $c$ given in the graph.
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{\it Solution:}  The given graph is a pdf because the total area
between the graph is finite and the function is not always 
increasing (and does not approach 1 as $x \to +\infty$).
The value of $c$ is determined from the requirement that the 
total area is $1$.  Using area formula for rectangles:
$1 = 2(2c) + c = 5c$, so $c = 1/5$.
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