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\centerline{Mathematics 136, section 2 -- Advanced Placement Calculus}
\centerline{Solutions for Practice Problems for Exam 2}
\centerline{October 31, 2009}
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\noindent
{\it Sample Exam} -- Note the real exam will contain different questions,
and possibly different types of questions.  The actual exam will be about this length.
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\noindent
I. In an alternate universe, the ``super-attractive'' force
exerted by two objects of mass $m_1,m_2$ 
is {\it directly} proportional to the square of the distance between
them: $F = S m_1m_2 r^2$, where $S$ is a ({\it positive}) constant and $r$ is the 
distance between the masses.  Two 1kg masses are fixed 
at points $x = -1$ and $x = 3$ respectively along a straight line
in this alternate universe.  At what location $x$ along that line should
a third 2kg mass be placed to {\it minimize} the sum of the super-attractive
forces exerted on it by the two unit masses?
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\noindent
{\it Solution:}  The total force felt by a 2kg mass placed at $x$ is 
$$F(x) = 2S(x+1)^2 + 2S(x - 3)^2.$$
We have $F'(x) = 4S(x + 1 + x - 3) = 4S(2x - 2)$.  This is zero
when $x = 1$.  Since the second derivative is $F''(x) = 8S > 0$,
this is a minimum by the second derivative test.  {\it Comment:  This 
is probably one of the easiest problems of this type that we have
seen, once the function to be minimized is set up.  Also be prepared
for more difficult problems like some of the suggested practice 
problems on the review sheet.}
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\noindent
II. (15) Evaluate $\lim_{x\to\infty} \left(1 - {3\over x}\right)^x$. 
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\noindent
{\it Solution:}  This is a $1^\infty$ indeterminate form.  So as
in several problems from Problem Set 4, we take logarithms first, 
then rearrange to $0/0$ form:
$$\eqalign{
\ln\left(\left(1 - {3\over x}\right)^x\right) &= x \ln\left(1 - {3\over x}\right)\cr
&= {\ln\left(1 - {3\over x}\right) \over {1\over x}}\cr}$$
Now apply L'Hopital's Rule.  The ratio of the derivatives is
$$
{ {1\over 1 - {3\over x}} \cdot {3\over x^2} \over {-1\over x^2}} =
{-3\over 1 - {3\over x}}. 
$$
In the limit as $x \to \infty$, this goes to $-3$.  Therefore,
$\lim_{x\to\infty} \left(1 - {3\over x}\right)^x = e^{-3}$.
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\noindent
III. (10)  In a car moving at 90 ft/sec, the driver 
suddenly saw an obstacle 400 feet ahead and braked
to a stop in 10 seconds.  The car's velocity was recorded
by a sensor in its onboard computer every two seconds:
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$$\matrix{t\ (sec) & 0 & 2 & 4 & 6 & 8 & 10\cr
     v(t)\ (ft/sec) & 90 & 75 & 50 & 25 & 5 & 0 \cr}$$
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Can you say for sure from the information
here whether the car hit the obstacle or not?  Explain, using 
the left- and right-hand Riemann sums for $v(t)$.
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\noindent
{\it Solution:}  The total distance traveled is $s = \int_0^{10} v(t)\ dt$ (feet).
The left-hand Riemann sum approximation with $n = 5$ has $\Delta t = 2$, so
$$s \doteq v(0) \Delta t + \cdots + v(8) \Delta t = 180 + 150 + 100 + 50 + 10 = 490.$$
The right-hand Riemann sum approximation is
$$s \doteq v(2) \Delta t + \cdots + v(10) \Delta t = 150 + 100 + 50 + 10 + 0 = 310.$$
Since $v(t)$ is (apparently) decreasing on the whole interval, the 
left-hand sum is an overestimate and the right-hand sum is an underestimate.
Since $400$ is included in the interval between the two 
estimates, we {\it cannot say for sure}.  It is 
possible, but the car may also have stopped short of the obstacle.
It would depend on what happened between the times at which the velocity 
values are known.  
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\noindent
IV. 
\item{A)} (10) Using the definition of the definite integral (not the 
Evaluation Theorem), determine $\int_1^2 x^2 + 3x + 1\ dx$.  
(Possibly useful information:
$$\sum_{i=1}^n i = {n(n+1)\over 2} \qquad \sum_{i=1}^n i^2 = {n(n+1)(2n+1)\over 6}.)$$ 
Check your answer using the Evaluation Theorem.
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\item{} {\it Solution:}  We subdivide $[1,2]$ into $n$ equal subintervals
with $\Delta x = 1/n$.  The right endpoints are $x_i = 1 + i/n$ so
setting up the Riemann sum and using the summation formulas above,
$$\eqalign{
\int_1^2 x^2 + 3x + 1\ dx &= \lim_{n\to\infty} \sum_{i=1}^n\left((1 + i/n)^2 + 3(1 + i/n) + 1\right)(1/n)\cr
&=\lim_{n\to\infty} \sum_{i=1}^n \left(5 + 5i/n + i^2/n^2\right)(1/n)\cr
&=\lim_{n\to\infty} \left((5/n) \sum_{i=1}^n 1 + (5/n^2) \sum_{i=1}^n i + (1/n^3)\sum_{i=1}^n i^2\right)\cr
&=\lim_{n\to\infty} \left( 5 + {5\over 2} + {5\over 2n} + {1\over 3} + {1\over 2n} + {1\over 6n^2}\right)\cr
&= {47\over 6}.\cr}$$
Checking by the Evaluation Theorem:
$$\eqalign{
\int x^2 + 3x + 1\ dx 
&= \left. {x^3\over 3} + {3x^2\over 2} + x \right|_1^2\cr
&={8\over 3} + 6 + 2 - {1\over 3} - {3\over 2}- 1\cr 
&= {16 + 36 + 12 - 2 - 9 - 6\over 6}\cr
&={47\over 6}.\cr}$$
\item{B)} (10) What is ${d\over dx} \int_1^{x^2} {\sin(t)\over t}\ dt$?
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\item{} {\it Solution:}  By the first part of the FTC and the chain rule,
this is 
$${\sin(x^2)\over x^2}\cdot 2x = {2\sin(x^2)\over x}.$$
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\noindent
V.  Methods of integration. In B,C,D you may use the table of integrals provided.
\item{A)} (10) Using integration by parts, show that
$$\int x^n \cos(ax)\ dx = {x^n\over a} \sin(ax) - {n\over a}\int x^{n-1} \sin(ax)\ dx.$$ 
\item{} {\it Solution:}  Let $u = x^n$ and $dv = \cos(ax)\ dx$. Then
$du = n x^{n-1}\ dx$ and $v = {1\over a}\sin(ax)$.  So applying the parts formula
$\int u\ dv = uv - \int v\ du$:
$$\int x^n \cos(ax)\ dx = {x^n\over a} \sin(ax) - {n\over a}\int x^{n-1} \sin(ax)\ dx$$
as desired.
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\item{B)} (10)  $\int \tan(\sqrt{x})/\sqrt{x}\ dx$
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\item{} {\it Solution:}  Make the preliminary substitution $u = \sqrt{x}$.  Then
$du = {1\over 2\sqrt{x}}\ dx$, so by \# 12 in the table, or 
another substition, the integral is 
$$2\int \tan(u)\ du = 2 \ln|\sec(u)| + C = 2\ln|\sec(\sqrt{x})| + C.$$
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\item{C)} (10)  $\int (3x + 2)/(x^2 + 8x + 7)\ dx$
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\item{} {\it Solution:}  We apply the partial fractions method here since
$x^2 + 8x + 7 = (x + 1)(x + 7)$:
$$\eqalign{
{3x + 2 \over (x + 1)(x + 7)} &= {A\over x + 1} + {B \over x + 7}\cr
3x + 2 &= A(x + 7) + B(x + 1)\cr}$$
Setting $x = -1$ gives $-1 = 6A$ so $A = {-1\over 6}$.  Then $x = -7$ gives
$-19 = -6B$, so $B = {19\over 6}$.  The integral is 
$$\int {{-1\over 6} \over x + 1} + {{19\over 6} \over x + 7}\ dx = 
{-1\over 6}\ln|x+1| + {19\over 6}\ln|x + 7| + C.$$

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\item{D)} (10) $\int (4 - x^2)^{-3/2}\ dx$
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\item{} {\it Solution:}  This is the form of \# 38 in the table with $a = 4$.  So 
the integral is 
$${x \over 4\sqrt{4 - x^2}} + C.$$
\bye 


