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\def\sol{\bigskip\item{} {\it Solution:}\ }
\centerline{Mathematics 134 -- Intensive Calculus for Science 2}
\centerline{Solutions for Final Examination -- May 10, 2006}
\bigskip
\noindent
I.  
\item{A)}  (10) Give a precise statement of the Fundamental Theorem
of Calculus (both parts).
\sol
Part 1:  Let $f$ be continuous for $a\le x \le b$, and let
$F$ be an antiderivative of $f$.  Then $\int_a^b f(x)\ dx = F(b) - F(a)$.

\item{} Part 2:   Let $f$ be continuous for $a\le x \le b$, and let
$F(x) = \int_a^x f(t)\ dt$.  Then $F(x)$ is an antiderivative of $f(x)$
(in other words $F'(x) = f(x)$.
\bigskip
\item{B)}  (15) The following graph shows $y = f(x)$.  Let $F$
be the antiderivative of $f$ with $F(0) = 0$ and $F$ continuous. Sketch 
the graph $y = F(x)$.
\sol
By the given information and the FTC, we have 
$16 = \int_0^2 f(x)\ dx = F(2) - F(0)$, so $F(2) = 16 + F(0) = 16$.
Similarly, $-7 = \int_2^3 f(x)\ dx = F(3) - F(2)$, so $F(3) = -7 + F(2) = 
-7 + 16 = 9$.  The graph of $F$ should be concave up where 
$f = F'$ is increasing ($0 < x < 1/2$), then concave down the rest of the way
to $x = 3$ (since $f = F'$ is decreasing).
$F$ has a local maximum at $x = 2$ (a rather ``flat'' one).  
\bigskip
\noindent
II.  Compute each of the following integrals.  You may use
the table of integrals anywhere on these.  If you do, say
which table entry you are using.
\bigskip
\item{A)} (10)  $\int 3x^6 - 4\sqrt{x} + \sin(x)\ dx$
\sol
By basic rules, the integral is
$${3\over 7} x^7 - {8\over 3}x^{3/2} - \cos(x) + C$$
\medskip
\item{B)} (10)  $\int {x^3 \over x^4 + 1}\ dx$ 
\sol
Let $u = x^4 + 1$.  Then $du = 4x^3\ dx$, so the form is 
$${1\over 4} \int {1\over u}\ du = {1\over 4}\ln|u| + C = 
{1\over 4}\ln|x^4 + 1| + C$$
\medskip
\item{C)} (15)  $\int x^2 e^{-9x}\ dx$
\sol
Integrating by parts twice (or using \# 14 in the table with $p(x) = x^2$
and $a = -9$), we have
$$\eqalign{\int x^2 e^{-9x}\ dx 
&= {-x^2\over 9} e^{-9x} + {1\over 9}\int 2x e^{-9x}\ dx\cr
&= {-x^2\over 9} e^{-9x} + {1\over 9}\left({-2x\over 9} e^{-9x} + {1\over 9}\int 2 e^{-9x}\ dx\right)\cr
&= {-x^2\over 9} e^{-9x} - {2x\over 81}e^{-9x} - {2\over 729}e^{-9x} + C}$$
\medskip
\item{D)} (10)  $\int {1 \over \sqrt{4x^2 + 3^2}}\ dx$
\sol
This equals $\int {1 \over \sqrt{(2x)^2 + 3^2}}\ dx$
Using \# 29 in the table, after a preliminary subsitution $u = 2x$ ($du = 2dx$),
and $a = 3$:
$$\eqalign{
&= {1\over 2}\int {1 \over \sqrt{u^2 + 3^2}}\ du\cr
&= {1\over 2}\ln\left|u + \sqrt{u^2 + 3^2}\right| + C\cr
&= {1\over 2}\ln\left|2x + \sqrt{4x^2 + 9}\right| + C\cr}$$
\medskip
\item{E)} (15)  $\int {x+1\over x^2 + 5x + 6}\ dx$
\sol
By partial fractions:  ${x+1 \over (x+2)(x+3)} = {A\over x+2} + {B \over x+3}$.
Clearing denominators, $x+1 = A(x+3) + B(x+2)$.  Setting $x = -2$, 
$A = -1$.  Setting $x = -3$, $B = 2$.  So
$$\int {x+1\over x^2 + 5x + 6}\ dx = \int {-1\over x+2} + {2\over x+3}\ dx
=-\ln|x+2| + 2\ln|x+3| + C$$
(This can be checked by \# 27 in the table with $a = -2, b = -3, c = d = 1$.)
\bigskip
\noindent
III.  Let $R$ be the region bounded by $y = x$, the $x$-axis,
and $x = 1$, $x = 4$.
\bigskip 
\item{A)} (15)  Find the volume of the solid obtained by rotating
$R$ about the line $y = -2$.
\sol
Cross-sections by planes perpendicular to the $x$-axis are
washers with inner radius 2, outer radius $x + 2$, so 
$$V = \int_1^4 \pi(x+2)^2 - \pi(2)^2\ dx = \pi\int_1^4 x^2 + 4x\ dx = 
\pi \left(x^3/3 + 2x^2\right|_1^4 = 51\pi.$$
\medskip
\item{B)} (10)  A thin plate has the shape of the region
$R$ ($x,y$ in cm) and density $\delta(x) = x^{-1/2}$ grams/cm${}^2$.
Find its total mass.
\sol
$$M = \int_1^4 x^{-1/2}\cdot x\ dx = \int_1^4 x^{1/2}\ dx
= {2\over 3} x^{3/2}|_1^4 = {14\over 3}$$
\bigskip
\noindent
IV.  At a particular location in Natick on the Mass Pike, a sensor
was set up to measure the passage of traffic.  The measurements
made were used to derive a probability density function for the
quantity $x = $ time gap between successive cars (in minutes).  The results
gave the following formula as a good fit for the pdf:
$p(x) = 11(1-x)^{10}$ if $0 < x < 1$, and zero otherwise.
\bigskip
\item{A)} (10)  Show that $p$ satisfies the usual property for 
a probability density function:  $\int_0^1 p(x)\ dx = 1$.
\sol
In the integral $\int_0^1 11(1 - x)^{10}\ dx$, let $u = 1-x$
and $du = -dx$.  Changing the limits of integration into their
$u$-equivalents, then reversing the limits of integration 
and introducing another $-$ sign, we get
$$\int_{u=1}^{u=0} -11u^{10}\ du = \int_0^1 11u^{10}\ du = u^{11}|_0^1 = 1.$$ 
\item{B)} (15)  What is the probability that the time gap
between successive cars is between $x = .1$ minute and $x = .2$ minute?
\sol
Using the same substitution as in part A, this probability is 
$$\int_{.1}^{.2} 11(1-x)^{10}\ dx 
= \int_{u =.9}^{u=.8} -11u^{10}\ du = \int_{u =.8}^{u=.9} 11u^{10}\ du
=u^{11}|_{.8}^{.9} \doteq .228$$
\bigskip
\noindent
V.  
\item{A)} (10) Using the definition of Taylor polynomials, compute the
Taylor polynomial of degree $n = 3$ for $f(x) = \sqrt{1 + 2x}$ at $a = 0$.  
\sol
We have 
$$\eqalign{f(0) &= 1\cr
f'(x) = {1\over 2}(1 + 2x)^{-1/2}\cdot 2 = (1+2x)^{-1/2} \Rightarrow f'(0) &= 1\cr
f''(x) = {-1\over 2}(1 + 2x)^{-3/2}\cdot 2 = -(1+2x)^{-3/2} \Rightarrow f''(0) &= -1\cr
f'''(x) = {3\over 2}(1 + 2x)^{-5/2}\cdot 2 = 3(1+2x)^{-1/2} \Rightarrow f'''(0) &= 3\cr}$$
So the Taylor polynomial of degree 3 is 
$$1 + x - {1\over 2}x^2 + {3\over 6}x^3 = 1 + x - {1\over 2}x^2 + {1\over 2}x^3$$
\item{B)} (10) Use our shortcut methods to check your work in part A.
\sol
Substituting $u=2x$ in the binomial series with $p = 1/2$:
$$1+{1\over 2}(2x) + {\left({1\over 2}\right)\left({-1\over 2}\right)\over 2}(2x)^2
+ {\left({1\over 2}\right)\left({-1\over 2}\right)\left({-3\over 2}\right)\over 6}(2x)^3
= 1 + x - {1\over 2}x^2 + {1\over 2} x^3$$
\item{C)}  (5) Use your polynomial from part A to compute an approximation
to $\sqrt{1.2}$.  What is the error in your approximation?
\sol
We take $x = .1$ so 
$$\sqrt{1.2} = \sqrt{1 + 2(.1)} \doteq 1 + (.1) - (.1)^2/2 + (.1)^3/3 = 1.09550$$
Using a calculator, the exact value is about $1.09544$, so the error is about $.00006$.
\bigskip
\noindent
VI. (15) Solve for $y$ by separation of variables:
${dy\over dx} = \cos(x)(1 + y^2)$ with  $y(0) = 1$.
\sol
After separation, ${dy \over 1+y^2} = \cos(x)\ dx$.  Integrating
both sides, we get 
$$\arctan(y) = \sin(x) + C \Rightarrow y = \tan(\sin(x) + C).$$  
Then from the initial condition, 
$1 = \tan(\sin(0) + C)$, so $C = \pi/4$.  The solution is
$$y = \tan(\sin(x) + \pi/4)$$
\bigskip
\noindent
VII.  An avian flu epidemic has broken out in 
Birdsburgh, a large city with total population 10 million.  
Write $N$ for the number of people
who have been infected, as a function of time.  The Birdsburgh
Public Health department determines that:
$${dN\over dt} = k N (10 - N),\leqno(1)$$
or in words: the rate of change of $N$ is proportional to
the product of $N$ and the number of people not yet infected, 
where $N$ is in millions of people, $t$ in weeks, $k$ a positive
constant.
\item{A)} (10) Which of the following slope field plots matches (1)?
Explain how you can tell.
\sol
The one that matches is slope field B.  Note that A has an equilibrium
at $N = 1$ which doesn't match the equation.  (Or, note that
since it is given $k > 0$, the slope values should be positive
for $0 < N < 10$.  Only B satisfies that.)
\bigskip
\item{B)} (10) For what value of $k$ is $N(t) = 10/(1 + 1000e^{-t})$
a solution of (1)?
\sol
For this $N(t)$, by the chain rule
$${dN\over dt} = {10000 e^{-t}\over (1 + 1000e^{-t})^2}$$
The right side is 
$$\eqalign{
kN(10 - N) &= k \cdot {10\over 1 + 1000e^{-t}}\cdot \left(10 - {10\over 1 + 1000e^{-t}}\right)\cr
 &=  k \cdot {10\over 1 + 1000e^{-t}}\cdot {10000 e^{-t}\over 1 + 1000e^{-t}}\cr
 &= {100000k e^{-t} \over   (1 + 1000e^{-t})^2}\cr}$$
Comparing the two, we see $100000k = 10000$, so $k = .1$.
\bigskip
\item{C)}  (5)  If the epidemic proceeds according to the 
function $N(t)$ in part B, 
how many weeks will pass before the 
number of infected people reaches 1 million?
\sol
We solve 
$$1 =  10/(1 + 1000e^{-t}) \Rightarrow 1 + 1000e^{-t} = 10 \Rightarrow
e^{-t} = 9/1000$$
So $t = -\ln(9/1000) = 4.71053$ weeks.  
\bigskip
\noindent
{\it Extra Credit} (20) 
The hull of a boat is 20 feet long.  At a distance
$s$ feet from the bow (the front), the cross section of the part
of the hull below $y = 0$ (the waterline) has the shape of the region 
in the $xy$-plane below $y = 0$ and above 
the parabola $y = ax^2 - b$, where $a,b$ are given in the following table:
$$\matrix{s & 5 & 10 & 15 & 20 \cr
          a & 2 & 3 & 4 & 5 \cr
          b & 2 & 3 & 4 & 4 \cr}$$
Estimate the {\it volume\/} enclosed by the hull below the water line.
\sol
Think of the usual approach for computing volumes  by slicing:
$V = \int A(s)\ ds$.
Since we only have the information at 5-foot intervals, we will 
compute $A(s)$ for $s = 5, 10, 15$ and use a left-hand sum approximation
for the integral of $A(s)$.   $A(5)$ is the area between $y = 2x^2 - 2$
and the $x$-axis:
$$A(5) = \int_{-1}^1 2x^2 - 2\ dx = -8/3$$
Then $A(10)$ is the area between $y = 3x^2 - 3$
and the $x$-axis:
$$A(10) = \int_{-1}^1 3x^2 - 3\ dx = -4$$
And $A(15)$ is the area between $y = 4x^2 - 4$
and the $x$-axis:
$$A(15) = \int_{-1}^1 4x^2 - 4\ dx = -16/3$$
The left-hand sum approximation for the volume of the hull below
the waterline is
$$V \doteq (8/3)\cdot 5 + (4)\cdot 5 + (16/3)\cdot 5 = 60$$
(cubic feet).  (Note: we got rid of the signs in the answers for
$A(5)$, etc. because the negatives just tell us the area is
below the $x$-axis.)
\bye