Math 152 diary, spring 2007 

In reverse order: the most recent material is first. 
Wednesday, February 14  (Lecture #9) 

Last time: an apology
Several students told me that I talked about too much last time, and
things seemed much too complicated. This is unfortunate, but here is
an outline of what I hoped students would get from Monday's lecture:
Exam preparation
A graph, an area, and an integral: their behavior for large x
Consider the function y=1/x^{2} for x>2. It "starts" at
(2,1/4), and is decreasing, concave up, and has limit 0 as
x>infinity. A part of the graph is shown to the right. Let's
consider the area under this curve from 2 to A, where A is a large
positive number. This area can be computed with a definite
integral:
_{2}^{a}1/x^{2} dx=1/x]_{2}^{A}=1/A+1/2 (the plus
comes from two minus signs!)
The integrand is x^{2} so its antiderivative is [1/(1)]x^{1}=1/x.
Now as A>infinity, the area>1/2 (since 1/A>0). The following
language and notation are used:
The improper integral _{2}^{infinity}1/x^{2} dx
converges and its value is 1/2. (So the limit of the "finite"
definite integrals exists, and that limit is a finite number.)
Another graph, an area, and an integral: their behavior for
large x
Consider the function y=1/sqrt(x) for x>2. It "starts" at
(2,1/sqrt(2)), and is decreasing, concave up, and has limit 0 as
x>infinity. A part of the graph is shown to the right. Let's
consider the area under this curve from 2 to A, where A is a large
positive number. This area can be computed with a definite
integral:
_{2}^{a}1/sqrt(x) dx=2sqrt(x)]_{2}^{A}=2sqrt(A)2sqrt(2) (the plus
comes from two minus signs!)
The integrand is x^{1/2} so its antiderivative is [1/(1/2)]x^{1/2}=2sqrt(x).
Now as A>infinity, the area>infinity, since sqrt(A) will also get
very large. The following language and notation are used:
The improper integral _{2}^{infinity}1/sqrt(x) dx
diverges. (Here the word "diverges" is used to cover various situations: the integral could grow without any bound, which is what happens here, or the integral could somehow oscillate as an integral of sine would. So "diverges" means that the integral does not converge.)
Visual intuition?
I love pictures, and I really like to work to develop my
"intuition". Intuition is, in most cases, just mounds of experience
with examples whose results are appropriately understood. I don't
think the pictures of the two graphs convey any useful intuition. At
least, I can't see, staring at the first graph, that the total area,
all the way "out" to infinity, is finite while the area under the
curve in the second graph has infinite area. The contrast between
convergence and divergence seems to be more sensitive than a fast
glance might detect. Here we have got to compute in detail to
understand what happens.
Comparison facts
Look at, say, 1/[x^{2}+98x^{67}+83]. The values of
this function when x>2 are certainly less than that of
1/x^{2}. So I bet that the following is true: if
0<g(x)<f(x) for x>a, and if _{a}^{infinity}f(x)dx is a convergent improper
integral, then _{a}^{infinity}g(x)dx is also a convergent
improper integral. That's because a smaller chunk of a region which
has finite area should also have finite area. And, actually the
smaller integral should be less than the larger integral, so, here, I
know that _{2}^{infinity}1/[x^{2}+98x^{67}+83]dx
converges and its value is less than 1/2.
A similar result is true the other way. If 0<f(x)<g(x) for x>a, and if _{a}^{infinity}f(x)dx diverges, then _{a}^{infinity}g(x)dx must also diverge.
No other valid comparisons are generally true. That is, if something is bigger than a convergent integral, you can't decide using that information alone if the bigger integral converges or diverges. Similarly, if something is smaller than a divergent integral, the nature (divergent/convergent) of the smaller integral can't be decided with only that information.
The exponential probability distribution
Many reallife phenomena are described using something called the
exponential probability distribution. For example, the probable lifetime of a lightbulb could be described with it. See here for more information.
Here I will just talk about lightbulbs. Here is a first attempt to
be precise. If the probability that a light bulb will fail in t
minutes is proportional to e^{Ct} then the lifespan of the
bulb is said to have an exponential probability distribution. More
specifically, if t=0 is NOW, and t_{1}<t_{2} are
later times, then then the probability of lightbulb failure between
the times t_{1} and t_{2} is proportional to _{t1}^{t2}e^{Ct}dt.
That is, that portion of the area to the right which is shaded
blue represents the chance of a lightbulb burning out during that
particular time interval.
What is the total probability from NOW to FOREVER?
Every (real!) lightbulb is going to fail some time. Probabilities of
an event range from 0 to 1, where an event which is certain
will have probability 1. Since any lightbulb will fail between t=0
(NOW) and t=infinity (FOREVER), we should consider the (improper)
integral _{0}^{infinity}e^{Ct}dt. Let me analyze this carefully.
Suppose A is a large positive number. Let's compute _{0}^{A}e^{Ct}dt=(1/C)(e^{Ct})]_{0}^{A}=(1/C)e^{CA}+(1/C)e^{0}=(1/C)(1/C)e^{CA}.
The antiderivative has a (1/C) factor because when e^{Ct} is
differentiated, the Chain Rule produces a multiplicative C, and the
(1/C) cancels this. Now what happens to (1/C)(1/C)e^{CA} as
A gets large, A>infinity? e^{CA} (with a negative sign,
with A and C positive) must go to 0 (this is exponential
decay). Therefore the improper integral _{0}^{infinity}e^{Ct}dt converges,
and its value is 1/C. But, wait: since every lightbulb fails,
shouldn't this be 1? Yes, surely. Let's fix this up. The key is
"proportional to". We should multiply the function e^{Ct} by
a constant so that the improper integral will turn out to be 1. The
computation we just did shows that the constant should be C. So the
probability distribution is actually Ce^{Ct}.
The {expectationmeanaverage} of an exponential probability
distribution
Let me try to discuss something a little bit harder, which I didn't do
well at in class. First, some background with (maybe) some easier
ideas. We could imagine a population of, say, bugs. Maybe there are
three types of bugs. One type, the yellow spotted bug, has a lifespan
of 20 days, and forms 30% of the population. Another type, the blue
striped bug, has a lifespan of 50 days and is 45% of the
population. Finally, the pink long bug has a lifespan of 80 days and
is the remaining 25% of the population. What is the average lifespan
of this bunch of bugs? Well, it isn't the average of the 3 lifespans
(the sum of 20, 50, and 80 divided by 3) since that doesn't take into
account the varying proportions of the bug types. If you think about
it, the average in this case is
20·(.3)+50·(.45)+80·(.25): a sort of weighted
sum. You should convince yourself that this is the correct number.
What about lightbulbs? What proportion of a lightbulb population will
"die" at time t? Well, that proportion is about
Ce^{Ct}dt. The appropriate weighted sum in this case
multiplied the proportion by t and adds it all up with an integral:
_{0}^{infinity}tCe^{Ct}dt. If we
compute this integral, we maybe can get some idea of when an average
lightbulb dies. This quantity is called the mean or the expectation.
We can compute tCe^{Ct}dt using integration by parts. If u=t then
dv=Ce^{Ct}dt and du=dt and v=e^{Ct}. Therefore
uvv du is
te^{Ct}e^{Ct}dt=te^{Ct}(1/C)e^{Ct}.
The definite integral from t=0 to t=A is
te^{Ct}(1/C)e^{Ct}]_{0}^{A}=Ae^{CA}(1/C)e^{CA}{(1/C)e^{0}}.
What happens at A>infinity? Well, e^{CA} certainly goes to
0 (radioactive decay!). But the term Ae^{CA} has a sort of
conflict. Although the exponential decays, certainly A>infinity?
Which factor "wins"? Exponential decay is faster than any degree of
polynomial growth, actually, so the limit is 0. You certainly can see
this with L'Hopital's rule:
lim_{t>infinity}te^{Ct}=lim_{t>infinity}t/[e^{Ct}]=lim_{t>inifinity}1/[Cexp(Ct)]=0.
So the integral from 0 to A, as A>infinity, approaches a limit,
which is {(1/C)e^{0}}=(1/C). This is the average lifespan of
a lightbulb. Incidentally, if we want to check that this is a valid
model, we can look at sample lifespans, and this can be used to
identify the value of the parameter C. The average lifespan written on
a package I just examines is 750 hours.
Other things ...
People who study statistics are interested in more details about life
and lightbulbs than are described here. For example, they may want to
know how dispersed the lifespans are around the mean. That is,
do all of the lightbulbs tend to die out right around 1/C, or is there
a considerable amount of variability? Various numbers measure this,
including variance and standard deviation. They all need computation
of _{0}^{infinity}t^{2}Ce^{Ct}dt, a
different improper integral.
Another kind of improper integral
The integrals we've looked at are called improper because their
domains are infinite. But there is another collection of integrals
which are also labeled improper because something goes wrong in their
ranges: the function to be integrated becomes infinite. Here are some
simple examples.
Moving electrons: inverse square law
Suppose Fred and Stanley are electrons. Fred sits at the origin of the
xaxis, x0, and Stanley is at x=2. It is known that electrons repel
each other with a force which is inversely proportional to the
distance between them. Therefore (this is a math class, and I'm giving
up on units, all units have numerical value 1) if Fred stays at 0 and
Stanley is at x, the force between them has magnitude 1/x^{2}.
How much work is needed to move Stanley from x=2 to x=1? Realize that
work is still force·distance, but here the force will vary. We
could chop up the interval [1,2] into little pieces of distance,
dx. In a tiny distance the force will be (approximately) constant.
Pushing Stanley that dx distance with magnitude 1/x^{2} gives
me a (piece of) work 1/x^{2} dx. The total work will be
the sum of these pieces of work, or _{1}^{2}1/x^{2} dx.
As was pointed out by several students,
there are some silly and probably wrong things here. The force
repels, and I should add work from 2 to 1. So really I should be
computing an integral from 2 to 1 of 1/x^{2} dx. I think
this is correct. Sigh. There are two minus signs which cancel out and
I get the integral above. For more authentic physical computations,
please see more authentic professors of physics.
I can compute this integral: _{1}^{2}1/x^{2} dx=1/x]_{1}^{2}=1/2(1/1)=1.
But suppose I move Stanley from x=2 to x=a, where a is a small
positive number, so Stanley will be very close to Fred. What's the
work done? We just need a small change in the computation:
_{a}^{2}1/x^{2} dx=1/x]_{a}^{2}=1/2(1/a)=1/a1/2.
Suppose I am willing to "expend" any (finite) amount of work. Can I
move Stanley to Fred's position? The answer is "No" because the limit
as a>0^{+} of 1/a1/2 is infinity. Therefore we say that the
improper integral _{0}^{2}1/x^{2} dx diverges. The
integral is improper because the function 1/x^{2} is certainly
unbounded on its domain. The integral diverges because we would
like to use a limit to define it, and there is no finite value for
this limit.
Moving "electrons": another inverse law
Suppose we are in another universe, and here electrons obey an inverse
cube root law for the force between them. So if they are separated by
a distance of x, the force has magnitude 1/x^{1/3}. In this
case it turns out that we can move Stanley to Fred. Here is the
computation, done fairly carefully.
_{0}^{2}1/x^{1/3} dx=lim_{a>0+}_{a}^{2}1/x^{1/3} dx=lim_{a>0+}(3/2)x^{2/3}]_{a}^{2}=lim_{a>0+}(3/2)2^{2/3}(3/2)a^{2/3}=(3/2)2^{2/3}.
The positive power on a (that is, the a^{2/3})
after antidifferentiation makes the limit of the term with a equal to 0.
Language
So there are convergent and divergent improper integrals of this type,
also. And there are similar comparison facts. And, darn it, the
phenomenon here is also more subtle than can be visualized with a
simple picture. A casual attempt to graph y=1/x^{2} and
y=1/x^{1/3} from (close to) x=0 until x=2 will show curves
that seem superficially much the same. As x gets closer to 0, these
curves go UP. It is not at all clear to me that one of them has
"finite area" and the other has "infinite area".
QotD
Does _{1}^{infinity}1/x^{3} dx
converge? If it does, what is its value?
Binary entropy function
I did not have time to discuss the improper integral _{0}^{1}x ln(x)dx. It does converge
(draw a graph [maybe with a machine!]). This function occurs in the
study of information transmission and the fact that the integral which
seems rather improper (hey: ln(x)>infinity as x>0^{+}!)
has finite area is a very useful fact.
Monday, February 12  (Lecture #8) 

Just a few more integrals, #2
x ln(2x+1) dx
Integrate by parts with u=ln(2x+1) and dv=x dx. Then
du=[1/(2x+1)]2 dx and v=(1/2)x^{2} so that
u dv=uvv du gives us
(1/2)x^{2}ln(2x+1)x^{2}/(2x+1)dx.
Now we need to divide x^{2} by 2x+1.
(1/2)x(1/4)  2x+1 ) x^{2} x^{2}+(1/2)x  (1/2)x (1/2)x(1/4)  (1/4)The result has quotient (1/2)x(1/4) with remainder (1/4). Therefore x^{2}/(2x+1)dx= (1/2)x(1/4)+[(1/4)/(2x+1)]dx which is (1/4)x^{2}(1/4)x+(1/8)ln(2x+1) (the extra 2 making the 8 comes from the 2x+1 inside the ln).
Just a few more integrals, #3
[1/{sqrt(x)+3}]dx
Here try sqrt(x)=t so x=t^{2} and dx=2t dt. Therefore
[1/{sqrt(x)+3}]dx=
[1/{t+3}]2t dt. Now 2t/(t+3) can be
divided, and the result is 2+(6)/(t+3). The antidervative of this is
2t6ln(t+3)=2sqrt(x)6ln(sqrt(x)+3).
These integrals, almost "at random", to me are maybe too small to type into a computer algebra program, even if I am lazy. I think Math 152 students should be able to do these computations by hand.
Numerical approximation of definite integrals
Many if not most definite integrals which arise in real computations
involve either functions defined by such complicated formulas that FTC
can't be used, or they involve functions which are only known through
data points (various measurements). In both of these cases, numerical
approximation of the definite integral's value is useful. Along with
this, however, is some knowledge of how accurate the approximation
is. (Or else, as I remarked in class, we might as well report "17" as
the approximate value of everything!)
What I would like students to get out of this material is what the
formulas are for various approximation schemes, and some idea of the
error size for these approximation methods.
The simplest idea
I want to estimate _{a}^{b}f(x) dx. I will break up [a,b]
into n equal pieces. The ends of the subintervals will be called
x_{0}=a, x_{1}, x_{2}, ..., x_{n1},
and x_{n}=b, with x_{i}x_{i}=(ba)/n. I will
approximate the area in each "chunk" by the height at the lefthand
endpoint of the subinterval, which is f(x_{i}) as i runs from
0 to n1. There will be n of these rectangles. I will call this the
Left Hand Rule (the word "Rule" is frequently used in numerical
approximations, and comes from an antique use of the word meaning "a
prescribed mathematical method for performing a calculation or solving
a problem."). So this is the sum:
(f(x_{0})+f(x_{1})+f(x_{3})+...+f(x_{n1}))({ba]/n).
The (ba)/n multiplies all the terms since the width of the
subrectangles is the same. There are n of these rectangles.
The error in one panel: weird integration by parts
I tried to show what the error term is like for one "panel" (a word
that's used to refer to a part of the picture). So I considered
_{S}^{T}f(x) dx
and tried to relate this sum to f(S)(TS). I used integration by parts, with one very strange part:
_{S}^{T}f(x) dx u dv = uv  v du u=f(x) du=f´(x) dx dv=dx v=xTNotice that xT is a legal (?) choice for x, since the derivative (with respect to x) of xT is 1. Now the uv term is f(x)(xT)]_{S}^{T} which is f(T)(TT)f(S)(ST) which is f(S)(TS). The weird choice of v is selected to make this come out. So we start with the true area from S to T, and then get the rectangular area, and the other term, _{S}^{T}f´(x)(xT)dx, is the error.
The error?
We can't compute the error exactly (otherwise we could do the original
problem exactly!) so what's needed is some efficient method
(easy to use!) to overestimate this error. I'll discard minus signs,
etc. Mostly people are interested in absolute values. And suppose I
know that f´(x) is overestimated by a number I'll call
M_{1}. Then the integral which is the error term will be
overestimated by
_{S}^{T}(Tx)dx. This is the area of a triangle whose
height is TS and whose width is TS, so its value is
(1/2)(TS)^{2}. The original error term, _{S}^{T}f´(x)(xT)dx is therefore less
than or equal to [M_{1}/2](TS)^{2}.
How can we use that in our original problem? There the difference,
TS, is always (ba)/n. But there are n "panels" so there might be n
contributions of error. Therefore the total error is at most
n·[M_{1}/2][(ba)/n]^{2}. If you square things,
then one power of n cancels top and bottom. The result is:
[M_{1}(ba)^{2}]/(2n). Here M_{1} is
any convenient overestimate of the size of the first derivative
on the interval [a,b]. What's important here is the n downstairs, on
the bottom of the error estimate. As n gets very large, this means
that the Error will >0. This is good, since we certainly want more
and more accurate approximations. I'll discuss some real computations
in a few more minutes.
The Trapezoid Rule
We can try to use other geometric objects to approximate the
integral. The simplest improvement might be trapezoids instead of
rectangles. So we divide the interval up again into n equal parts
again. The trapezoid rule approximates the j^{th} area by the
(average of the bases)·(height)=(1/2)[f(x_{j})+f(x_{j+1})]h.
A picture of a trapezoid and its error is shown to the right.
Each trapezoid contributes half of the height of the vertical sides,
but notice that each inside edge (except for those over x_{0}
and x_{n}) is a member of two trapezoids. Therefore the
insides are weighted (1/2)+(1/2)=1 and the two outside edges are just
weighted (1/2) each. This is the trapezoid rule approximation:
((1/2)f(x_{0})+f(x_{1})+f(x_{3})+...+f(x_{n1})+(1/2)f(x_{n}))({ba]/n).
The error can be analyzed just with what we know now, but the details
are complicationed. I did this
analysis in another course. The result is that the error is always
at most [M_{2}(ba)^{3}]/(12n^{2}). Here
M_{2} is any convenient overestimate of f´´(x)
(the second derivative) on [a,b]. The second derivative measures how
far the graph bends from being a straight line, so it is natural that
when it is large, the area under y=f(x) is far away from being the
area under a collection of straight line segments.
Simpson's rule
Simpson's rule approximates y=f(x) by parabolic arcs:
y=Ax^{2}+Bx+C. There are 3 "free" variables in the second
degree quadratic. Such a function can interpolate three points (the
side of the trapezoid is linear, and interpolates two points).
An example
I gave the points (0,3) and (1,5) and (2,1) and we tried to find
y=Ax^{2}+Bx+C which goes "through" all these points. If x=0,
y=C so since (0,3) is on the parabola, then C=3. The other two points
give enough information t0o deduce the values of A and B. If fact,
finding B and C were the QotD.
It is a remarkable fact, and certainly not obvious, that the area under a parabolic arc which passes through three equally (horizontally!) spaced points has a simple formula. For example, if the parabolic arc passes through (x,y_{0}) and (x+h,y_{1}) and (x,y_{2}), then the area under it (the shaded region) turns out to be (h/3)[1y_{}+4y_{1}+1y_{2}]
The h/3 and 1 4 1 pattern are "famous". If you want to see a proof of this remarkable fact, please look at p.523 of your textbook. Simpson's rule is gotten by interpolating triples of points furnished by the function y=f(x).
The picture to the right is my attempt to show such interpolation. Again we can ask
So what's the formula?
Here is what happens. The number of subdivisions must be even
to use Simpson's rule. The panels contribute to the definite integral
in pairs.
[(ba)/(3n)][1f(x_{0})+4f(x_{1})+1f(x_{2}]
(panels #1 and #2)+
[(ba)/(3n)][1f(x_{2})+4f(x_{3})+1f(x_{4}]
(panels #3 and #4)+
[(ba)/(3n)][1f(x_{4})+4f(x_{5})+1f(x_{6}]
(panels #5 and #6)+
[(ba)/(3n)][1f(x_{6})+4f(x_{7})+1f(x_{8}]
(panels #7 and #8)+
Again, there are some coincidences. Everything gets multiplied by
[(ba)/(3n)]. The boundary nodes only appear once. The interior nodes,
after you add everything up and notice coincidences, appear with
weights of 4 2 4 2 4 2 4. The 4's and
2's alternate. This is the formula for Simpson's rule used with n
subdivisions (n must be an even integer) on the interval [a,b] with
the function, f(x):
f(x_{1})+f(x_{n})+SUM_{1<j<n, j even}4f(x_{j})
+SUM_{1<j<n, j odd}2f(x_{j})
all multiplied by [(ba)/3n].
Of course on a computer which does binary arithmetic, multiplication
by 2 and 4 is fast and easy (just shifting).
The error estimate
Simpson's rule with n pieces on the interval [a,b] has the error bound
K(ba)^{5}/(180n^{4}. Here K is some overestimate of
the fourth derivative of f(x).
Some numbers
All of this is too darn abstract, I feel. I gave lots of numbers in
class, maybe too many. Here are those numbers again. I computed
approximations to two examples. One was _{1}^{2}[1/x]dx=ln(x)]_{1}^{2}=ln(2)=ln(1)=ln(2)0=ln(2),
which I "know". The other is _{0}^{1}[4/{1+x^{2}}]dx=4 arctan(x)]_{0}^{1}=4 arctan(1)4 arctan(0)=4(Pi/4)4(0)=Pi. Here
are the results, with the three different rules and with a varying
number of pieces in the partition.
_{1}^{2}(1/x) dx, approximately 0.6931471806  

n=  Left Hand Rule  Trapezoid Rule  Simpson's Rule 
10 100 1,000 10,000 100,000 
0.7187714032 0.6956534302 0.6933972431 0.6931721811 0.6931496806 
0.6937714035 0.6931534305 0.6931472430 0.6931471810 
0.6931473748 0.6931471808 0.6931471803 
_{0}^{1}(4/{1+x^{2}) dx, approximately 3.141592654 (Pi)  
n=  Left Hand Rule  Trapezoid Rule  Simpson's Rule 
10 100 1,000 10,000 100,000 
3.239925989 3.151575986 3.142592487 3.141692652 3.141602654 
3.139925989 3.141575988 3.142592487 3.141592652 
3.141592652 3.141592653 
Things to think about
Although in theory any one of these computational schemes can give you
any accuracy desired, in practice there are other
considerations. First, you can worry about the amount of computational
time. Function evaluations take time on real machines. The lowest row
(n=100,000) took about half a minute. Second, numbers are represented
on real machines using floating point techniques, and arithmetic using
floating point numbers passes along and increases errors, so reducing
the number of divisions, multiplications, etc. is definitely a good
idea. For both of these reasons, and some others, the table should
convince you that the Trapezoid Rule and, especially, Simpson's Rule,
are worthwhile. They require almost the same amount of computation,
just a little more bookkeeping. But the n^{2} and especially
the n^{4} "downstairs" in the error estimates make these Rules
worthwhile.
Please try some of the problems in the textbook. I got the numbers above using Maple. For example, here is how to create one of the entries in the table:
> with(Student[Calculus1]): > evalf(ApproximateInt(1/x,x=1..2,method=trapezoid,partition=1000)); 0.6931472431
Wednesday, February 7  (Lecture #7) 

Polynomials
Polynomials are probably the first functions anyone thinks about. So I suggest the following polynomial:
((3+(5+8x^{3})^{46})^{37}.
Certainly this is a polynomial. It has degree 5,106. I believe
that I can differentiate it with very little difficulty (two uses of
the Chain Rule). How about antidifferentiation? Well, goodness, we
know how to do that ... but we know how to do that easily when this
polynomial is presented nicely, as a sum of constants multiplied by
nonnegative integer powers of x. When I typed a request to integrate
something like this into Maple on either
my home or work computers, all I succeeded in doing was freezing the
processors because, I suspect, the program wanted to expand the
presentation and then antidifferentiate. Much storage and processing
was needed to do that.
Rational functions
A rational function f(x) is a quotient, P(x)/Q(x), where the top and
bottom are both polynomials. The aim is not to get a nice
representation of such a function. The bestknown representation
is called partial fractions. My aim is to describe the method
of changing "small" examples of rational functions into their partial
fraction representation. I'll also comment, as we go through the
steps, about the computational difficulty of doing this representation
in the "real world". It is true that every rational function has a
unique partial fraction representation, just as every polynomial has a
unique standard representation as a sum of multiples of powers of
x. That there is always a partial fraction representation and there is
exactly one such representation is a theorem. The proof takes a while
and is not part of this course. Describing the process and the type of
result to expect is enough.
The textbook presents the partial fraction representation as an integration strategy. It is certainly such a strategy, because once the rational function is written in its partial fraction representation, antidifferentiation is straightforward. So I will discuss antidifferentiating the pieces, also. But there are practical reasons one might want the rational functions decomposed. I mentioned in class that the repulsive force between two classical charged objects (electrons?) is rational (inverse square). You could imagine that there's a complicated force law which is rational and has singularities (where the bottom of the rational function is 0 and the top is not). I might want to write this as a sum of different forces each with singularities in only one location. That is always possible and is a consequence of the partial fraction representation.
Back to the example
We wrote [(5x7)/(x^{2}2x3)] as a sum of constants
multiplying (x3)^{1} and (x+1)^{1} and then got
specific values of the constants. As you'll see this is a typical
partial fractions example. But there's one preparatory step we might
have to do which is not included in this example. What is that?
Suppose I want to look at, say, [(x^{3}+4x1)/(x^{2}2x3)]. My preparation would consist of the following:
x^{1}+2  x^{2}2x3  x^{3}+0x^{2}+4x1 x^{3}2x^{2}3x  2x^{2}+7x+1 2x^{2}4x6  11x+7So x+2 is the quotient and 11x+7 is the remainder. We can write the original fraction as:
x^{3}+4x1 11x+7  = x+2 +  x^{2}2x3 x^{2}2x3The polynomial is something we know and understand, and it is correctly "packaged" in a standard way. From now on I'll only consider proper rational fractions where the degree of the top is less than the degree of the bottom.
Partial fractions, Step 0
If the input is P(x)/Q(x) and is not proper, then divide the top by
the bottom, and rewrite the result as Quotient+Remainder/Q(x). Here
Quotient will be a polynomial, and Remainder/Q(x) will be a proper
rational fraction. Pass the "Remainder/Q(x)" on to the other steps.
Computational effort of Step 0
This is straightforward, easy to program, and doesn't take much time
or space.
Partial fractions, Step 1, sort of
This step is: given P(x)/Q(x),
find the factors of Q(x). I'll need to say more about this, but look:
6 5 4 3 2 A := 30 x + 179 x  3859 x  6591 x + 43369 x + 23500 x  113652 > factor(A); (x  2) (5 x  11) (x  9) (2 x + 7) (3 x + 41) (x + 2)Actually this example is totally silly. I created A by multiplying the factors which are shown. In reality, factoring is a very difficult problem.
Most of the protocols which guarantee privacy and security in web transactions rely, ultimately, on the difficulty of factoring, even the difficulty of finding the prime factors of positive integers. The most wellknown algorithm (RSA, for RivestShamirAdelman) is based on the following problem: suppose you know that a positive integer is the product of two primes. What are the primes? Well, 6 is 2 times 3, so ... except that the numbers are hundreds of decimal digits long, and there is no known feasible way of finding the integer factors. In the case of polynomials, the problems are, if anything, even more difficult, as you will see. The types of factors can vary.
Partial fractions, Step 2, sort of
Write a symbolic sum based on the factorization of the bottom. If we wanted to rewrite (11x+7)/(x^{2}2x3), which is
(11x+7)/[(x3)(x+1)], the appropriate symbolic sum is
A B  +  x3 x+1So there is one term for each factor.
Partial fractions, Step 3, sort of
Find values of the constants in the symbolic sum. In this specific
simple (toy!) example, since
A B A(x+1)+B(x3)  +  =  x3 x+1 (x3)(x+1)is supposed to be equal to (11x+7)/(x^{2}2x3), we know that
Another way of finding A and B so that 11x+7=A(x+1)+B(x3) is to
"expand" the righthand side, getting Ax+A+Bx3B=(A+B)x+(A3B). Then
we can look at the coefficients of x and the constant term (the
coefficient of x^{0}) to getr
11=A+B
7=A3B
This is a system of two linear equations in two unknowns and
there are many ways to solve such things. (Hey: A=10 and B=1 are the
solutions of this system!)
Another example
The previous example ended with 11x+7 divided by
x^{2}2x3. The bottom (denominator?) here can be factored
easily as (x3)(x+1). Let me change the example by changing
the bottom. Let's look at [(11x+7)/{(x3)(x+1)^{2}]. Now
one of the factors has a higher power than 1. Some people say that 1
is a root of the bottom polynomial which has multiplicity 2. In this
case, the symbolic sum has to be modified. There is one term for each
power going up from 1 to the multiplicity. So here:
11x+7 A B C  =  +  +  (x3)(x+1)^{2} x3 x+1 (x+1)^{2}This is step 2, the symbolic sum. Step 3 requires that specific values be found for A and B and C. What I do here is good for small examples. So I would combine the fractions on the righthand side, using the least common denominator. You need to be careful about this. I have made mistakes lots of times.
A B C A(x+1)^{2}+B(x3)(x+1)+C(x3)  +  +  =  x3 x1 (x1)^{2} (x3)(x+1)^{2}Compare the top of the result here with the top of [(11x+7)/{(x3)(x+1)^{2}.
The pieces are:
5/2 5/2 1  +  +  x3 x+1 (x+1)^{2}Can we integrate the result? Yes. The first piece gives 5/2ln(x3) and the second piece gives 5/2ln(x+1). The last part is (x+1)^{2} which gives (1)(x+1)^{1}.
If (xroot)^{multiplicity} is a factor of the bottom, then in Step 2 there will be a bunch of parts, (various constants)/(xroot)^{integer}, with the integer going from 1 to multiplicity. For example, the Step 2 response to the following:
x^43x^2+5x7 A B C D E F  =  +  +  +  +  +  (x+5)^{2}(x7)^{4} x+5 (x+5)^{2} x7 (x7)^{2} (x7)^{3} (x7)^{4}And this would lead to a system of 6 linear equations in 6 unknowns. By the way, Step 3, solving the linear equations, turns out to be computationally quite straightforward. Here I don't mean doing things by hand, but on a machine. Very big sytems (tens of thousands of linear equations) are solved efficiently on computers frequently.
Yet another example
Well, there is one more wrinkle. Consider the rational function
11x+7  x^{3}+2x^{2}2x12We need to factor x^{3}+2x^{2}2x12. As a hint, I mentioned that if x=2, then 2^{3}+2·2^{2}2(2)+12=8+8412=0. You may remember from high school that if x=2 is a root of x^{3}+2x^{2}2x12=0, then x2 is a factor x^{3}+2x^{2}2x+12. So:
x^{2}+4x+6  x2  x^{3}+2x^{2}2x12 x^{3}2x^{2}  4x^{2}2x12 4x^{2}8x  6x12 6x12  0So the quotient is x^{2}+4x+6 and the remainder is 0.
In Math 152 we are supposed to deal only with real numbers. Therefore x^{2}+4x+6 can't be factored any further. This sort of polynomial is called an irreducible quadratic.
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Partial fractions, back to Step 2 and Step 3
Here is how to handle an irreducible quadratic in a slightly simpler
case. I'll show you the symbolic pieces, and then solve for them.
11x+7 A Bx+C  =  +  (x3)(x^{2}+1) x3 x^{2}+1So when there is an irreducible quadratic then an unknown linear term needs to be put on top. If we combine terms and look at the tops, the result is
Therefore
11x+7 4 4x1  =  +  (x3)(x^{2}+1) x3 x^{2}+1QotD
11x+7 4 4x 1  =  +  +  (x3)(x^{2}+1) x3 x^{2}+1 x^{2}+1As answer, I expected 4ln(x3) and arctan(x) for the first and third terms. The middle term can be done with u=x^{2}+1 so du=2x dx, with the result 2ln(u)=2ln(x^{2}+1). And +C of course.
Monday, February 5  (Lecture #6) 

The tools for today
The alert student will see that there is an extra "functional
equation". You will soon see why. Also I will exchange the previous
variable, x, for a new variable, , which your text prefers here.
Derivative formulas  Functional equations 

d
d d d  cos(2)=(cos())^{2}(sin())^{2} sin(2)=2sin()cos() (sec())^{2}=(tan())^{2}+1 (tan())^{2}=(sec())^{2}1 
Area of part of a circle
I continued with the example introduced at the end of the previous
lecture.
_{0}^{1/2}sqrt(1x^{2})dx.
If I want to use FTC to compute this, I should find an antiderivative
for sqrt(1x^{2}). Last time I tried to motivate the
following substitution: x=sin(). Then dx=cos()d and
sqrt(1x^{2})=sqrt(1sin()^{2})=sqrt(cos()^{2})=cos(). Therefore:
sqrt(1x^{2})dx=[cos()]^{2}d.
We use the double angle formula from the toolbox:
[cos()]^{2}d=(1/2)[1+cos(2)]d=(1/2)[+(1/2)sin(2)]+C
The extra 1/2 comes from the Chain Rule worked backwards on the
"inside" part of cos(2). Now I'd like to return to xland by
writing everything in terms of x instead of . Since
sin()=x, we know that arcsin(x)=. That allows us to
translate the first part of the answer back to x's. But what about
sin(2)? Here is where I need the new formula in the toolbox:
sin(2)=2sin()cos(). Each of these I already know in
terms of x's because x=sin() and cos() is above (look!): it
is sqrt(1x^{2}). And we have a formula:
sqrt(1x^{2})dx=(1/2)[arcsin(x)+x sqrt(1x^{2})]+C.
There is cancellation of one of the (1/2)'s with the 2 coming from
sine's double angle formula.
We are not yet done, since we asked for a definite integral.
_{0}^{1/2}sqrt(1x^{2})dx=(1/2){arcsin(x)+x sqrt(1x^{2})}]_{0}^{1/2}.
Now arcsin(0)=0 and the other term is also 0 at 0, so the lower limit
gives a contribution of 0. The upper limit has
(1/2){arcsin(1/2)+(1/2)sqrt(1(1/2)^{2})}. With some thought
we might recall that arcsin(1/2) is Pi/6, and also "compute" that
sqrt(1(1/2)^{2})=sqrt(3)/2. Therefore the definite integral
is
(1/2)(Pi/6)+(1/2)(1/2)(sqrt(3)/2).
Geometric solution
Actually you can "see" these numbers in the picture.
The triangle has a base whose length is 1/2. Since
the formula for the upper semicircle is y=sqrt(1x^{2}), the
height of the triangle is sqrt(3)/2. The area of the triangle must
then be (1/2)(1/2)(sqrt(3)/2).
The base acute angle of the triangle is Pi/3, so the
circular sector has inside angle Pi/6. The area of a circular sector
is (1/2)(angle {in radians!)radius^{2}. Since the radius here
is 1, the area of the circular sector is (1/2)(Pi/6).
Maple's version
First the indefinite integral, and then the definite integral.
> int(sqrt(1x^2),x); 2 1/2 x (1  x )  + 1/2 arcsin(x) 2 > int(sqrt(1x^2),x=0..1/2); 1/2 3 Pi  +  8 12
Area under another curve
I want to compute _{}0^{1}1/sqrt(1+x^{2})dx. The value of
this integral is the area under y=sqrt(1+x^{2})dx between 0
and 1. A picture of this area is shown to the right. The curve has
height 1 at x=0, and then decreases to height sqrt(1/2) at x=1. Since
sqrt(1/2) is about .7, I know that the definite integral, always
between (Max. value)·(Interval length) and
(Min. value)·(Interval length), is between 1 and
.7.
It is always nice to know some size estimate of things to be
computed. That gives us at least a rough way of checking on the
methods and the result.
How to do it
Look in the toolbox and see
(sec())^{2}=(tan())^{2}+1. If I look at
sqrt(x^{2}+1) then I think of trying x=tan(), so
dx=[sec()]^{2} and sec()=srqt(x^{2}+1).
Then
1/sqrt(1+x^{2})dx becomes
[1/sec()][sec()]^{2}d which is
sec() d. We officially know this integral
because of the ludicrous computation done
earlier. Its value is ln(sec()+tan())+C. We can get back to
xland using what we already know, so the antiderivative we need is
ln(sqrt(x^2+1)+x)+C.
The definite integral computation then becomes:
ln(sqrt(x^2+1)+x)]_{0}^{1}=ln(sqrt(2)+1)+ln(1+0)=ln(sqrt(2)+1)
since ln(1)=0. And ln(sqrt(2)+1) is about .88137, certainly between .7
and 1.
Maple's version
You may ask for an antiderivative. Look at the result, which is
slightly surprising:
> int(1/sqrt(x^2+1),x); arcsinh(x)What is this? The function called "arcsinh" is an inverse function to one of the hyperbolic functions. The hyperbolic functions are discussed in section 3.9 of the text. They are frequently just as interesting and relevant to describing and solving problems as the more commonly used trig functions. The theory is totally parallel. The trig functions are connected to the circle, x^{2}+y^{2}=1. The hyperbolic functions are connected to x^{2}+y^{2}=1 or y^{2}=1+x^{2}. There is a Maple instruction which "translates" the inverse hyperbolic function into things we are supposed to understand. Here it is:
> convert(arcsinh(x),ln); 2 1/2 ln(x + (x + 1) )The definite integral is recogizable, except for a use of ln(A)=ln(1/A) where A=sqrt(2)1. (You can check that 1/[sqrt(2)1] is the same as sqrt(2)+1!) The next instruction finds an approximate numerical value of the previous answer.
> int(1/sqrt(x^2+1),x=0..1); 1/2 ln(2  1) > evalf(%); 0.8813735879
A third definite integral
I wanted to evaluate _{0}^{1}[sqrt(x^{2}1)/x^{3}]dx. Because
I see sqrt(x^{2}1) and the toolbox contains the equation
(tan())^{2}=(sec())^{2}1 I will "guess" at
the substitution x=sec(), which gives
dx=sec()tan()d and
sqrt(x^{2}1)=tan(). With this know, I can rewrite the
integral.
[sqrt(x^{2}1)/x^{3}]dx becomes [tan()/sec()^{3}]sec()tan()d.
Then cancel everything you can. The result is [cos()]^{2}d. We
already considered this integral, The key observation was:
(cos(x))^{2}
can be replaced by (1/2)(1+cos(2x)): the degree is halved, but the
function gets more complicated.
so
[cos()]^{2}d=
(1/2)[1+cos(2)]d=(1/2)[+(1/2)sin(2)}+C=(1/2)[+sin()cos()]+C
using the double angle formula for sine, also in today's toolbox.
Back to xland
The translation back turns out to be interesting and more involved
than similar previous transactions. We know that sec()=x and we
want to know what sine and cosine of are in terms of x. One way
some people use is drawing a right triangle with one acute angle equal
to , and with sides selected so that sec() is x. Since (with
some effort) we know that secant is ADJACENT/HYPOTENUSE, and we can
think that x is x/1, well the triangle must be like what is
pictured. Then Pythagoras allows us to get the oppositite side, as the
square root of the difference of the squares of the hypotenuse and the
adjacent side. From the triangle we can read off
sin()=sqrt(x^{2}1)/x and cos()=1/x. Then
(1/2)[+sin()cos()]+C becomes
(1/2)[arcsec(x)+sqrt(x^{2}1)/x^{2}]+C. As I explained
in class, arcsec is a fairly loathsome function (yes, this is a
value judgement about a morally neutral function). In fact,
when I asked my silicon friend, the reply was the following:
> int(sqrt(x^21)/x^3,x); 2 3/2 2 1/2 (x  1) (x  1) 1     1/2 arctan() 2 2 2 1/2 2 x (x  1)That is, Maple does not want to deal with arcsec at all. Yes, if you really wish, there is a convert instruction to get the arcsec version. With some effort, you can look at the triangle and see what arcsec is using arctan.
Are we stupid?
We're really not done with the problem. I asked for a definite
integral. Here is what happened to that inquiry on a machine:
> int(sqrt(x^21)/x^3,x=0..1); infinity II wanted to do this specifically to show you the result. The machine will do (try to do?) what you ask. This is a sort of silly answer, but the question is, indeed, sort of ridiculous. The integrand is [sqrt(x^{2}1)/x^{3}]. On the interval [0,1], the bottom has x^{3}. I bet that as x>0^{+}, something weird may happen: the integrand gets very, very large. That explains the infinity in the result. Later in the course we will see how to assign certain integrals finite values even when bad things happen to the integrand, but that procedure won't apply in this case. Where does the I come from? Well, look at x's inside [0,1]. Those x's make x^{2}1 negative and the integrand is requesting the square root of a negative number. Although we are not supposed to discuss complex numbers in this course, the machine believes we want a complex number computation, and it emits (?) an I to indicate this. So a silly question gets an appropriately silly answer.
A final integral with a square root
The last antiderivative of this type I looked at was something like:
sqrt(x^{2}+6x+7)dx.
Here
the novelty is the 4x term. Some algebra which you have likely seen
before can be used to change this to a form we can handle.
Getting rid of the x term
We are "motivated" by the expansion:
(x+A)^{2}=x^{2}+2Ax+A^{2}. We will
complete the square. So:
x^{2}+6x+7=x^{2}+2(3x)+7=x^{2}+2(3x)+3^{2}3^{2}+7=(x+3)^{2}9+7=(x+3)^{2}2.
Now make the substitution u=x+3 with du=dx. Then
sqrt(x^{2}+6x+7)dx
becomes
sqrt(u^{2}2)dt. We can
sort of handle this with a trig substition but there is 2 instead of
1.
Start with (tan())^{2}=(sec())^{2}1 and multiply by 2 to get 2(tan())^{2}=2(sec())^{2}2. I "guess" that I would like to try t=sqrt(2)sec(). Then t^{2}=2(sec())^{2}, so t^{2}2 is 2(tan())^{2}. Also dt=sqrt(2)sec()tan() d. The integral in land is sqrt[2(tan())^{2}]sqrt(2)sec()tan() d which is 2sec()[tan()]^{2}d. As I said in class, I got bored here. The previous methods can handle this. The details are complicated and offer lots of opportunity for error. Look below for a final answer.
Etc.
> int(sqrt(x^2+6*x+7),x); 2 1/2 (2 x + 6) (x + 6 x + 7) 2 1/2   ln(x + 3 + (x + 6 x + 7) ) 4
Really really simple algebra
From eighth grade we know:
3 7 3(x+5)+7(x+2) 10x+29  +  =  =  x+2 x+5 (x+2)(x+5) x^{2}+7x+10Therefore we can write: [10x+29/x^{2}+7x+10]dx=3ln(x+2)+7ln(x+5)+C. This antidifferentiation is maybe not obvious. So let me try something a bit more adventurous.
An antiderivative
Can we find an antiderivative of [5x7/x^{2}2x3]? We can try
to imitate what was just done ("reverse engineering"). Surely
x^{2}2x3=(x3)(x+1). And if we want to write:
5x7 A B A(x+1)+B(x3)  =  +  =  (x3)(x+1) x3 x+1 (x3)(x+1)then 5x7 should be the same as A(x+1)+B(x3). Well, if 5x7=A(x+1)+B(x3), then, for example, if x=1, I know that 5(1)7=A(0)+B(13) or 12=4B and B=3.
QotD
What is the value of A (my suggestion: use another magic number for
x)? After A is found, write down (very little computation should be
required!) a formula for [5x7/x^{2}2x3]dx.
Wednesday, January 31  (Lecture #5) 

The integral of ln(x) dx is x*ln(x)  x + C First, let u = ln(x), dv = dx, du = 1/x dx, and v = x. Then by integrating by parts, the integral of ln(x) dx = x*ln(x)  the integral of x*(1/x) dx. The integral of x*(1/x) dx equals x, so the answer is x*ln(x)  x + C. Bo Hye S. Lee Section 05So the integral of ln x is x ln xx+C.
I thank Bo Hye S. Lee and Michael Yang and Alex C. Scheller and Joe Corry and Glenn Davis and Kenneth Kong and Erica Choi and Abisola Oluwo and Kyle K. Vu for their wonderful work. They all responded with the correct answer. All will get ... whatever.
The clinic this Sunday will be held from 2 to 5 PM (in the afternoon). I strongly urge you to attend and do some problems. The "facilitator" has great experience helping students and is good at it (probably better than I am, darn it!). The course material is rapidly getting denser (!) and practice is really useful.
I mentioned that, according to the local rules, students could get a quiz tomorrow (in their recitations). I then had a remarkable twitch in my right eye (which, interestingly enough, was taken apart last year [really] in a remarkable operation).
Today I will slip behind the official syllabus, because I just think we need more time. I will do one more integration by parts, an amazing result, and then start what's in 7.2. The official syllabus has me also covering 7.3, but I will merely hint at what's in 7.3.
General topic: integrals of powers of trig functions
The easier one: sines and cosines
I want to describe how to find (sin(x))^{A}(cos(x))^{B}dx if A and B are
nonnegative integers. So I will concentrate of how to do this by
hand. You need to know this for exams in Math 152. If I were
describing a method to be implemented by a program, I'd say other
things (more below). y intention is to concentrate on moderately
efficient methods for relatively small A and B.
Sine & Cosine Tools 
d
 1=(cos(x))^{2}+(sin(x))^{2} 

d
 cos(2x)=(cos(x))^{2}(sin(x))^{2} 
Odds
If either A or B is odd things are easy. I will do an example with A=4 and B=5:
(sin(x))^{4}(cos(x))^{5}dx.
I don't remember if this is exactly what I did in class. Here "borrow"
one of the cosine's to live with the dx: cos(x) dx. This suggests
to me that I can use u=sin(x) so du=cos(x)dx. Now (sin(x))^{4}
becomes u^{4}. But we also have (cos(x))^{4} to change
into u's. Remember (the tools above) that
(cos(x))^{2}=1(sin(x))^{2}. We need to be careful
about exponents, but
(cos(x))^{4}=((cos(x))^{2})^{2}=(1(sin(x))^{2})=(1u^{2})^{2}
Therefore with u=sin(x), the integral changes
(sin(x))^{4}(cos(x))^{5}dx=u^{4}(1u^{2})^{2}du.
We have changed an integral involving powers of sine and cosine into
an integral which is just a polynomial. We can expand the square,
multiply, integrate, and then change back to x's:
u^{4}(1u^{2})^{2}du=u^{4}(12u^{2}+u^{4})du=u^{4}2u^{6}+u^{8}du=(1/5)u^{5}(2/7)u^{7}+(1/9)u^{9}+C=(1/5)(sin(x))^{5}(2/7)(sin(x))^{7}+(1/9)(sin(x))^{9}+C
This is the easy case!
Evens
In one lecture I did a sane example, and in the other lecture, the one
I did was rather poorly chosen (too darn elaborate!). But if I had to
do one of these integrals "by hand" with both
A and B even I would try the strategy shown in the example below. I'll do A=4
and B=0 here (they are both nonnegative even integers):
(sin(x))^{4}dx.
I can't just borrow one of the sine's to create a u as we
did before. The result will not give me a simple polynomial to
integrate  I'll have some mess with square roots in it, and things
will just be complicated. Here is what I would do. This is a trick
which sort of halves the power, but at the expense of making other
things more complicated.
Let's consider (sin(x))^{4}. It is, of course, ((sin(x))^{2})^{2}. The trick is to use the other column of the tools in an appropriate way. Well, we can subtract the equations. Take a look:
1 = (cos(x))^{2}+(sin(x))^{2}  cos(2x) = (cos(x))^{2}(sin(x))^{2} 1cos(2x) = 2(sin(x))^{2}Therefore (sin(x))^{2} becomes (1/2)(1cos(2x)) and (sin(x))^{4} becomes {(1/2)(1cos(2x))}^{2} and this can be multiplied out to get (1/4){12cos(2x)+(cos(2x))^{2}}.
Now what? We still have (cos(2x))^{2}. We need a version of the same trick, but dealing with a cosine squared term. Again consider the second column of the tools. We can add the equations. The (sin(x))^{2} will cancel, and there will be two (cos(x))^{2}. That is:
1 = (cos(x))^{2}+(sin(x))^{2} + cos(2x) = (cos(x))^{2}(sin(x))^{2} 1+cos(2x) = 2(cos(x))^{2}Therefore (cos(x))^{2} can be replaced by (1/2)(1+cos(2x)): the degree is halved, but the function gets more complicated. But we need to deal with (cos(2x))^{2}. So the change the equation (cos(x))^{2}=(1/2)(1+cos(2x)) by substituting 2x for x everywhere. Then we have (cos(2x))^{2}=(1/2)(1+cos(4x)).
(1/4){12cos(2x)+(cos(2x))^{2}} becomes
(1/4){12cos(2x)+(1/2)(1+cos(4x)}=(1/4)(1/2)cos(2x)+(1/8)+(1/8)cos(4x).
Remember that the goal of all this is to get something that we can
antidifferentiate, and, finally, we can with do that here. The
antiderivative is almost easy, if we remember that the chain rule
makes dividing by the multiplier of the x necessary. So the antiderivative is
(1/4)x(1/4)sin(2x)+(1/8)x+(1/32)sin(4x)+C.
Maple results compared to those just obtained
So I asked Maple to find the
antiderivatives. Below are the responses, together with what I got.
Function  Maple's antiderivative  What I got 

(sin(x))^{4}(cos(x))^{5}  (1/9)(sin(x))^{3}(cos(x))^{6}(1/21)sin(x)(cos(x))^{6} +(1/105)(cos(x))^{4}sin(x)+(4/315)(cos(x))^{2}sin(x) +(8/315)sin(x)  (1/5)(sin(x))^{5}(2/7)(sin(x))^{7}+(1/9)(sin(x))^{9}+C 
(sin(x))  (1/4)(sin(x))^{3}cos(x)(3/8)cos(x)sin(x)+(3/8)x  (1/4)x(1/4)sin(2x)+(1/8)x+(1/32)sin(4x)+C. 
It turns out that these actually are exactly the same functions. You can graph them and the results "overlay" one another. The algebraic differences can be irritating.
Warning!
Sometimes the functions you get can actually be different. Let me show
you a very artificial but simple example. How can we find an
antiderivative of x(x+1)? Well, a sane human being would multiply and
get x^{2}+x and then integrate to get
(1/3)x^{3}+(1/2)x^{2} (+C). A crazy person could do the following: integrate by parts. So:
x (x+1)dx=(1/2)x(x+1)^{2}(1/2)(x+1)^{2}dx=(1/2)x(x+1)^{2}(1/6)(x+1)^{3}. udv = uv  vdu u=x du=dx dv=(x+1)dx v=(1/2)(x+1)^{2}Therefore (1/3)x^{3}+(1/2)x^{2} and (1/2)x(x+1)^{2}(1/6)(x+1)^{3} must both be antiderivatives of x(x+1). Please notice that the first answer has value 0 when x=0 and the second answer has value 1/6 when x=0, so these are distinct functions! Is there a problem? Actually, no. Here theory (if you remember MVT) says you can have infinitely many (not just 2!) distinct antiderivatives of one function, provided they differ by a constant. And, indeed, the algebra says:
The two graphs are parallel to each other. Of course this example is maybe quite silly, but if you use computer algebra systems, you've always got to remember that "different" answers can both be valid! A computer program can't always be relied on to give sensible (?) answers.
Reduction formula
Here is probably what Maple uses on powers of sine, say.
Since
(sin(x))^{n}dx=
(sin(x))^{n1} cos(x)dx we can do this:
(sin(x))^{n1} cos(x)dx=(sin(x))^{n1}{cos(x)}(n1)(cos(x))(sin(x))^{n2}cos(x)dx udv = uv  vdu u=(sin(x))^{n1} du=(n1)(sin(x))^{n2}cos(x)dx dv=sin(x)dx v=cos(x)But the integral you get "out" is =(sin(x))^{n1}{sin(x)}(n1)(sin(x))^{n2}(cos(x))^{2}dx) and since (cos(x))^{2} is the same as 1(sin(x))^{2}, you can "solve" for the original integral. This does work. I think the result is:
Now secants and tangents
I want to describe how to find (sec(x))^{A}(tan(x))^{B}dx if A and B are
nonnegative integers. So I will concentrate of how to do this by
hand. You need to know this for exams in Math 152. Again, if I were
describing a method to be implemented by a program, I'd say other
things. Here I will definitely only look at
small A and B because things get very messy rapidly.
What's needed for these functions
Secant & Tangent Tools 
d

(sec(x))^{2}=(tan(x))^{2}+1 

d

(tan(x))^{2}=(sec(x))^{2}1 
Early examples
(sec(x))^{2}dx=tan(x)+C
(tan(x))^{2}=(sec(x))^{2}1 dx
=tan(x)x+C
I hope that you see even these "low degree" examples have a bit of novelty. Even worse are the first powers!
tan(x) dx=[sin(x)/cos(x)]dx=(1/u)du=ln(u)+C=ln(cos(x))+C=ln(sec(x))+C
where I rewrote tan(x) as sin(x) over cos(x), and then used the
substitution u=cos(x) with du=sin(x) dx and then got a log and
then pushed the minus sign inside the log by taking the reciprocal of
the log's contents! Horrible.
Much worse, much much worse, is the following:
This is not at all clear to me. This "computation" is probably the single most irritating (because of the lack of motivation) in the whole darn course. The history (I don't have a convenient reference) of this fact is that it was discovered as a result of the construction of certain numerical tables used for ocean navigation. This fact is sort of absurd.
Another example
(tan(x))^{4}dx=
(tan(x))^{2}(tan(x))^{2}dx=
(tan(x))^{2}{(sec(x))^{2}1}dx=
(tan(x))^{2}(sec(x))^{2}(tan(x))^{2}dx=
(1/3)(tan(x))^{3}(tan(x)x))+C.
The first "chunk" came from realizing that
(tan(x))^{2}(sec(x))^{2} is the derivative of tangent
multiplied by tangent squared, and so a substitution u=tan(x) changes
this to u^{2} and the result is (1/3)u^{3} (and then
back to x). The second part comes from the previous integral of
(sec(x))^{2}.
And another
NO, no ... I'm getting tired. You try some, and do the homework.
The ideas
All of the algebraic work is to get these trig powers to somehow
change to polynomials or other very simple functions. But:
Why look at these things?
Look at the upper half of the unit circle in the plane. This is the
graph of y=sqrt(1x^{2}). Suppose we want to find the area
"under" this curve from 0 to 1/2. This is not a horribly complicated
desire. Then we would need to compute _{0}^{1/2}sqrt(1x^{2}) dx. Nothing
in this course so far allows us to use FTC here. We need to find an
antiderivate of sqrt(1x^{2}). Well, the "natural" this is to
look for a substitution. Well, let's try to guess a substitution which
will interact well with the square root. If x=? then let's call
sqrt(1?^{2})=!. I don't like square roots, so let's square
this: 1?^{2}=!^{2}. I don't like minus signs, so let
me get rid of that one:
1=?^{2}+!^{2}
If we knew a nice pair of functions whose squares add up to 1, then
they would be natural candidates for ? and !. Indeed, we do know such
functions. I'll finish this example next time.
QotD
What is (sin(x))^{2}(cos(x))^{3}dx?
Monday, January 29  (Lecture #4) 

The product rule for derivatives states that if f(x) and g(x) are
differentiable functions, then the product function,
(f·g)(x)=f(x)g(x), is also differentiable, and its derivative
is given by the following:
(f·g)´(x)=f´(x)g(x)+f(x)g´(x).
Let's integrate this equation. But, wait, if we integrate
(f·g)´(x) we'll just get f(x)g(x), the product of the
functions (the integral of a derivative is the original function). So:
f(x)g(x)=f´(x)g(x)dx+f(x)g´(x)dx.
One version of integration by parts occurs if we put the second term
on the right on the other side of the equation, with a minus sign, of
course. So:
f´(x)g(x)dx=f(x)g(x)f(x)g´(x)dx.
What's above is the total theoretical content of this lecture (!) and of lots of other lectures. Clever choices of the functions turn out to be what's important. So let me immediately do
Example 0
Let's compute x sin(x)dx. I
don't happen to know an antiderivative, so maybe what I will do is try
to fit the "template" of the lefthand side of the equation above:
f´(x)g(x)dx. I will "choose"
f´(x) to be sin(x) and g(x) to be x. Then I know that
g´(x)=1 and f(x)=cos(x). The whole equation
f´(x)g(x)dx=f(x)g(x)f(x)g´(x)dx
becomes
x sin(x)dx=x[cos(x)][cos(x)]1 dx.
So we have "traded in" the hard (difficult?) integral, x sin(x)dx, for an easier integral, [cos(x)]1 dx, with a "penalty" of x[cos(x)]. We can compute the easier integral:
[cos(x)]1 dx=sin(x)+C
and now we package things together and get:
x sin(x)dx=x[cos(x)][sin(x)]+C.
There's several remarks to make. First, we can usually check
integration results quite easily, just by differentiating. So:
d/dx of x cos(x)+sin(x)+C is cos(x)+(x)[sin(x)]+cos(x)+0.
The derivative of the first term "gives birth" to the first two terms
of the result (using the product rule) and then the derivative of the
second term just happens (no, not at all!) to cancel one of the first
terms, and, of course, the +C differentiates to 0. An alert student
will surely see that we are just doing in reverse what was called
integration by parts, so the verification should not be too
surprising. I won't bother with further verifications because, as you
will see, they tend to be rather tedious.
One rather superficial but irritating aspect of this computation is that there are many minus signs. You will see, when using integration by parts, lots and lots and lots of minus signs. They will be common, and every human being will make errors handling them. Please don't make too many.
The notation I've shown you is quite clumsy, and almost everyone uses
a more compact way of writing things. Here is the integration by parts
formula, as everyone uses it:
u dv=uvv du
In order to use this, we will need to complete the information below:
u=______ du=_______
dv=_______ v=______
In what we just did, u=x and dv=cos(x), so du=dx and v=sin(x)dx.
Let me show some more examples.
Example 1
x e^{x}dx. Here:
u dv=uvv du
In order to use this, we will need to complete the information below:
u=x du=dx
dv=e^{x}dx v=e^{x}
Therefore, u dv=uvv du becomes
x e^{x}dx=x e^{x}e^{x}dx=x e^{x}e^{x}+C.
The second integral is easier. Next:
Example 2
x^{2} e^{x}dx. Here maybe there are more
choices for the formula u dv=uvv du but the idea is
to take a hard integral and ... well, make it easier: maybe not easy,
that's not always possible, but frequently we can make it
easier. Here's some choices:
u=x^{2} du=2x dx
dv=e^{x}dx v=e^{x}
So we get:
x^{2} e^{x}dx=x^{2} e^{x}2x e^{x}dx
We need to compute 2x e^{x}dx which is 2x e^{x}dx. My goodness! Example 1 told us that this integral is
x e^{x}e^{x}, so we can complete the integration:
x^{2} e^{x}dx=x^{2} e^{x}2x e^{x}dx=x^{2} e^{x}2[x e^{x}e^{x}]+C.
Example 3
x^{3} e^{x}dx. I bet we can
... well, this is where I began to get bored in class. We could take
u=x^{3} and dv=e^{x}dx and ... there will be a
transition to an easier integral. Or we could be more systematic, and
more symbolic. Here:
Example n
Suppose n is a positive integer, and we want to analyze x^{n} e^{x}dx
using integration by parts. Choose parts as follows:
u=x^{n} du=nx^{n1}dx
dv=e^{x}dx v=e^{x}
Then u dv=uvv du becomes
x^{n}e^{x}dx=x^{n}e^{x}nx^{n1}e^{x}dx
I pulled the n out of the second integral because there's no x in the
n, so (relative to dx) it is a constant. This sort of formula is
called a reduction formula. Here is how I might use it,
if I had to do the following computation by hand:
(7x^{5}3x^{2}+8)e^{x}dx=
I'll first break things up and take out constant multiplies:
I would hope that you will not have to do this sort of thing "by hand". But, as I mentioned in class, you can use it to check the "shape" of the answer. Here are some Maple questions and responses:
> int((7*x^53*x^2+8)*exp(x),x); 2 3 4 5 (838 + 846 x  423 x + 140 x  35 x + 7 x ) exp(x) > int(7*x^53*x^2+8*exp(x),x); 6 7 x 3   x + 8 exp(x) 6The first question and response is the correct one. And, hey, the answer does have the right shape: it is a polynomial multiplied by exp(x) (this is e^{x} in this version of Maple notation). But, golly, if I just forgot the parentheses, look at the difference in the answer. If I knew what to expect, maybe I could correct my own error more easily.
Example 17
Arctan is an interesting function. It is the inverse to the
tangent function on the interval between Pi/2 and Pi/2. The domain of
arctan is all numbers, and its range is the open interval between
Pi/2 and Pi/2. It is always increasing. A picture is shown to the
right. What's the area "under" y=arctan(x) between 0 and 1? This can
be computed if we know about _{0}^{1}arctan(x)dx, so I need an antiderivative of arctan(x). Since
this is the integration by parts lecture, probably (sigh!) we will use
that method. So:
u dv=uvv du and the first integral is arctan(x)dx.
There really aren't many choices of parts here. I bet we should choose u=arctan(x). Once this choice is made, all of the others are forced. So dv must be dx, and v=x. We can get du if we remember the derivative of arctan(x). So du=[1/(1+x^{2})]dx. I mention that after some experience with integration by parts, sometimes you can look ahead and see if the resulting integral is actually "easier". Well, here is the result:
arctan(x)dx=x arctan(x)x=[1/(1+x^{2})]dx.
What about x=[1/(1+x^{2})]dx? If you are familiar with substitutions, then this won't be difficult. Let me use the letter w, since u is already in the parts formula.
To compute x=[1/(1+x^{2})]dx, take w=1+x^{2}. Then dw=2x dx, and we only need x dx, so (1/2)dw=x dx. Then:
x=[1/(1+x^{2})]dx=(1/2)(1/w) dw=(1/2)ln(w)+C=(1/2)ln(1+x^{2})+C.
We need to put this back into the previous computation.
arctan(x)dx=x arctan(x)(1/2)ln(1+x^{2})+C. But we wanted a definite integral:
_{0}^{1}arctan(x)dx=x arctan(x)(1/2)ln(1+x^{2})]_{0}^{1}.
Now when x=0,
0 arctan(0)=0 and ln(1+0^{2})=ln(1)=0. So plugging in 0
gets us 0. Plug in 1: arctan(1)(1/2)ln(2). Now arctan(1) can be given
in terms of tranditional constants, and it is Pi/4. And actually,
(1/2)ln(2)=ln(sqrt(2)). So the area under arctan between 0 and 1 is
Pi/4 minus the natural log of the square root of 2. This is so
... silly that it almost makes the computation worthwhile. Pi/4 is
about .78539, and ln(sqrt(2)) is about .34657, so the area seems to be
about .43882. The picture to the left is supposed to show this chunk
of arctan inside the unit square. Is about 40% of the area under the
curve? You decide.
An interesting example ...
For my last example, I looked at
_{0}^{1}x sin(nx)dx where n is some
large integer. This integral arises in all sorts of applications. If
you imitate what we did in the very first example we tried today (but
are careful with the n!) the antidervative can be computed. So:
If u=x, then dv=sin(nx)dx. So du=dx, and v=(1/n)cos(nx). The (1/n)
occurs because when you differentiate the cos(nx), you get sin
(that's where the minus sign in v comes from) but you also get an n
(chain rule) from the nx on the inside. So to get rid of that we
multiply by 1/n. Therefore:
x sin(nx)dx=x[cos(nx)][(1/n)]cos(nx)dx=(1/n)x cos(nx)+[1/n]cos(nx)dx=(1/n)x cos(nx)+[1/n][1/n]sin(nx)+C.
The second 1/n comes from the antiderivative of cos(nx), for the same reason as the first 1/n appeared. If you desperately wish, we can check this with a silicon friend:
> int(x*sin(n*x),x); sin(n x)  x cos(n x) n  2 nand this is the same as what we got. I do think it is really necessary, though, to do a large number of the calculations by hand, because you need to get some feeling for the result.
I wanted a definite integral, though. So:
_{0}^{1}x sin(nx)dx=(1/n)x cos(nx)+[1/n^{2}]sin(nx)]_{0}^{1}. Again, if you plug in
x=0 both terms are 0 (hey: this is not always true  look at the
previous QotD!). Plug in x=1, and the result is
cos(n)/n+[sin(n)/n^{2}]. It turns out that people will rarely
be interested in the specfic values of this integral, but rather in
the asymptotic behavior as n gets large. Here are some pictures:
x sin(10x) on [0,1]  x sin(20x) on [0,1]  x sin(40x) on [0,1]  x sin(80x) on [0,1] 

You should see that as n increases, the wiggling gets more and more rapid, and the amplitude (height, both + and ) is always trapped between x and x (since sine is between 1 and +1). Sometimes x is called the envelope of these curves (it is what the curves are sort of "packaged" inside, after all). The more and more rapid wigglings will tend to cancel each other out in the definite integral (area below the xaxis is negative, remember!). So the geometry tends to suggest that the net area over [0,1] goes to 0 as n goes to infinity. The algebraic form of the answer we got, cos(n)/n+[sin(n)/n^{2}], repeats this. The top in both cases (sine and cosine of n) is something between 1 and +1, while the bottom in both cases, goes to infinity. The result will certainly approach 0. In the context of vibrations, where many such integrals arise, this means there is less and less energy at high frequencies in a certain kind of impulse. You'll see more of this later.
There's an interesting problem of regarding the display of these curves. I can imagine what x sin(1,000x) looks like on [0,1], but I don't know of any display (handheld or whatever) which will show the graph accurately. If you are curious, you should try both on a graphing calculator and on Maple trying to graph some of these curves, and compare the result to what the curves "really" are. There's sort of a conflict between the pixels and sampling space and the actual values.
QotD
Compute x ln(x)dx. (Again,
this is not a random silly function. I mentioned in 151 that the
integrand function occurs in studying "entropy", the amount of
information certain kinds of communication channels can carry ("binary
symmetric channels": aren't words great!).
This integral is not too hard if you select the correct parts.
Wednesday, January 24  (Lecture #3) 

Work
which is physics which is something I know little about. I have been
told that Work=Force·Distance. Also I have been told that units
matter, and the most generally used units for Distance, Force, and
Work are meters, newtons, and joules. I will use as my units feet,
pounds, and footpounds. Sigh. So lifting 10 pounds for 5 feet does 50
footpounds of work. Huh. The only thing wrong is that the
abbreviation for pound is lb.
Pulling a chain up a cliff
I have an iron chain which weighs 3 pounds per foot and is 100 feet
long. It hangs from the edge of a tall cliff. How much work is
required to pull the chain to the top of the cliff?
Here's a picture of the chain. (Drawing the pictures is the fun part
for me.) We can imagine the chain's length being broken up into tiny
pieces, and then we would need to lift the pieces up the cliff. Let's
see: suppose we have a piece which is x feet from the top of the
cliff, and a tiny piece of length dx is imagined there. Then the
weight of that tiny piece is 3 dx (the 3 above is actually a sort
of linear density). To lift just that piece to the top of the cliff
needs x·3 dx amount of work. But the whole chain is made
up of these pieces, and so we need to add up this amount of work, and
(due to the way I set this up) we should take the integral from the
top (x=0) to the bottom (x=100) of the chain. This will get the total
work:
_{0}^{100}3x dx=(3/2)x^{2}]_{0}^{100}=(3/2)(100)^{2}.
Springs
Many real world springs obey Hooke's Law over "a portion of their
elastic range". This means that the distortion of the spring from its
equilibrium length is in a direction opposite to the impressed force
and has length directly proportional to the force. This is, more or
less, "F=kx", where F is the force and x is the distortion from
equilibrium and k is a constant, frequently called the spring
constant. So twice the weight on a spring will distort it twice as
much, but you probably should not assume the same for, say, ten
thousand times as much weight. Here's a typical Hooke's law/work
problem.
Pulling a spring
In equilibrium a spring is 2 feet long. A weight of 10 pounds
stretches it to 2.5 feet. How much work is needed to stretch the
spring from 3 feet long to 5 feet long?
Since 10 pounds changes the length of a spring by .5 feet, we know
that 10=k(.5) so that the spring constant, k, must be 20. Now consider
the various stages of the spring as it goes from the start position
(when the length is 3 and the distortion of 1) to the end position
(when the length is 5 and the distortion is 3). I'll call the
distortion, x. Perhaps we could consider an intermediate position. If
we pull the spring just a little bit more (change the length from x to
x+dx) we won't change the force very much. The force needed in that
intermediate position is 20x. The additional distance we're stretching
the spring is dx, so the "piece of work" we're doing is
20x dx. To get the total work we need to add up the pieces of
work from 3 feet long (when x=1) to 5 feet long (when x=3).
_{1}^{3}20x dx=10x^{2}]_{1}^{3}=10(3^{2})10(1^{2}).
Caution When I do these problems, an
easy mistake to make is to confuse the spring length with the
distortion from equilibrium. Hooke's law applies to the distortion, so
that is what must be considered.
Emptying a pool
A pool has a rectangular top, 10 feet by 30 feet. The pool is 1 foot
deep at one of the edges which is 10 feet long, and is 8 feet deep at
the other edge which is 10 feet long. The depth varies linearly
between these edges. The pool is filled with water but the top of the
water level is 1 foot below the top of the pool. How much work is
needed to pump out the water in the pool (that is, to the top of the
pool?).
An oblique view of the pool is shown to the right. I hope that this picture corresponds to what your view of the description in words above. We need to raise the water to the top of the pool. To do this, we need some information about the force needed (the weight of the water).
The density of water is about 62.5 pounds per cubic foot. One student cleverly asked if water near the bottom of the pool would have a higher density. Generally, of course, stuff near the bottom compresses, but it turns out that water as rather low compressibility and we can accept the 62.5 as valid for all of the water in the pool.
Although the lovely picture above pleases me (artistically!), a more
reasonable view might be sideways. I will put the origin of a
coordinate system for depth at the bottom of the pool (certainly there
may be one or two other reasonable places to put it). Then I will look
at a typical intermediate slice of the water volume at height x from
the bottom of the pool. The slice will have thickness dx. The reason
to look at this is that all of the water in that slice will need to be
lifted the same distance to the top of the pool. So this method of
organizing the computation allows me to put the distance into one part
of the problem, and then concentrate on the force (the weight of the
slice) in another part of the problem. But now we need to think about
the volume of the slice. It is dx thick, and I hope you can see that
the crosssection of the slice is a rectangle. It goes entirely across
the 10 foot width of the pool, and what varies in the slice is the
length, which I labeled L in the diagram. Similar triangles tell me
that L/x=30/7 so that L=(30/7)x. The volume of a slice is
(10)(30/7)x dx, so that the weight of a slice is
(62.5)(10)(30/7)x dx. This slice needs to be lifted to the top
of the pool (not just the top of the water!) and this distance is
1+(7x)=8x (I wrote it this way to emphasize that 7x is the distance
to the top of the water, and 1 more foot to get to the top of the
pool). So the amount of work needed is (8x)(62.5)(10)(30/7)x dx. To
get the total work I need to add up the work from the bottom of the
water (x=0) to the top of the water (x=7):
_{0}^{7}(8x)(62.5)(10)(30/7)x dx=(62.5)(10)(30/7)_{0}^{7}8xx^{2}dx=(62.5)(10)(30/7)
(4x^{2}x^{3}/3)]_{0}^{7}=(62.5)(10)(30/7)(4(7^{2})7^{3}/3).
Since I don't much care about the answer, all I said about it in class
was it would be useful to check if it were positive (otherwise
draining pools would solve the world's energy problems in a really
creative and "green" fashion). And, yeah, 4(7^{2})=196 and
7^{3}/3=343/3, so the answer is positive.
Some comments on solutions of the work problems
The methods of solution are reasonable and the selection of problems
that I showed you was carefully structured. In the first problem (the
chain), the force was constant (3dx) and the distance varied. In the
second problem, the distance was constant (dx!) and the force
varied. In the third problem, both the distance (8x) and the force
(the slice's weight) varied, and I organized the problem so that I
took advantage of the geometry.
Definition of the average value of a function
The average value of a function f(x) defined on an interval [a,b] is
(
_{a}^{b}f(x) dx)/(ba).
I'll discuss why this definition is reasonable but first a very simple
example.
Example
Let's compute the average value of 1/x^{2} on [1,4]. So: _{1}^{4}(1/x^{2})dx=1/x]_{1}^{4}=1/4(1/1)=3/4. (That's
because 1/x^{2} is x^{2} which has antiderivative
{1/(1)}x^{1}=1/x. The average value is 3/4 divided by 41=3, so the average value is 1/4.
To the right is a graph of y=1/x^{2} on [1,4] together with a
horizontal line, y=1/4, the average value. Does it "look" right? I
hope so.
I made an error ... In class I made an incorrect assertion about areas which I won't repeat here, but look at the picture: the area beneath the curve and above the line (the line and the curve meet at x=2) is _{1}^{2}(1/x^{2})(1/4)dx is 1/4 and that is certainly not half of the area under the curve. So a horizontal line at the average height does not split the area under the curve in half.
A continuous function on an interval and its average value
The average line does intersect the curve, though, always, if
the function is continuous. This is true because a continuous
function, f, on an interval [a,b] has a max, M, and a min, m, and its
integral must be between M(ba) and m(ba) so the average value
(divide by ba) must be between M and m. And the Intermediate Value
Theorem guarantees that y=f(x) must have this value at least once
somewhere in the interval.
Samples and sample means
If you model a physical process (or a computer algorithm) by f(x) for
certain values of x, say in [a,b], you might be interested in checking
the outputs or the running time or ... lots of things. So you might
sample the function f(x) on [a,b] a finite number of times:
x_{1}, x_{2}, ..., x_{n}. Then you'd get
results f(x_{1}), f(x_{2}), ..., f(x_{n}). You
might want to analyze these results statistically and hope that the
information you get would represent, somehow, information about "all"
of the values of f(x) on [a,b]. The average of these n outputs is
called the sample mean.
An example
Maple has a fairly good "random" number
generator. That is, the properties of the generator are known and have
been investigated systematically, and they satisfy many of the
statistical criteria that are desired for randomness. I asked for 30
random numbers between 1 and 4. The list began with 1.198304612 and
continued with 1.640439652 and then with 3.984266209 ... (27 more
numbers like this!). Then I computed the mean of the values of
f(x)=1/x^{2} for these 30 numbers. This sample mean was
0.2241511003. Remember, the average value for f(x) on [1,4] is 1/4, or
.25. The picture to the right shows a graph of 1/x^{2} on
[1,4], the 30 points created, and two lines. The top line is the
average value of 1/x^{2} on the interval (y=.25) and the lower
line is y=the sample mean. Please realize that the picture has been
distorted (the curve previously shown is correct) so that you can see
the two lines and the sample points more clearly.
Random samples and what can be hoped ...
If we take many, many samples on [a,b], we can hope (random? What does
"random" mean?) that these samples are distributed fairly evenly over
the interval. So then look at this:
The sample mean Multiply top and bottom by ba and rearrange algebraically f(x_{1})+f(x_{2})+...+f(x_{n}) f(x_{1})+f(x_{2})+...+f(x_{n}) (ba)  = · = n n (ba) f(x_{1})·[(ba)/n]+f(x_{2})·[(ba)/n]+...+f(x_{n})·[(ba)/n]  baIf you now look closely at the top, I hope that you can "see" a Riemann sum for f on [a,b]: well, at least you should see values of f at sample points, and the length of a subinterval of [a,b] when [a,b] is split up into n equal subintervals. We can hope that almost the sample points are in the correct subinterval.
Connection between sample means and average value
It turns out that, if the sample points are chosen uniformly at
random over the interval [a,b], then the sample means will almost
always > the average value of the function (as defined above, with
the definite integral). To actually verify this takes some effort
because you need to understand what random and uniform and ... please
learn some probability and statistics.
Folding back on itself: how to approximate an integral
The previous result is true, and it has been used in very cute
reversed fashion. That is, one can compute sample means and then use
the sample means to estimate the definite integral. That is, if you
wanted to know the value of _{a}^{b}f(x) dx numerically,
approximately, take a large number of samples of f(x) in the interval
[a,b] (uniformly, randomly) and take their average. Multiply the
result by the length of the interval, ba. The result is an
approximation to the value of the integral.
An example
Maple has a fairly efficient (fast and
satisfies some wellknown criteria for randomness) random number
"generator". I used it to try to compute the integral of x^{3}
over the interval [5,8]. The "true value" of this integral is
867.95. Here are Monte Carlo approximations for specific "flips" (?)
or choices of random points in the interval. What I asked Maple to do is the following: select some "random"
numbers between 5 and 8, say x_{1}, x_{2}, ...,
x_{n}. Then compute this sum:
x_{1}^{3}+x_{2}^{3}+...x_{n}^{3}.
Then divide this by n, and multiply by 3, the length of the
interval. The results of this experiment are below. Please realize
that if I ran this experiment another time I would likely get
different results (!).
# of points  Reported approximation 

10  1007.29154 
100  815.78669 
1,000  850.26269 
10,000  868.39938 
100,000  867.07575 
1,000,000  867.77362 
10,000,000  867.66869 
QotD I remarked that Maple had reported that the average value of x^{2}sqrt{4+5x^{3}} on [0,1] was 38/45. I asked students to give convincing evidence that this was true.
I expected the following: since 10=1, the value of the integral
of x^{2}sqrt{4+5x^{3}} on [0,1] should be 38/45. I
gave this hint: u=4+x^{3}. With that hint and the use
of substitution we see:
du=15x^{2}dx, so (1/15)du=x^{2}dx and x^{2}sqrt{4+5x^{3}}dx=(1/15)u^{1/2}du=(1/15)(1/[3/2])u^{3/2}+C=(2/45)(4+5x^{3})^{3/2}+C.
Therefore _{0}^{1}x^{2}sqrt{4+5x^{3}}dx=(2/45)(4+5x^{3})^{3/2}]_{0}^{1}=(2/45)9^{3/2}(2/45)4^{3/2}=(2·27)/45(2·8)/45=38/45.
Comments Some errors were made because students assumed that the antiderivative had to be 0 at the lower limit, where x=0. Of course (2/45)(4+5x^{3})^{3/2} is not 0 when x=0. A substantial number of students showed they were not able to complete an antiderivative using the substitution method, even after being told what the substitution was. On their papers I wrote, "Please review substitution as done in Math 151." This is serious advice and a real warning, since what we will do in this course (and what students need to know) is more complicated methods of finding antiderivatives. Those students should review by doing 5 to 10 such problems from last semester.
Monday, January 22  (Lecture #2) 

Clinics, student lists, workshops ...
The purpose of the clinics is to help
people do textbook problems and to study the course material. The
major purpose of the student lists is to
help people connect and form study groups, to help them work with the
problems in the course. The major purpose of the workshop writeups is to help students get
familiar with the demands of written technical exposition, on a small
scale.
Before ...
The volume of a general solid of revolution is discussed in the previous lecture, along with a verification of
the formula for the volume of a right circular cone.
Volume of a sphere
To get the volume of a sphere, let's choose a fairly simple profile
curve. A circle of radius R centered at the origin is
x^{2}+y^{2}=R^{2}. If we solve for y, we get
y=+/sqrt(R^{2}x^{2}). We only need the upper semicircle revolved around the xaxis in order to get a whole sphere of radius R, so we just need the + sign: f(x)=sqrt(R^{2}x^{2}). We will add the slices from the left, x=R, to the right, x=R, and this should be the volume of the sphere. The general formula, _{left}^{right}Pi f(x)^{2}dx becomes
Pi_{R}^{R}(sqrt(R^{2}x^{2}))^{2}dx
and (luckily?) the square root cancels with a square, and the result can
be evaluated by FTC:
Pi
_{R}^{R}R^{2}x^{2}dx=
Pi(R^{2}x(x^{3}/3))]_{R}^{R}=Pi(R^{3}(R^{3}/3)Pi(R^{2}(R)((R)^{3}/3)). This
is (after the minus signs are correctly considered) equal to
(4/3)Pi R^{3}, which is the textbook formula for the
volume of a sphere.
Volume of a parabolic chunk
Consider now the region in the plane bounded by y=x^{2}, x=1
and x=2, and the xaxis. This is sort of a parabolic chunk. Suppose we
take this region and revolve it around the line y=3. This is a line
parallel to the xaxis but three units below it. We get some sort of
weird solid. The volume of the solid can be computed with only some
slight adjustment of ideas. We slice the solid by planes perpendicular
to the axis, only now the slices will not have crosssectional areas
which are discs or coins, but a bit more complicated.
Take a thin vertical slice of the region, a slice which is dx thick. Then revolve that slice around the line y=3. We'll get (with only a little bit of lying at the edge) an object which looks like a washer. The "official" name for the crosssectional area, a region between two circle which have the same center, is an annulus. What is the area of such a region?
Suppose the radius of the inside, smaller circle is r_{in},
and the radius of the outside, larger circle is r_{out}. The
area surrounded by the big circle is
Pi (r_{out})^{2} and the area inside the smaller
circle is Pi (r_{in})^{2}. Therefore the area of
the annulus is the difference between these:
Pi (r_{out})^{2}Pi (r_{in})^{2}.
In the specific case we are considering here, the outer radius is the
vertical distance from y=3 to y=x^{2}. This is
x^{2}+3 or, as some people would like me to write,
x^{2}(3). So r_{out}=x^{2}3. And
r_{in} is the distance from x=0 to x=3, and this is always
3. We can get dV, a thin slice of the volume, by multiplying the area
of the annulus by the thickness, dx. The result is:
Pi (x^{2}3)^{2}Pi (3)^{2}dx.
Please note the contents of the following information.
An arithmetic advertisement  

(357)^{2}=127,449 (886)^{2}=784,996 (886)^{2}(357)^{2}=657,547  886357=529 (529)^{2}=279,841 
Notice that (886)^{2}(357)^{2} does not equal (886357)^{2}. Parentheses are important. 
We add up the slices from left (x=1) to right (x=2). The total volume
is V=_{1}^{2}Pi (x^{2}3)^{2}Pi (3>)^{2}dx.
We can pull out the constant factor of Pi. Also,
(x^{2}3)^{2}=(x^{2})^{2}2·x^{2}·3+3^{2}=x^{4}6x^{2}+9, so the 9's will cancel, and the volume we want is:
Pi_{1}^{2}x^{4}6x^{2}dx. FTC
handles this:Pi((x^{5}/5)(6x^{3}/3))]_{1}^{2} which is
Pi(2^{5}/5(6·2^{3}/3))Pi((1^{5}/5(6·1^{3}/3)). (I'm
not going to "simplify" this!)
Volume of a torus
A torus (doughnut!) is a region is space which is gotten by revolving
a circle and its inside around an external line. To the right is a
picture of a torus. There's a circle of radius r, which is revolved
around an axis (a vertical line in the picture) so that the center of
the circle is at distance R from the the axis. I'd like to compute the
volume of this region. This computation is certainly not quite as
simple or toylike as the previous examples. I would like to use the
methods we've developed to compute the volume of this torus.
We need to get an appropriate "profile". We can use essentially any curves we want to get the torus. Here are some choices which turn out to make the volume quite computable. Make the xaxis be the axis of symmetry. Create the surface of the torus by taking a circle of radius r centered at (0,R) on the yaxis. The equation of that circle is x^{2}+(yR)^{2}=r^{2}. If we take slices perpendicular to the xaxis which are dx thick, then the rotated slice is a washer. There is an inner radius, r_{in}, and an outer radius, r_{out}. The piece of volume, dV, is the crosssectional area multiplied by the thickness. The crosssectional area is the difference between the areas of the circles: Pi r_{out}^{2}Pi r_{in}^{2}.
We need to find the formulas for r_{out} and
r_{in}. These can be gotten from the equation
x^{2}+(yR)^{2}=r^{2}: they are the values of
y which correspond, respectively, to the upper and lower
semicircles. So let's get y:
x^{2}+(yR)^{2}=r^{2} becomes
(yR)^{2}=r^{2}x^{2} and
yR=+/sqrt{r^{2}x^{2}} and finally
y=R+/sqrt{r^{2}x^{2}}.
Here the + corresponds to r_{out} and the 
corresponds to r_{}. The crosssectional area,
Pi r_{out}^{2}Pi r_{in}^{2},
becomes:
Pi [R+sqrt{r^{2}x^{2}}]^{2}+Pi [Rsqrt{r^{2}x^{2}}]^{2}
When some algebra is done, there are R^{2} terms which cancel,
and also r^{2}x^{2} terms which cancel. What remains
is (being careful of the minus signs!)
4Pi R sqrt{r^{2}x^{2}}.
This formula is the crosssectional area, and we need to multiply it
by the thickness, dx, and add up these slices from left to right in
the diagram shown. But left to right here is r to +r. So the volume
of the torus is:
_{r}^{+r}4Pi R sqrt{r^{2}x^{2}}dx.
The 4Pi R is a constant and can be pulled out of the integral
(there's no x in it!). But what remains is _{r}^{+r}sqrt{r^{2}x^{2}}dx. But
now, honestly, we've got a problem. I would like to use FTC,
but (at this stage of our "technology") I don't know an antiderivative
of sqrt{r^{2}x^{2}}. So we seem to be stuck.
A trick helps ...
But if we somehow happen to recognize a definite integral, maybe we
can "recognize" its value. And _{r}^{+r}sqrt{r^{2}x^{2}}dx
represents the area "under" y=sqrt{r^{2}x^{2}} as x
goes from r to +r. But what's the curve
y=sqrt{r^{2}x^{2}}? It is part of
x^{2}+y^{2}=r^{2}. This is a circle of radius
r centered at the origin. And the definite integral exactly describes
the area of the upper half of that circle, so we do know its value,
half of Pi r^{2}.
Back to the volume of the torus:
It is supposed to be 4Pi R
multiplied by _{r}^{+r}sqrt{r^{2}x^{2}}dx,
so the volume of the torus is
(4Pi R)((1/2)Pi r^{2}, or
2Pi^{2}Rr^{2}.
As I remarked in class, this is sort of an appropriate formula for a
volume. It multiplies three length measurements, so dimensionally the
formula is correct for a volume. It does have a strange multiplicative
constant, the 2Pi^{2}. But even that can be explained (read
on!).
Theorem of Pappus (balancing squashing and
stretching)
Here's how you could think about it. The circular disc we are
revolving about the axis "clearly" has a center of gravity at the
center of the circle. This center of gravity is being moved itself
around a circle of radius R, so the center of gravity is moving a
total distance of 2Pi R (the circumference of a circle of radius
R). The area inside the circle is Pi r^{2}. Maybe, just
maybe, "on average", the whole volume that the circular disc sweeps
out is the are multiplied by the length that the center of gravity
moves. That's (2Pi R)(Pi r^{2}), so this
"explanation" shows why we get the constant 2Pi^{2}
multiplying the dimensionally appropriate Rr^{2}.
This idea of looking at the center of gravity, and hoping that the squashing closer to the axis of symmetry is balanced out by the stretching farther away from the axis of symmetry, is actually correct but is not at all clear to me. It is part of an idea called the Theorem of Pappus. Some references can be found in this Wikipedia article. A good explanation is better saved for a calculus course in more than 1 dimension.
The volume of a chunk around the other way
Suppose we take the region bounded by bounded by y=x^{2}, x=1
and x=2, and the xaxis, which we considered earlier, and revolve it
around the yaxis. We can analyze the volume of this region by taking
slices perpendicular to the axis of symmetry, the yaxis. Now the
slices will be dy thick. We should try to write everything in terms of
y. We will now get washers in the y direction, and the integral giving the volume will be this:
_{bottom}^{top}
Pi (r_{out})^{2}Pi (r_{in})^{2}dy.
Well, the bottom is y=0 and the top is y=4. r_{out} is
relatively easy: it will always be 2. But r_{in}, which must
be written as a function of y, is more difficult. Look at the
picture. If y is between 0 and 1, r_{in} is 1, and if y is
between 1 and 4, r_{in} is sqrt(y) (I got this by solving for
y in the equation y=x^{2} and selecting the positive solution
since we're looking at the curve in the first quadrant). So r_{in} is a piecewise function. We certainly could compute the integral, but I would have to split it into two pieces.
This seemes clumsy. Sometimes it could also be difficult or essentially impossible  solving a equation (finding the inverse of a function) can be difficult. So let us see another strategy.
Try to slice dx!
Suppose we take a dx slice. This is a very thin vertical strip whose
distance from the yaxis is x. We could try to see what happens when
this is revolved around the yaxis, and analyze the resulting
solid. This turns out to be possible, and then the results are added
up with a definite integral.
The magic scissors
Scissors seem to
have been invented about 3500 years ago. My attempt to draw scissors
in class was met by considerable derision ("contemptuous laughter") so
I won't attempt a picture here. But the idea of the thin shell method
(well, I think of it as a ribbon) is to take the dx slice which has
height H, and revolve it around the yaxis at distance R. Then cut the
ribbon with the magic scissors, and unroll it or flatten it out (as I
mentioned in class, yes, there is certainly some distortion, and I
will address why the distortion can be neglected later in this
course). If you flatten the ribbon, the result is just about a
rectangular solid. The dimensions of the solid are inherited from the
ribbon. The thickness is dx, the height is H, and the length is
2Pi R, the circumference that the slice is moved. So the volume,
dV, that we want is 2Pi RH dx. If you go back to the
original picture with the parabola, you can see that H is
f(x)=x^{2}. R is the distance to the yaxis, which is actually
x, sort of weird but true. Therefore we know that
dV=2Pi xf(x) dx.
Let's compute the total volume now. Well, V=_{left}^{right}dV, so:
V=_{x=1}^{x=2}2Pi x(x^{2})dx=2Pi_{x=1}^{x=2}x^{3}dx=2Pi(x^{4}/4)]_{1}^{2}=2Pi(2^{4}/4)2Pi(1^{4}/4).
6.3: #2
In the first lecture I had enough time to do this textbook
problem. Please take a look at it. Here f(x)=sin(x^{2}), and
the integrand for the volume of the solid involved
2Pi x sin(x^{2})dx. The "extra" x is just what's
needed (using the substitution u=x^{2}) to find an
antiderivative so that FTC can be used successfully.
The Question of the Day (QotD)
I draw the picture shown on the board, and asked students to answer
the following questions:
1. Write an integral for the volume which results when the region
bounded by y=x(1x) and the xaxis is rotated around the xaxis.
I hoped that most students would answer this using the formula
V=_{left}^{right}Pi f(x)^{2}dx with
left=0 and right=1 and f(x)=x(1x).
Answer _{0}^{1}Pi (x(1x))^{2}dx.
2. Write an integral for the volume which results when the region
bounded by y=x(1x) and the xaxis is rotated around the yaxis.
Here the simplest answer would use the technique and formula just
discussed.
_{left}^{right}
2Pi RH dx. Now H is x(1x) and R is x, and left=0 and right=1.
Answer _{0}^{1}2Pi x(x(1x))dx.
Wednesday, January 17  (Lecture #1) 


The lecturer will support student work in the course in and out of the classroom. There will be weekly meetings of Math 152 clinics. Students are urged to use these and all opportunities to support their own work.
Background needed
The minimal background needed for this course is adequate knowledge of
the material of Math 151. In particular, students should know the
definitions and basic interpretations and uses of the derivative and
integral, and should be able to work with the standard functions,
knowing, where possible, their domains, ranges, graphs, derivatives,
and antiderivatives. We'll need the Fundamental Theorem of Calculus
(FTC) and the Mean Value Theorem (MVT).
What's this course about?
The previous course had some great ideas (MVT and FTC). This course
"exploits" these ideas but Warning! this
course definitely has more lengthy and intricate computations than
Math 151. An outline of the major topics of Math 152 includes the
following:
Volumes by slicing: a pyramid A pyramid has a square base which is 10 feet on each side. The peak (vertex?) of the pyramid is 30 feet above the center of the square. Compute the volume of the pyramid. The approach here is to slice up the volume into simpler pieces, approximate the pieces, add up the approximations, take the limit, and recognize the result as a definite integral. Finally, compute the definite integral, which in this "toy" case will be straightforward. Here is what I did with more detail. The slices will be taken perpendicular to the axis of symmetry, which in this case is determined by a line segment from the center of the base square to the vertex of the pyramid.
 
 
 
Here if we look at a side view of the pyramid, and label "everything"
I hope you will see some similar triangles. If we consider the common
ratios of height/base, then
s/[30h]=(10)/(30) so that s=(1/3)([30h].
As I mentioned, you can (very roughly!) check this by looking at
the "extreme" values of h. When h=0 (the bottom) then s=(1/3)(30)=10,
which is certainly true. When h=30 (the top) then s=(1/3)(0)=0,
again true.
Therefore the volume becomes:
V=_{bottom}^{top}s^{2}dh=_{h=0}^{h=30}{(1/3)(30h)}^{2}dh.
We can compute this using FTC. Probably the most elementary way
is to write
{(1/3)(30h)}^{2}=(1/9)(90060h+h^{2}). Then
antidifferentiate, and the resulting value is
(1/9){900h(60/2)h^{2}+(1/3)h^{3}}]_{0}^{30}, and the h=0
term gives us nothing, and the h=30 term gives
(1/9){900·30(60/2)(30)^{2}+(1/3)(30)^{3}}
Another way to compute
{(1/3)(30h)}^{2}dh is to use the substitution
u=(1/3)(30h) so that du=(1/3)dh and 3du=dh. Then
{(1/3)(30h)}^{2}dh=(1/3)u^{2}du=(1/3){u^{3}/3}+C=
(1/3){((1/3)(30h))^{3}/3}+C. Wow!
The value of the definite integral is then
(1/3){((1/3)(30h))^{3}/3}]_{h=0}^{h=30} and now the h=30
part gives 0 while the h=0 part is ((1/3){((1/3)(30))^{3}/3}).
The two numerical results are the same! This is not at all
obvious to me.
Volumes by slicing: squares over a semicircle
A semicircular region in the xyplane is defined by taking the inside
of a circle of radius 5 centered at the origin which is in the
halfplane with y>=0. A solid has that region as base, and the
crosssections of the solid which are perpendicular to the xaxis are
squares, with one side of the square on the xyplane.
Compute the volume of the solid.
Here is a picture of the base of the solid. The boundary of the base
is the "upper" half of a circle of radius 5 centered at (0,0) and a
line segment whose endpoints are (5,0) and (5,0). The heavy green line segment is a side of a square perpendicular to the xyplane and this square will be part of the solid.  
Here is an oblique (tilted) view of the base which also shows the square slice by a plane perpendicular to the base. As the line segment in the base moves from the left to the right, the square slice changes in size, from very small to large to very small.  
Now this is my attempt to "sketch" a picture of the whole solid in
space. I have tried also to show the edges of a few representative
square slices perpendicular to the xaxis.
As I mentioned in class, the volume of this solid turns out to be quite computable even though a sketch of it is, to me, difficult. I asked in class how many flat sides this object had and I was told that it had two flat sides and two curvy sides. Clear? Not very ... 
I would like to write everything in terms of x. Well, the bounds on the integral, left and right, are determined by the values of x, and these are 5 and 5 (the semicircle). What about s? The semicircle is part of a circle of radius 5 centered at (0,0), and the circle is describe algebraically by x^{2}+y^{2}=5^{2}. If you look at the first picture of the base, you should see that the square side, s, is actually the distance from the xaxis to the upper semicircle, and this distance is y with y>=0. So x^{2}+y^{2}=5^{2} becomes y^{2}=5^{2}x^{2} and y=+/sqrt(5^{2}x^{2}). We take the + sign because we need the upper semicircle, so s=sqrt(5^{2}x^{2}).
V=
_{left}^{right}s^{2}dx=
_{x=5}^{x=5}sqrt(5^{2}x^{2})^{2}dx.
The square root is canceled by the squaring, and we compute:
_{x=5}^{x=5}5^{2}x^{2} dx=
5^{2}x(1/3)x^{3}]_{x=5}^{x=5}=2(5^{3}(1/3)5^{3}).
Unless you bribe me or we need the result in some other form, I
won't "simplify" and you may do the same. Remember, every time you
touch some piece of arithmetic, there is some chance of breaking it
and getting the wrong result. So don't touch it if you don't need
to!
Volumes by slicing: triangles over a semicircle
A semicircular region in the xyplane is defined by taking the inside
of a circle of radius 5 centered at the origin which is in the
halfplane with y>=0. A solid has that region as base, and the
crosssections of the solid which are perpendicular to the yaxis are
isoceles right triangles with the hypotenuse of the triangle on the xyplane.
Compute the volume of the solid.
The base is the same region in the plane, with its boundary being
the "upper" half of a circle of radius 5 centered at (0,0) and a
line segment whose endpoints are (5,0) and (5,0). The heavy red line segment is a hypotenuse of an isoceles right triangle which is perpendicular to the xyplane, and this triangle will be part of the solid.  
Here is an oblique (tilted) view of the base which also shows the triangular slice by a plane perpendicular to the base. As the line segment in the base moves from the bottom to the top, the triangular slice changes in size, from very large (relatively!) to very small.  
Now this is my attempt to "sketch" a picture of the whole solid in
space. I have tried also to show the edges of a few representative
square slices perpendicular to the xaxis.
Here's my attempt to sketch this solid. I found this solid more difficult to sketch than the previous one. Again, the volume of this solid turns out to be quite computable. This imaginary (?) object has two flat sides and two curvy sides. Again, not very clear ... 
Now if you look back at the first picture of the semicircle in this
discussion, you should be convinced that the total volume, V, is given
by V=
_{bottom}^{top}s^{2}/4 dy.
I want to write everything in terms of y. The bottom in the integral
bound corresponds to y=0 and the top in the integral bound corresponds
to y=5. What about s? In a typical slice where y is between 0 and 5, s
goes from the left side to the right side. These "sides" are
determined by x. The equation which connects x and y is again
x^{2}+y^{2}=5^{2} so that
x=+/sqrt(5^{2}y^{2}). But here we actually need
"both" x's, since one gives the left and one gives the right. s is
x_{right}x_{left}. And
x_{right}=+sqrt(5^{2}y^{2}) and
x_{left}=sqrt(5^{2}y^{2}), so that
s=2sqrt(5^{2}y^{2}). Now back to computing V.
V= _{y=0}^{y=5}s^{2}/4 dy= _{0}^{5}{2sqrt(5^{2}y^{2})}^{2}/4 dy. Miraculously (not really) the square and square root cancel (and here so do "2^{2}" and "/4") so that the volume is _{0}^{5}5^{2}y^{2}) dy= 5^{2}y(1/3)y^{3}]_{0}^{5}=5^{3}(1/3)5^{3}.
Reality?
Questions could be asked about how real the shapes and solids
considered above are. I think they are very conceivable, and could
quite easily be "real". Consider the works shown of these two
architects, who are worldfamous.
A view of the
Guggenheim Bilbao Museum Architect: Frank Gehry 
A view of the BMW Central Building in Leipzig Architect: Zaha Hadid 

Revolving regions to create solid objects
Suppose y=f(x) is a positive (or at least nonnegative) function
defined on the interval [a,b]. Then the xaxis, the lines x=a and x=b,
and the part of the curve over [a,b] define a region in the plane. We
could then revolve that region around the xaxis, The resulting object
is called a solid of revolution. Such objects occur very
frequently in physical and geometric problems. Essentially any object
with an axis of symmetry comes from this solid of revolution
setup. The corresponding solid of revolution is shown far right  a
sort of vaseshaped object. Sometimes I will call y=f(x) the
profile curve of the solid.
Some simple examples of solids of revolution
Simple functions give some pictures of nice solids. A positive
constant function on an interval turns into a cylinder. A straight
line segment with one end point 0 makes a right circular cone. The
word "circular" here means that the cross sections are circles, and
the word "right" here means that the axis is perpendicular to the
cross sections. So, more precisely, the cylinder shown here is a right
circular cylinder.
And, of course, a semicircle which is the upper semicircle of a circle centered on the xaxis makes a sphere. 
Volume of a solid of revolution
Slice the solid by planes perpendicular to the axis of symmetry. In
our setup, this is the xaxis. Imagine that the slice is dx thick, and
the radius of the crosssectional area is r. Then dV=Pi r^{2}dx,
and V=
_{left}^{right}Pi r^{2}dx=_{a}^{b}Pi f(x)^{2}dx because the radius of the
circle is y=f(x), and left is x=a and right is x=b.
Volume of a cone
Lots of books tell me that the volume of a (right circular) cone is
Pi r^{2}h, where r is the radius of the base of the cone
and h is the height of the cone.
I think that the simplest profile curve which turns into the cone is a
line segment whose endpoints are the origin and (h,r). Therefore
f(x)=(Constant)x, and since f(h)=r, we know (Constant)h=r, so
Constant=r/h. Therefore the general formula for the volume of a solid
of revolution around the xaxis, with a profile curve f(x), which is
_{a}^{b}Pi f(x)^{2}dx, becomes in this case
V=
_{0}^{h}Pi (rx/h)^{2}dx=
_{0}^{h}Pi r^{2}x^{2}/h^{2} dx. FTC then gives Pi r^{2}x^{3}/(3h^{2})]_{0}^{h}=Pi r^{2}h^{3}/(3h^{2})=(1/3)Pi r^{2}h. The answer agrees with the formula in the books.
Warning
This note is directed both to students and to me. I will try
very diligently to cover approximately the same material in the two
lectures I give each Monday and each Wednesday. The record above shows
my failure on the first day! It shows what I did in the first
lecture. In the second lecture, I did not get to the material
on solids of revolution. I am sorry, and I will try harder in the
future.
Maintained by greenfie@math.rutgers.edu and last modified 2/17/2007.