Math 152 diary, spring 2007
In reverse order: the most recent material is first.

Wednesday, February 14 (Lecture #9)
In the snow, our love for mathematics ...
I am happy students were able to attend today's class. The weather wasn't really too horrible. It was just not so nice in the context of this remarkably mild winter.

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:

• Many quantities are computed with definite integrals. If the function involved is complicated, or if the information you have about the function consists of data points, then the definite integral must be approximated numerically.
• There are a variety of strategies to compute approximate values to definite integrals. I displayed a few of these, and there are many more. The strategies vary in their description (the formulas) and in the error estimates which accompany any good numerical technique.
• The strategies described all have descriptions and error estimates on the formula sheet for the first exam. What students should note, and what is important in practice, is how the powers of n differ in the error estimates for the different methods. Rather minor changes in assembling the approximate value can make huge differences (both absolute and relative) in the error estimates.

Exam preparation

A graph, an area, and an integral: their behavior for large x
Consider the function y=1/x2 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:
2a1/x2 dx=-1/x]2A=-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 2infinity1/x2 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:
2a1/sqrt(x) dx=2sqrt(x)]2A=2sqrt(A)-2sqrt(2) (the plus comes from two minus signs!)
The integrand is x-1/2 so its antiderivative is [1/(1/2)]x1/2=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 2infinity1/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/[x2+98x67+83]. The values of this function when x>2 are certainly less than that of 1/x2. So I bet that the following is true: if 0<g(x)<f(x) for x>a, and if ainfinityf(x)dx is a convergent improper integral, then ainfinityg(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 2infinity1/[x2+98x67+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 ainfinityf(x)dx diverges, then ainfinityg(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 real-life phenomena are described using something called the exponential probability distribution. For example, the probable life-time 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 t1<t2 are later times, then then the probability of lightbulb failure between the times t1 and t2 is proportional to t1t2e-Ctdt.
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 0infinitye-Ctdt. Let me analyze this carefully.
Suppose A is a large positive number. Let's compute 0Ae-Ctdt=-(1/C)(e-Ct)]0A=-(1/C)e-CA+(1/C)e0=(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 0infinitye-Ctdt 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 {expectation|mean|average} 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-Ctdt. The appropriate weighted sum in this case multiplied the proportion by t and adds it all up with an integral: 0infinitytCe-Ctdt. 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-Ctdt using integration by parts. If u=t then dv=Ce-Ctdt and du=dt and v=-e-Ct. Therefore uv-v du is -te-Ct--e-Ctdt=-te-Ct-(1/C)e-Ct. The definite integral from t=0 to t=A is -te-Ct-(1/C)e-Ct]0A=-Ae-CA-(1/C)e-CA-{-(1/C)e0}.
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:
limt-->infinityte-Ct=limt-->infinityt/[eCt]=limt-->inifinity1/[Cexp(Ct)]=0.
So the integral from 0 to A, as A-->infinity, approaches a limit, which is -{-(1/C)e0}=(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 0infinityt2Ce-Ctdt, 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 x-axis, x-0, 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/x2.

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/x2 gives me a (piece of) work 1/x2 dx. The total work will be the sum of these pieces of work, or 121/x2 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/x2 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: 121/x2 dx=-1/x]12=-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:
a21/x2 dx=-1/x]a2=-1/2-(-1/a)=1/a-1/2.
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/a-1/2 is infinity. Therefore we say that the improper integral 021/x2 dx diverges. The integral is improper because the function 1/x2 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/x1/3. In this case it turns out that we can move Stanley to Fred. Here is the computation, done fairly carefully.
021/x1/3 dx=lima-->0+a21/x1/3 dx=lima-->0+(3/2)x2/3]a2=lima-->0+(3/2)22/3-(3/2)a2/3=(3/2)22/3.
The positive power on a (that is, the a2/3) 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/x2 and y=1/x1/3 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 1infinity1/x3 dx converge? If it does, what is its value?

Binary entropy function
I did not have time to discuss the improper integral 01-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, #1
[1/(ex+1)]dx
If t=ex, then dt=exdx so dt/ex=dx and dt/t=dx. Therefore [1/(ex+1)]dx=[1/(t+1)](1/t)dt. Now:
[1/(t+1)](1/t)=A/(t+1)+B/t so [1/(t+1)](1/t)=[At+B(t+1)]/[(t+1)t] and 1=At+B(t+1).
Now t=0 gives B=1 and t=-1 gives 1=A(-1) so A=-1. Therefore:
[1/(t+1)](1/t)dt=(-1/t)+(1/{t+1})dt==ln(t)+ln(t+1)+C=-ln(ex)+ln(ex+1)+C.

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)x2 so that u dv=uv-v du gives us (1/2)x2ln(2x+1)-x2/(2x+1)dx.
Now we need to divide x2 by 2x+1.

```      (1/2)x-(1/4)
-----------------
2x+1 ) x2
x2+(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 -x2/(2x+1)dx= (1/2)x-(1/4)+[(1/4)/(2x+1)]dx which -is (1/4)x2-(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=t2 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 2t-6ln(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 abf(x) dx. I will break up [a,b] into n equal pieces. The ends of the subintervals will be called x0=a, x1, x2, ..., xn-1, and xn=b, with xi-xi=(b-a)/n. I will approximate the area in each "chunk" by the height at the left-hand endpoint of the subinterval, which is f(xi) as i runs from 0 to n-1. 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(x0)+f(x1)+f(x3)+...+f(xn-1))({b-a]/n).
The (b-a)/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 STf(x) dx and tried to relate this sum to f(S)(T-S). I used integration by parts, with one very strange part:

```    STf(x) dx
u dv =  uv  - v du
u=f(x)  du=f´(x) dx
dv=dx     v=x-T```
Notice that x-T is a legal (?) choice for x, since the derivative (with respect to x) of x-T is 1. Now the uv term is f(x)(x-T)]ST which is f(T)(T-T)-f(S)(S-T) which is f(S)(T-S). 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, -STf´(x)(x-T)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 M1. Then the integral which is the error term will be overestimated by ST(T-x)dx. This is the area of a triangle whose height is T-S and whose width is T-S, so its value is (1/2)(T-S)2. The original error term, |STf´(x)(x-T)dx| is therefore less than or equal to [M1/2](T-S)2. How can we use that in our original problem? There the difference, T-S, is always (b-a)/n. But there are n "panels" so there might be n contributions of error. Therefore the total error is at most n·[M1/2][(b-a)/n]2. If you square things, then one power of n cancels top and bottom. The result is: [M1(b-a)2]/(2n). Here M1 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 jth area by the (average of the bases)·(height)=(1/2)[f(xj)+f(xj+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 x0 and xn) 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(x0)+f(x1)+f(x3)+...+f(xn-1)+(1/2)f(xn))({b-a]/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 [M2(b-a)3]/(12n2). Here M2 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=Ax2+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=Ax2+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,y0) and (x+h,y1) and (x,y2), then the area under it (the shaded region) turns out to be (h/3)[1y+4y1+1y2]

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.
[(b-a)/(3n)][1f(x0)+4f(x1)+1f(x2] (panels #1 and #2)+
[(b-a)/(3n)][1f(x2)+4f(x3)+1f(x4] (panels #3 and #4)+
[(b-a)/(3n)][1f(x4)+4f(x5)+1f(x6] (panels #5 and #6)+
[(b-a)/(3n)][1f(x6)+4f(x7)+1f(x8] (panels #7 and #8)+
Again, there are some coincidences. Everything gets multiplied by [(b-a)/(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(x1)+f(xn)+SUM1<j<n, j even4f(xj) +SUM1<j<n, j odd2f(xj) all multiplied by [(b-a)/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(b-a)5/(180n4. 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 12[1/x]dx=ln(x)]12=ln(2)=ln(1)=ln(2)-0=ln(2), which I "know". The other is 01[4/{1+x2}]dx=4 arctan(x)]01=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.

12(1/x) dx, approximately 0.6931471806
n=Left Hand RuleTrapezoid RuleSimpson'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

01(4/{1+x2) dx, approximately 3.141592654 (Pi)
n=Left Hand RuleTrapezoid RuleSimpson'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

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 n2 and especially the n4 "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)
Nice representions of functions
Functions can be studied as the (possibly!) interesting objects that they are, but it is likely that various kinds of functions will be used by most students in the class to model physical situations, or to match data sets. Usually people try to use familiar functions first, and to use formulas for these functions. In the setting of calculus, we differentiate and integrate functions. Later, in courses following calculus which many students will take, other things will be done to functions (Laplace transforms, Fourier transforms, ...). It is important to get nice representations of functions, representations which make it easier to do things with the functions. This is maybe too darn abstract. Let's look at a particular example.

Polynomials
Polynomials are probably the first functions anyone thinks about. So I suggest the following polynomial:
((3+(5+8x3)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 non-negative 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 e-x-p-a-n-d 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 best-known 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 [(5x-7)/(x2-2x-3)] as a sum of constants multiplying (x-3)-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, [(x3+4x-1)/(x2-2x-3)]. My preparation would consist of the following:

```           x1+2
---------------
x2-2x-3 | x3+0x2+4x-1
x3-2x2-3x
--------------
2x2+7x+1
2x2-4x-6
----------
11x+7```
So x+2 is the quotient and 11x+7 is the remainder. We can write the original fraction as:
``` x3+4x-1              11x+7
---------- = x+2 +  --------
x2-2x-3             x2-2x-3```
The 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 well-known algorithm (RSA, for Rivest-Shamir-Adelman) 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)/(x2-2x-3), which is (11x+7)/[(x-3)(x+1)], the appropriate symbolic sum is

``` A     B
--- + ---
x-3   x+1```
So 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(x-3)
--- + --- = --------------
x-3   x+1     (x-3)(x+1)```
is supposed to be equal to (11x+7)/(x2-2x-3), we know that
11x+7=A(x+1)+B(x-3)
but then (x=-1) 11(-1)+7=B(-4) so -4=-4B and B=1. And (x=3) 11(3)+7=A(4) so A=40/10=10.

Another way of finding A and B so that 11x+7=A(x+1)+B(x-3) is to "expand" the right-hand side, getting Ax+A+Bx-3B=(A+B)x+(A-3B). Then we can look at the coefficients of x and the constant term (the coefficient of x0) to getr
11=A+B
7=A-3B
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 x2-2x-3. The bottom (denominator?) here can be factored easily as (x-3)(x+1). Let me change the example by changing the bottom. Let's look at [(11x+7)/{(x-3)(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
---------- = --- + --- + ------
(x-3)(x+1)2  x-3   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 right-hand 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(x-3)(x+1)+C(x-3)
--- + --- + ------ = ----------------------------
x-3   x-1   (x-1)2           (x-3)(x+1)2```
Compare the top of the result here with the top of [(11x+7)/{(x-3)(x+1)2.
11x+7=A(x+1)2+B(x-3)(x+1)+C(x-3)
I want any method of getting values for A and B and C which make this equation correct. "Theory" guarantees that there are such values, and that these values are unique. I can try magic numbers for x again.
x=-1 gives 11(-1)+7=A·0+B·0+C(-4) so that C=1.
x=3 gives 11(3)+7=A(4)2+B·0+C·0 so that A=40/16=5/2.
There doesn't seem to be a magic number which will produce B. Here are some strategies I use for such examples:
• One strategy
Take x=0. This isn't magic, but it simplifies things. Then the equation becomes
11·0+7=A(1)2+B(-3)(1)+C(-3). Since we know A and C we get 7=(5/2)-3B+1(-3) and 7-(5/2)+3=-3B and 15/2=-3B so B=-5/2.
• Another strategy
Look at the x2 coefficient on both sides of the equation 11x+7=A(x+1)2+B(x-3)(x+1)+C(x-3).
On the left there are no x2's, so the result is 0. On the right (be careful!) there are A x2's and also B x2's (there aren't any from the C term). Therefore 0=A+B and since A is 5/2, B is -5/2.
Thank goodness these are the same. No one was watching and checking me. I needed three tries to get everything to work out correctly!

The pieces are:

```5/2   -5/2     1
--- + ---- + -----
x-3    x+1   (x+1)2```
Can we integrate the result? Yes. The first piece gives 5/2ln(x-3) and the second piece gives -5/2ln(x+1). The last part is (x+1)-2 which gives (-1)(x+1)-1.

If (x-root)multiplicity is a factor of the bottom, then in Step 2 there will be a bunch of parts, (various constants)/(x-root)integer, with the integer going from 1 to multiplicity. For example, the Step 2 response to the following:

```x^4-3x^2+5x-7     A       B       C       D       E       F
------------- = ----- + ----- + ----- + ----- + ----- + -----
(x+5)2(x-7)4     x+5    (x+5)2   x-7    (x-7)2  (x-7)3  (x-7)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
-------------
x3+2x2-2x-12```
We need to factor x3+2x2-2x-12. As a hint, I mentioned that if x=2, then 23+2·22-2(2)+12=8+8-4-12=0. You may remember from high school that if x=2 is a root of x3+2x2-2x-12=0, then x-2 is a factor x3+2x2-2x+12. So:
```       x2+4x+6
---------------
x-2 | x3+2x2-2x-12
x3-2x2
------------
4x2-2x-12
4x2-8x
-------------
6x-12
6x-12
-----
0```
So the quotient is x2+4x+6 and the remainder is 0.
Therefore x3+2x2-2x-12=(x-2)(x2+4x+6). What are the roots of x2+4x+6? The quadratic formula indicates that the discriminant, here 42-4·1·6, is important. The discriminant is -10. The roots are complex!

In Math 152 we are supposed to deal only with real numbers. Therefore x2+4x+6 can't be factored any further. This sort of polynomial is called an irreducible quadratic.

Yes, COMPLEX NUMBERS are much
easier
to work with than real numbers!

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
----------- = ----- + ------
(x-3)(x2+1)    x-3     x2+1```
So 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
11x+7=A(x2+1)+(Bx+C)(x-3).
Again, any method that gets the answer is a good method. So:
x=3 gives 11·3+7=A(32+1)+(Bx+C)0 so 40=10A so A=4.
x=0 gives 7=A(1)+(B·0+C)(-3) so 7=A-3C so since A=4, 7=4-3C and C=-1.
Consider the x2 coefficients: 0=A+B, so since A=4, B=-1.

Therefore

```   11x+7        4     -4x-1
----------- = ----- + ------
(x-3)(x2+1)    x-3     x2+1```
QotD
Integrate this stuff. I remarked that if students found there were working hard on this, probably they were doing it wrong. As a hint, a small rewriting:
```   11x+7        4      -4x      -1
----------- = ----- + ------ + -----
(x-3)(x2+1)    x-3     x2+1     x2+1```
As answer, I expected 4ln(x-3) and -arctan(x) for the first and third terms. The middle term can be done with u=x2+1 so du=2x dx, with the result -2ln(u)=-2ln(x2+1). And +C of course.

Monday, February 5 (Lecture #6)
Finding eagles
As I mentioned, I had a happy Saturday hiking around a reservoir about 35 miles away. I saw a bald eagle, and was excited (a huge bird, with wing span is 6 to 7 feet). The area was beautiful, and did not look like a state with a population density >1,100 people per square mile. Here is more information about eagles in New Jersey.

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
--- sin() = cos()
d
d
--- cos() = -sin()
d
d
--- sec() = sec()tan()
d
d
--- tan() = (sec())2
d
1=(cos())2+(sin())2

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.
01/2sqrt(1-x2)dx.
If I want to use FTC to compute this, I should find an antiderivative for sqrt(1-x2). Last time I tried to motivate the following substitution: x=sin(). Then dx=cos()d and sqrt(1-x2)=sqrt(1-sin()2)=sqrt(cos()2)=cos(). Therefore:
sqrt(1-x2)dx=[cos()]2d.
We use the double angle formula from the toolbox:
[cos()]2d=(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 x-land 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(1-x2). And we have a formula:
sqrt(1-x2)dx=(1/2)[arcsin(x)+x sqrt(1-x2)]+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.
01/2sqrt(1-x2)dx=(1/2){arcsin(x)+x sqrt(1-x2)}]01/2.
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(1-x2), 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!)radius2. 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(1-x^2),x);
2 1/2
x (1 - x )
------------- + 1/2 arcsin(x)
2
> int(sqrt(1-x^2),x=0..1/2);
1/2
3       Pi
---- + ----
8      12```

Area under another curve
I want to compute 011/sqrt(1+x2)dx. The value of this integral is the area under y=sqrt(1+x2)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(x2+1) then I think of trying x=tan(), so dx=[sec()]2 and sec()=srqt(x2+1). Then 1/sqrt(1+x2)dx becomes [1/sec()][sec()]2d 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 x-land 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)]01=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, x2+y2=1. The hyperbolic functions are connected to -x2+y2=1 or y2=1+x2. 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 01[sqrt(x2-1)/x3]dx. Because I see sqrt(x2-1) 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(x2-1)=tan(). With this know, I can rewrite the integral.

[sqrt(x2-1)/x3]dx becomes [tan()/sec()3]sec()tan()d. Then cancel everything you can. The result is [cos()]2d. 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()]2d= (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 x-land
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(x2-1)/x and cos()=1/x. Then (1/2)[+sin()cos()]+C becomes (1/2)[arcsec(x)+sqrt(x2-1)/x2]+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^2-1)/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^2-1)/x^3,x=0..1);
infinity I```
I 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(x2-1)/x3]. On the interval [0,1], the bottom has x3. 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 x2-1 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(x2+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=x2+2Ax+A2. We will complete the square. So:
x2+6x+7=x2+2(3x)+7=x2+2(3x)+32-32+7=(x+3)2-9+7=(x+3)2-2.
Now make the substitution u=x+3 with du=dx. Then sqrt(x2+6x+7)dx becomes sqrt(u2-2)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 t2=2(sec())2, so t2-2 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()]2d. 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

``` 3     7    3(x+5)+7(x+2)     10x+29
--- + --- = ------------- = ----------
x+2   x+5    (x+2)(x+5)      x2+7x+10```
Therefore we can write: [10x+29/x2+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 [5x-7/x2-2x-3]? We can try to imitate what was just done ("reverse engineering"). Surely x2-2x-3=(x-3)(x+1). And if we want to write:

```   5x-7        A       B     A(x+1)+B(x-3)
---------- = ----- + ----- = -------------
(x-3)(x+1)    x-3     x+1     (x-3)(x+1)```
then 5x-7 should be the same as A(x+1)+B(x-3). Well, if 5x-7=A(x+1)+B(x-3), then, for example, if x=-1, I know that 5(-1)-7=A(0)+B(-1-3) 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 [5x-7/x2-2x-3]dx.

Wednesday, January 31 (Lecture #5)
The derivative of ln x is 1/x.
The integral of ln x is [I asked students to send me e-mail about this and offered .1 bonus point on the final exam for any correct answers.]
Here is one message I received:
```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 05```
So the integral of ln x is x ln x-x+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))Bdx if A and B are non-negative 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 & CosineTools d --- sin(x) = cos(x) dx 1=(cos(x))2+(sin(x))2 d --- cos(x) = -sin(x) dx 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))5dx.
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 u4. 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)=(1-u2)2
Therefore with u=sin(x), the integral changes
(sin(x))4(cos(x))5dx=u4(1-u2)2du.
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:
u4(1-u2)2du=u4(1-2u2+u4)du=u4-2u6+u8du=(1/5)u5-(2/7)u7+(1/9)u9+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 non-negative even integers):
(sin(x))4dx.
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
1-cos(2x) = 2(sin(x))2```
Therefore (sin(x))2 becomes (1/2)(1-cos(2x)) and (sin(x))4 becomes {(1/2)(1-cos(2x))}2 and this can be multiplied out to get (1/4){1-2cos(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){1-2cos(2x)+(cos(2x))2} becomes (1/4){1-2cos(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.

FunctionMaple's antiderivativeWhat 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))4sin(x)+(4/315)(cos(x))2sin(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))3cos(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 x2+x and then integrate to get (1/3)x3+(1/2)x2 (+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)2dx=(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)x3+(1/2)x2 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:
(1/2)x(x+1)2-(1/6)(x+1)3= (1/2)x(x2+2x+1)-(1/6)(x3+3x2+3x+1)= (1/2)x3+x2+(1/2)x-(1/6)x3-(1/2)x2-(1/2)x-(1/6)= (1/3)x3-(1/2)x2-(1/6)

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))ndx= (sin(x))n-1 cos(x)dx we can do this:

```(sin(x))n-1 cos(x)dx=(sin(x))n-1{-cos(x)}-(n-1)(-cos(x))(sin(x))n-2cos(x)dx
udv          =       uv          -        vdu
u=(sin(x))n-1         du=(n-1)(sin(x))n-2cos(x)dx
dv=sin(x)dx            v=-cos(x)
```
But the integral you get "out" is =(sin(x))n-1{-sin(x)}-(n-1)(sin(x))n-2(cos(x))2dx) 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:
(sin(x))ndx= (1/n)(sin(x))n-1{-cos(x)}+[(n-1)/n](sin(x))n-2dx
I wouldn't use this strategy by hand, but it is lovely for a program.

Now secants and tangents
I want to describe how to find (sec(x))A(tan(x))Bdx if A and B are non-negative 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.

 Secant &TangentTools d --- sec(x) = sec(x)tan(x) dx (sec(x))2=(tan(x))2+1 d --- tan(x) = (sec(x))2     dx (tan(x))2=(sec(x))2-1

Early examples
(sec(x))2dx=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:

1. Here is the beginning:sec(x) dx
2. Multiply the top and bottom by sec(x)+tan(x), which doesn't change the value since this is "just" an extremely weird way of writing 1: sec(x){[sec(x)+tan(x)]/[sec(x)+tan(x)]}dx
3. Rearrange algebraically a little bit: {[(sec(x))2+sec(x)tan(x)]/[sec(x)+tan(x)]}dx
4. Notice (hah!) that the derivative of sec(x)+tan(x) is sec(x)tan(x)+(sec(x))2 and that, therefore (hah hah!) if you take u=sec(x)+tan(x) then du=[sec(x)tan(x)+(sec(x))2]dx which is exactly the top, so the integral is (1/u)du which is ln(u)+C.
5. And now back to x's: the result is ln(sec(x)+tan(x))+C.
So, clearly ln(sec(x)+tan(x))+C is the integral of sec(x).

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))4dx= (tan(x))2(tan(x))2dx= (tan(x))2{(sec(x))2-1}dx= (tan(x))2(sec(x))2-(tan(x))2dx= (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 u2 and the result is (1/3)u3 (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(1-x2). 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 01/2sqrt(1-x2) dx. Nothing in this course so far allows us to use FTC here. We need to find an antiderivate of sqrt(1-x2). 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))3dx?

Monday, January 29 (Lecture #4)
Maintenance ...
• Workshops I graded two sections of the first workshop. The grades I gave ranged from 0 to 10 (more 10's than 0's, though). Students got 10's who gave good explanations of the computations and then provided evidence of the computations. Earning a 10 did not mean that a long essay was written. 10 points could certainly be earned with one page handed in. Students who provided no justification or explanation of their work were automatically limited to 5 points out of 10!
In this week's workshop (pumping out three tanks), I believe the conclusion of {most|least} work should be almost clear, and students will need to show their work and support their assertions with clear, simple sentences.
• QotD The last QotD was supposed to be a straightforward integration using substitution. I even gave a hint of the substitution. Students who had trouble with this are already (only 3 lectures in!) way behind. I think that what I discuss today will emphasize this.
We move today from what I think of as a conceptual part of the course (definite integrals can be used to think about ... many things [we'll return to this, though]) to a more computational part of the course: how to compute definite integrals. Certainly the most important methods are substitution, which reverses the chain rule, and integration by parts, which (sort of!) reverses the product rule. Today's lecture is entirely about integration by parts.

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 left-hand 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=uv-v 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 exdx. Here: u dv=uv-v du
In order to use this, we will need to complete the information below:
u=x   du=dx
dv=exdx   v=ex

Therefore, u dv=uv-v du becomes
x exdx=x ex-exdx=x ex-ex+C.
The second integral is easier. Next:

Example 2
x2 exdx. Here maybe there are more choices for the formula u dv=uv-v 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=x2   du=2x dx
dv=exdx   v=ex

So we get:
x2 exdx=x2 ex-2x exdx
We need to compute 2x exdx which is 2x exdx. My goodness! Example 1 told us that this integral is x ex-ex, so we can complete the integration:
x2 exdx=x2 ex-2x exdx=x2 ex-2[x ex-ex]+C.

Example 3
x3 exdx. I bet we can ... well, this is where I began to get bored in class. We could take u=x3 and dv=exdx 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 xn exdx using integration by parts. Choose parts as follows:
u=xn   du=nxn-1dx
dv=exdx   v=ex

Then u dv=uv-v du becomes
xnexdx=xnex-nxn-1exdx
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:
(7x5-3x2+8)exdx=
I'll first break things up and take out constant multiplies:

1. 7x5exdx=7(x5ex-5( x4ex-4x3ex-3(x2ex-2(xex-1(ex ) ) ) ) ))
2. -3x2dx=-3(x2ex-2(xex-1(ex) ))
3. +8exdx=8ex
And then I would add up everything and, oh yes, put on +C.

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^5-3*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^5-3*x^2+8*exp(x),x);
6
7 x     3
---- - x  + 8 exp(x)
6```
The 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 ex 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 01arctan(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=uv-v 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+x2)]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+x2)]dx.
What about x=[1/(1+x2)]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+x2)]dx, take w=1+x2. Then dw=2x dx, and we only need x dx, so (1/2)dw=x dx. Then:
x=[1/(1+x2)]dx=(1/2)(1/w) dw=(1/2)ln(w)+C=(1/2)ln(1+x2)+C.
We need to put this back into the previous computation.

arctan(x)dx=x arctan(x)-(1/2)ln(1+x2)+C. But we wanted a definite integral:
01arctan(x)dx=x arctan(x)-(1/2)ln(1+x2)]01.
Now when x=0, 0 arctan(0)=0 and ln(1+02)=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 01x 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
n```
and 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:
01x sin(nx)dx=-(1/n)x cos(nx)+[1/n2]sin(nx)]01. 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)/n2]. 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 x-axis 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)/n2], 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 (hand-held 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)
QotD and clinics
• I tried to address the purpose of the QotD (Question of the Day). There's no one purpose, but here are some candidates:
1. Attendance. So I can know who comes to the lecture and who doesn't, and perhaps, give some appropriate award to those who do.
2. Feedback to me. As I mentioned, the person talking to a group of people may sometimes not know how effective the intended communication is. If I give what I think as a straightforward problem related to the lecture, I may get some information about that effectiveness when I look at the results.
3. Feedback to you. Again, if the lecturer's supposedly straightforward problem repeatedly is misunderstood or badly done by a student, perhaps this signals the student to do more work. Feedback is important -- communication is important in any human relationship. Also, you can see how I grade, which might be useful in formal exams.
• The clinics have been augmented by another time and day (Tuesday). If you are serious about learning the material, these form an excellent opportunity to do homework with people who are professionally prepared and interesting in increasing your success. Sunday, when Ms. Panova, who has a great track record with the LRC, is working, should be especially appropriate for students (hey, it is hard to imagine a class conflicting with the Sunday clinic!).

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 foot-pounds. Sigh. So lifting 10 pounds for 5 feet does 50 foot-pounds 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:
01003x dx=(3/2)x2]0100=(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).
1320x dx=10x2]13=10(32)-10(12).
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 cross-section 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+(7-x)=8-x (I wrote it this way to emphasize that 7-x 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 (8-x)(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):
07(8-x)(62.5)(10)(30/7)x dx=(62.5)(10)(30/7)078x-x2dx=(62.5)(10)(30/7) (4x2-x3/3)]07=(62.5)(10)(30/7)(4(72)-73/3).
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(72)=196 and 73/3=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 (8-x) 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 ( abf(x) dx)/(b-a).
I'll discuss why this definition is reasonable but first a very simple example.

Example
Let's compute the average value of 1/x2 on [1,4]. So: 14(1/x2)dx=-1/x]14=-1/4-(-1/1)=3/4. (That's because 1/x2 is x-2 which has antiderivative {1/(-1)}x-1=-1/x. The average value is 3/4 divided by 4-1=3, so the average value is 1/4.
To the right is a graph of y=1/x2 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 12(1/x2)-(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(b-a) and m(b-a) so the average value (divide by b-a) 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: x1, x2, ..., xn. Then you'd get results f(x1), f(x2), ..., f(xn). 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/x2 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/x2 on [1,4], the 30 points created, and two lines. The top line is the average value of 1/x2 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 b-a
and rearrange algebraically
f(x1)+f(x2)+...+f(xn)     f(x1)+f(x2)+...+f(xn) (b-a)
--------------------  =  ---------------------·----- =
n                      n             (b-a)

f(x1)·[(b-a)/n]+f(x2)·[(b-a)/n]+...+f(xn)·[(b-a)/n]
--------------------------------------------------
b-a```
If 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 abf(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, b-a. The result is an approximation to the value of the integral.

An example
Maple has a fairly efficient (fast and satisfies some well-known criteria for randomness) random number "generator". I used it to try to compute the integral of x3 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 x1, x2, ..., xn. Then compute this sum: x13+x23+...xn3. 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 pointsReported approximation
101007.29154
100815.78669
1,000850.26269
10,000868.39938
100,000867.07575
1,000,000867.77362
10,000,000867.66869

Please notice that the results "seem" to be getting closer to the true value, but the approach is not very systematic. Actually the last answer is more distant than the preceding one. (A possible reason is accumulated inaccuracy because of the large numbers of floating point arithmetic operationcs which are needed.) I would use Monte Carlo methods only if other numerical or analytic approaches seem to be inadvisable. We will see later in the course many methods for computation of integrals. Further information about the Monte Carlo method.

QotD I remarked that Maple had reported that the average value of x2sqrt{4+5x3} on [0,1] was 38/45. I asked students to give convincing evidence that this was true.

I expected the following: since 1-0=1, the value of the integral of x2sqrt{4+5x3} on [0,1] should be 38/45. I gave this hint: u=4+x3. With that hint and the use of substitution we see:
du=15x2dx, so (1/15)du=x2dx and x2sqrt{4+5x3}dx=(1/15)u1/2du=(1/15)(1/[3/2])u3/2+C=(2/45)(4+5x3)3/2+C. Therefore 01x2sqrt{4+5x3}dx=(2/45)(4+5x3)3/2]01=(2/45)93/2-(2/45)43/2=(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+5x3)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 x2+y2=R2. If we solve for y, we get y=+/-sqrt(R2-x2). We only need the upper semicircle revolved around the x-axis in order to get a whole sphere of radius R, so we just need the + sign: f(x)=sqrt(R2-x2). 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, leftrightPi f(x)2dx becomes
Pi-RR(sqrt(R2-x2))2dx
and (luckily?) the square root cancels with a square, and the result can be evaluated by FTC:
Pi -RRR2-x2dx= Pi(R2x-(x3/3))]-RR=Pi(R3-(R3/3)-Pi(R2(-R)-((-R)3/3)). This is (after the minus signs are correctly considered) equal to (4/3)Pi R3, 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=x2, x=1 and x=2, and the x-axis. 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 x-axis 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 cross-sectional 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 cross-sectional 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 rin, and the radius of the outside, larger circle is rout. The area surrounded by the big circle is Pi (rout)2 and the area inside the smaller circle is Pi (rin)2. Therefore the area of the annulus is the difference between these:
Pi (rout)2-Pi (rin)2.

In the specific case we are considering here, the outer radius is the vertical distance from y=-3 to y=x2. This is x2+3 or, as some people would like me to write, x2-(-3). So rout=x2-3. And rin 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 (x2-3)2-Pi (3)2dx.
Please note the contents of the following information.

(357)2=127,449
(886)2=784,996
(886)2-(357)2=657,547
886-357=529
(529)2=279,841
Notice that
(886)2-(357)2 does not equal (886-357)2.
Parentheses are important.

We add up the slices from left (x=1) to right (x=2). The total volume is V=12Pi (x2-3)2-Pi (3>)2dx.
We can pull out the constant factor of Pi. Also, (x2-3)2=(x2)2-2·x2·3+32=x4-6x2+9, so the 9's will cancel, and the volume we want is: Pi12x4-6x2dx. FTC handles this:Pi((x5/5)-(6x3/3))]12 which is Pi(25/5-(6·23/3))-Pi((15/5-(6·13/3)). (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 x-axis be the axis of symmetry. Create the surface of the torus by taking a circle of radius r centered at (0,R) on the y-axis. The equation of that circle is x2+(y-R)2=r2. If we take slices perpendicular to the x-axis which are dx thick, then the rotated slice is a washer. There is an inner radius, rin, and an outer radius, rout. The piece of volume, dV, is the cross-sectional area multiplied by the thickness. The cross-sectional area is the difference between the areas of the circles: Pi rout2-Pi rin2.

We need to find the formulas for rout and rin. These can be gotten from the equation x2+(y-R)2=r2: they are the values of y which correspond, respectively, to the upper and lower semicircles. So let's get y:
x2+(y-R)2=r2 becomes
(y-R)2=r2-x2 and
y-R=+/-sqrt{r2-x2} and finally
y=R+/-sqrt{r2-x2}.

Here the + corresponds to rout and the - corresponds to r-. The cross-sectional area, Pi rout2-Pi rin2, becomes:
Pi [R+sqrt{r2-x2}]2+Pi [R-sqrt{r2-x2}]2
When some algebra is done, there are R2 terms which cancel, and also r2-x2 terms which cancel. What remains is (being careful of the minus signs!)
4Pi R sqrt{r2-x2}.

This formula is the cross-sectional 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+r4Pi R sqrt{r2-x2}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+rsqrt{r2-x2}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{r2-x2}. 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+rsqrt{r2-x2}dx represents the area "under" y=sqrt{r2-x2} as x goes from -r to +r. But what's the curve y=sqrt{r2-x2}? It is part of x2+y2=r2. 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 r2.

Back to the volume of the torus:
It is supposed to be 4Pi R multiplied by -r+rsqrt{r2-x2}dx, so the volume of the torus is
(4Pi R)((1/2)Pi r2, or 2Pi2Rr2.
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 2Pi2. 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 r2. 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 r2), so this "explanation" shows why we get the constant 2Pi2 multiplying the dimensionally appropriate Rr2.

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=x2, x=1 and x=2, and the x-axis, which we considered earlier, and revolve it around the y-axis. We can analyze the volume of this region by taking slices perpendicular to the axis of symmetry, the y-axis. 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:
bottomtop Pi (rout)2-Pi (rin)2dy.
Well, the bottom is y=0 and the top is y=4. rout is relatively easy: it will always be 2. But rin, which must be written as a function of y, is more difficult. Look at the picture. If y is between 0 and 1, rin is 1, and if y is between 1 and 4, rin is sqrt(y) (I got this by solving for y in the equation y=x2 and selecting the positive solution since we're looking at the curve in the first quadrant). So rin 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 y-axis is x. We could try to see what happens when this is revolved around the y-axis, 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 y-axis 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)=x2. R is the distance to the y-axis, 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=leftrightdV, so:
V=x=1x=22Pi x(x2)dx=2Pix=1x=2x3dx=2Pi(x4/4)]12=2Pi(24/4)-2Pi(14/4).

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(x2), and the integrand for the volume of the solid involved 2Pi x sin(x2)dx. The "extra" x is just what's needed (using the substitution u=x2) 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(1-x) and the x-axis is rotated around the x-axis.
I hoped that most students would answer this using the formula
V=leftrightPi f(x)2dx with left=0 and right=1 and f(x)=x(1-x).

2. Write an integral for the volume which results when the region bounded by y=x(1-x) and the x-axis is rotated around the y-axis.
Here the simplest answer would use the technique and formula just discussed.
leftright 2Pi RH dx. Now H is x(1-x) and R is x, and left=0 and right=1.

Wednesday, January 17 (Lecture #1)
A student information sheet was handed out. The following information was on the sheet.

 Student responsibilities for Math 152 An average of 8 hours per week outside of class studying Math 152 course material. Attending every lecture and every recitation. Active class participation, including asking questions when appropriate. Working on the homework problems in the syllabus. Careful preparation for exams, going over homework, lecture notes, and review problems. Reading appropriate sections of the textbook before and after each lecture.

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).

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:

1. The definite integral
The definite integral can be used to compute many interesting quantities. Some of these include area, volume, work, pressure, distance, center of gravity ... (parts of chapters 6, 8, and 10).
2. Computing and using the definite integral
Symbolic methods: integration by parts, partial fractions, substitutions (chapter 7).
Numerical methods: trapezoidal rule, Simpson's rule (chapter 7).
Differential equations and direction fields (chapter 9).
The remarks about technology on the local information page are relevant here. Hand computations can be elaborate.
3. Taylor's Theorem
In one variable calculus, MVT concerns the equation f´.;(c)=[f(b)-f(a)][b-a]. After relabeling and unfolding algebra, this becomes f(a+h)=f(a)+f´(c)h. There's also the linear approximation idea: f(a+h) is approximately f(a)+f´(a)h+Error. Here with graphical evidence we concluded that the sign of the second derivative explained why the error was likely to be an {over|under} estimate.
Taylor's Theorem includes the MVT and the linear approximation idea. Taylor's Theorem give a local approximation of a differentiable function by a polynomial, and also analyzes the error involved in the approximation. This leads to "infinite series" and a new method of describing functions (chapter 11).
 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. Sketch the volume as well as possible. The first picture is an oblique (tilted) view of the pyramid. Probably more useful is the second picture shown below. Slice. Here I show some of the many, many, many slices. They are very thin. I have drawn a side view, with the b axis along one side of the base square and the h axis going from the base to the vertex. The slices are a little bit of h apart, so I will call the vertical difference between them dh. I will let the notation of calculus help me in this computation. Approximate the volume of a typical slice. The first picture below attempts to be a true picture of the slice. Notice that the edges are slightly tilted compared to the slices. The approximation is a "rectangular parellelopiped": the analog in three dimensions of a rectangle in the plane. The thickness of the slice is dh, a little bit of h. The cross-sectional area is determined by the length of the edge. Suppose that the edge has length s. Then the volume of a slice, dV, is approximately s2dh.
1. Write a definite integral which is equal to the limit of the sum of the approximating slices. So the total volume, V, is bottomtopdV and this is bottomtops2dh.
2. Compute the integral, after writing everything in terms of the controlling variable. I use the phrase, "controlling variable", to mean that we should convert everything we can into formulas and information involving one variable. Here the candidate is the variable h, mostly because of the dh. So "bottom" becomes h=0 and "top" becomes "h=30". We need to write s in terms of h.

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/[30-h]=(10)/(30) so that s=(1/3)([30-h].
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=bottomtops2dh=h=0h=30{(1/3)(30-h)}2dh.
We can compute this using FTC. Probably the most elementary way is to write {(1/3)(30-h)}2=(1/9)(900-60h+h2). Then antidifferentiate, and the resulting value is (1/9){900h-(60/2)h2+(1/3)h3}]030, 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)(30-h)}2dh is to use the substitution u=(1/3)(30-h) so that du=-(1/3)dh and -3du=dh. Then
{(1/3)(30-h)}2dh=-(1/3)u2du=-(1/3){u3/3}+C= -(1/3){((1/3)(30-h))3/3}+C. Wow!
The value of the definite integral is then
-(1/3){((1/3)(30-h))3/3}]h=0h=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 xy-plane is defined by taking the inside of a circle of radius 5 centered at the origin which is in the half-plane with y>=0. A solid has that region as base, and the cross-sections of the solid which are perpendicular to the x-axis are squares, with one side of the square on the xy-plane. 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 xy-plane 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 x-axis. 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 ...
Computing the volume
I imagine that each slice is dx thick, and I'll call the side of the square, s. Then the volume dV of the slice is a piece of the volume of the solid, so dV=s2dx. And the total volume V of the solid is the Sum of the volumes of the slices. I add up and take a limit, and the result is a definite integral: V= leftrights2dx.

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 x2+y2=52. If you look at the first picture of the base, you should see that the square side, s, is actually the distance from the x-axis to the upper semicircle, and this distance is y with y>=0. So x2+y2=52 becomes y2=52-x2 and y=+/-sqrt(52-x2). We take the + sign because we need the upper semicircle, so s=sqrt(52-x2).

V= leftrights2dx= x=-5x=5sqrt(52-x2)2dx. The square root is canceled by the squaring, and we compute:
x=-5x=552-x2 dx= 52x-(1/3)x3]x=-5x=5=2(53-(1/3)53).
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 xy-plane is defined by taking the inside of a circle of radius 5 centered at the origin which is in the half-plane with y>=0. A solid has that region as base, and the cross-sections of the solid which are perpendicular to the y-axis are isoceles right triangles with the hypotenuse of the triangle on the xy-plane. 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 xy-plane, 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 x-axis. 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 ...
Computing the volume
Well, V will be a Sum of dV's, the volumes of the slices. Each of the dV's will be the thickness, dy, multiplied by the cross-sectional area. By the way, I did this problem in class to show that we can integrate dy and this won't hurt too much. What is the cross-sectional area? Here we need to think, but not too much.
Suppose the hypotenuse of an isoceles triangle has length s, and the hypotenuse is the base of the triangle, as shown. Then the perpendicular line (the altitude) from the top to the base has length s/2. This is because all of the acute angles shown are Pi/4, 45o, and so all of the triangles are also isoceles. Therefore the area of the whole triangle, which is one-half the base multiplied by the height, is s2/4.

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= bottomtops2/4 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 x2+y2=52 so that x=+/-sqrt(52-y2). But here we actually need "both" x's, since one gives the left and one gives the right. s is xright-xleft. And xright=+sqrt(52-y2) and xleft=-sqrt(52-y2), so that s=2sqrt(52-y2). Now back to computing V.

V= y=0y=5s2/4 dy= 05{2sqrt(52-y2)}2/4 dy. Miraculously (not really) the square and square root cancel (and here so do "22" and "/4") so that the volume is 0552-y2) dy= 52y-(1/3)y3]05=53-(1/3)53.

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 world-famous.

A view of the Guggenheim Bilbao Museum
Architect: Frank Gehry
A view of the BMW Central Building in Leipzig

Revolving regions to create solid objects
Suppose y=f(x) is a positive (or at least non-negative) function defined on the interval [a,b]. Then the x-axis, 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 x-axis, 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 vase-shaped 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 x-axis 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 x-axis. Imagine that the slice is dx thick, and the radius of the cross-sectional area is r. Then dV=Pi r2dx, and V= leftrightPi r2dx=abPi f(x)2dx 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 r2h, 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 x-axis, with a profile curve f(x), which is abPi f(x)2dx, becomes in this case V= 0hPi (rx/h)2dx= 0hPi r2x2/h2 dx. FTC then gives Pi r2x3/(3h2)]0h=Pi r2h3/(3h2)=(1/3)Pi r2h. 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.