The pieces are:

5/2   -5/2     1
--- + ---- + -----
x-3    x+1   (x+1)2

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 

Yet another example
Well, there is one more wrinkle. Consider the rational function

    11x+7
-------------
 x3+2x2-2x-12

       x2+4x+6
     ---------------
 x-2 | x3+2x2-2x-12
       x3-2x2
      ------------
        4x2-2x-12          
        4x2-8x  
        -------------
            6x-12
            6x-12
            -----
                0

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.

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Yes, COMPLEX NUMBERS are much
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Partial fractions, back to Step 2 and Step 3
Here is how to handle an irreducible quadratic in a slightly simpler case. I'll show you the symbolic pieces, and then solve for them.

   11x+7        A      Bx+C
----------- = ----- + ------
(x-3)(x2+1)    x-3     x2+1

Therefore

   11x+7        4     -4x-1
----------- = ----- + ------
(x-3)(x2+1)    x-3     x2+1

   11x+7        4      -4x      -1
----------- = ----- + ------ + -----
(x-3)(x2+1)    x-3     x2+1     x2+1

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.
₀^1/2sqrt(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.
₀^1/2sqrt(1-x²)dx=(1/2){arcsin(x)+x sqrt(1-x²)}]₀^1/2.
Now arcsin(0)=0 and the other term is also 0 at 0, so the lower limit gives a contribution of 0. The upper limit has (1/2){arcsin(1/2)+(1/2)sqrt(1-(1/2)²)}. With some thought we might recall that arcsin(1/2) is Pi/6, and also "compute" that sqrt(1-(1/2)²)=sqrt(3)/2. Therefore the definite integral is (1/2)(Pi/6)+(1/2)(1/2)(sqrt(3)/2).

Geometric solution
Actually you can "see" these numbers in the picture.
The triangle has a base whose length is 1/2. Since the formula for the upper semicircle is y=sqrt(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)

> convert(arcsinh(x),ln);
                                       2     1/2
                              ln(x + (x  + 1)   )

> 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)

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

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
From eighth grade we know:

 3     7    3(x+5)+7(x+2)     10x+29
--- + --- = ------------- = ----------
x+2   x+5    (x+2)(x+5)      x2+7x+10

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)

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.

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

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.

What's needed

Sine & Cosine Tools	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

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

(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.

Function	Maple's antiderivative	What I got
(sin(x))4(cos(x))5	-(1/9)(sin(x))3(cos(x))6-(1/21)sin(x)(cos(x))6 +(1/105)(cos(x))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

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)

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.

What's needed for these functions

Secant & Tangent Tools	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.

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 ₀^1/2sqrt(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?

• 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.

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 e^xdx. Here: u dv=uv-v du
In order to use this, we will need to complete the information below:
u=x   du=dx
dv=e^xdx   v=e^x
Therefore, u dv=uv-v du becomes
x e^xdx=x e^x-e^xdx=x e^x-e^x+C.
The second integral is easier. Next:

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

Example 3
x³ e^xdx. I bet we can ... well, this is where I began to get bored in class. We could take u=x³ and dv=e^xdx 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 e^xdx using integration by parts. Choose parts as follows:
u=xⁿ   du=nx^n-1dx
dv=e^xdx   v=e^x
Then u dv=uv-v du becomes
xⁿe^xdx=xⁿe^x-nx^n-1e^xdx
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)e^xdx=
I'll first break things up and take out constant multiplies:

1. 7x⁵e^xdx=7(x⁵e^x-5( x⁴e^x-4x³e^x-3(x²e^x-2(xe^x-1(e^x ) ) ) ) ))
2. -3x²dx=-3(x²e^x-2(xe^x-1(e^x) ))
3. +8e^xdx=8e^x

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

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

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.

• 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 ( _a^bf(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, ..., x_n. Then you'd get results f(x₁), f(x₂), ..., f(x_n). You might want to analyze these results statistically and hope that the information you get would represent, somehow, information about "all" of the values of f(x) on [a,b]. The average of these n outputs is called the sample mean.

An example
Maple has a fairly good "random" number generator. That is, the properties of the generator are known and have been investigated systematically, and they satisfy many of the statistical criteria that are desired for randomness. I asked for 30 random numbers between 1 and 4. The list began with 1.198304612 and continued with 1.640439652 and then with 3.984266209 ... (27 more numbers like this!). Then I computed the mean of the values of f(x)=1/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

Connection between sample means and average value
It turns out that, if the sample points are chosen uniformly at random over the interval [a,b], then the sample means will almost always --> the average value of the function (as defined above, with the definite integral). To actually verify this takes some effort because you need to understand what random and uniform and ... please learn some probability and statistics.

Folding back on itself: how to approximate an integral
The previous result is true, and it has been used in very cute reversed fashion. That is, one can compute sample means and then use the sample means to estimate the definite integral. That is, if you wanted to know the value of _a^bf(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, ..., x_n. Then compute this sum: x₁³+x₂³+...x_n³. Then divide this by n, and multiply by 3, the length of the interval. The results of this experiment are below. Please realize that if I ran this experiment another time I would likely get different results (!).

# of points	Reported approximation
10	1007.29154
100	815.78669
1,000	850.26269
10,000	868.39938
100,000	867.07575
1,000,000	867.77362
10,000,000	867.66869

QotD I remarked that Maple had reported that the average value of 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)u^1/2du=(1/15)(1/[3/2])u^3/2+C=(2/45)(4+5x³)^3/2+C. Therefore ₀¹x²sqrt{4+5x³}dx=(2/45)(4+5x³)^3/2]₀¹=(2/45)9^3/2-(2/45)4^3/2=(2·27)/45-(2·8)/45=38/45.

Comments Some errors were made because students assumed that the antiderivative had to be 0 at the lower limit, where x=0. Of course (2/45)(4+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.

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, _left^rightPi f(x)²dx becomes
Pi_-R^R(sqrt(R²-x²))²dx
and (luckily?) the square root cancels with a square, and the result can be evaluated by FTC:
Pi _-R^RR²-x²dx= Pi(R²x-(x³/3))]_-R^R=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 r_in, and the radius of the outside, larger circle is r_out. The area surrounded by the big circle is Pi (r_out)² and the area inside the smaller circle is Pi (r_in)². Therefore the area of the annulus is the difference between these:
Pi (r_out)²-Pi (r_in)².

In the specific case we are considering here, the outer radius is the vertical distance from y=-3 to y=x². This is x²+3 or, as some people would like me to write, x²-(-3). So r_out=x²-3. And r_in is the distance from x=0 to x=-3, and this is always 3. We can get dV, a thin slice of the volume, by multiplying the area of the annulus by the thickness, dx. The result is:
Pi (x²-3)²-Pi (3)²dx.
Please note the contents of the following information.

An arithmetic advertisement
(357)2=127,449 (886)2=784,996 (886)2-(357)2=657,547	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, r_in, and an outer radius, r_out. 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 r_out²-Pi r_in².

We need to find the formulas for r_out and r_in. 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 r_out and the - corresponds to r_-. The cross-sectional area, Pi r_out²-Pi r_in², 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:
_bottom^top Pi (r_out)²-Pi (r_in)²dy.
Well, the bottom is y=0 and the top is y=4. r_out is relatively easy: it will always be 2. But r_in, which must be written as a function of y, is more difficult. Look at the picture. If y is between 0 and 1, r_in is 1, and if y is between 1 and 4, r_in is sqrt(y) (I got this by solving for y in the equation y=x² and selecting the positive solution since we're looking at the curve in the first quadrant). So r_in is a piecewise function. We certainly could compute the integral, but I would have to split it into two pieces.

This seemes clumsy. Sometimes it could also be difficult or essentially impossible -- solving a equation (finding the inverse of a function) can be difficult. So let us see another strategy.

Try to slice dx!
Suppose we take a dx slice. This is a very thin vertical strip whose distance from the 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=_left^rightdV, so:
V=_x=1^x=22Pi x(x²)dx=2Pi_x=1^x=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=_left^rightPi f(x)²dx with left=0 and right=1 and f(x)=x(1-x).
Answer ₀¹Pi (x(1-x))²dx.

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.
_left^right 2Pi RH dx. Now H is x(1-x) and R is x, and left=0 and right=1.
Answer ₀¹2Pi x(x(1-x))dx.

The lecturer will support student work in the course in and out of the classroom. There will be weekly meetings of Math 152 clinics. Students are urged to use these and all opportunities to support their own work.

Background needed
The minimal background needed for this course is adequate knowledge of the material of Math 151. In particular, students should know the definitions and basic interpretations and uses of the derivative and integral, and should be able to work with the standard functions, knowing, where possible, their domains, ranges, graphs, derivatives, and antiderivatives. We'll need the Fundamental Theorem of Calculus (FTC) and the Mean Value Theorem (MVT).

What's this course about?
The previous course had some great ideas (MVT and FTC). This course "exploits" these ideas but Warning! this course definitely has more lengthy and intricate computations than Math 151. An outline of the major topics of Math 152 includes the following:

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.

3. Write a definite integral which is equal to the limit of the sum of the approximating slices. So the total volume, V, is _bottom^topdV and this is _bottom^tops²dh.
4. 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=_bottom^tops²dh=_h=0^h=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=0^h=30 and now the h=30 part gives 0 while the h=0 part is -(-(1/3){((1/3)(30))³/3}).
   The two numerical results are the same! This is not at all obvious to me.

Volumes by slicing: squares over a semicircle
A semicircular region in the 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= _left^rights²dx.

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= _left^rights²dx= _x=-5^x=5sqrt(5²-x²)²dx. The square root is canceled by the squaring, and we compute:
_x=-5^x=55²-x² dx= 5²x-(1/3)x³]_x=-5^x=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, 45^o, 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= _bottom^tops²/4 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 x_right-x_left. And x_right=+sqrt(5²-y²) and x_left=-sqrt(5²-y²), so that s=2sqrt(5²-y²). Now back to computing V.

V= _y=0^y=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 Architect: Zaha Hadid

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= _left^rightPi r²dx=_a^bPi 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 _a^bPi f(x)²dx, becomes in this case V= ₀^hPi (rx/h)²dx= ₀^hPi r²x²/h² dx. FTC then gives Pi r²x³/(3h²)]₀^h=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.

Maintained by greenfie@math.rutgers.edu and last modified 2/17/2007.