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, here is the Step 2 response to the following:

 x4-3x2+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

There is one more significantly different computational difficulty. I will discuss it on Wednesday, the next lecture, and complete the statement of the partial fraction algorithm.

And now another difficulty ...
Factoring polynomials is more difficult than we have seen so far. In addition to linear factors, there are polynomials like x²+1. This polynomial is never zero, so it can't be written as a product with any linear x–a (if x²+1 were equal to such a product, when x=a the result would be 0, which never happens). So let me show an example.

Yet another example
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

Integration
Let's integrate the pieces. A small amount of rewriting gives:

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

What is an irreducible quadratic?
An irreducible quadratic is a degree 2 polynomial which has no real root. So 5x²–9x+17 is an irreducible quadratic but 7x²–88x–20 is not. How do I know this? Well, the roots of Ax²+Bx+C=0 are given by the famous formula [(–B+/–sqrt{B²–4AC})/2A]. The roots are real exactly when the "stuff" under the square root sign in the formula is not negative. (Negative gives complex roots and I am not supposed to discuss these with you lest your brains explode.) The B²–4AC is usually called the discriminant. For 5x²–9x+17 the discriminant is (–9)²–4(5)(17) sure looks negative and for 7x²–88x–20, (–88)²–4(–20)7 certainly seems positive.
Geoemtrically, the graph of a quadratic is a parabola. The roots represent intersections with the x-axis. It is certainly possible that the parabola is high up (or high down?) and does not intersect the x-axis. In that case, the quadratic is irreducible (cannot be factored).

What we did so far ...
I've tried to explain how partial fractions splits up a rational function into a sum of pieces which are easier to understand and analyze. In the context of Math 152, the further work is antidifferentiation. So we've looked at Top(x)/Bottom(x). First we divided, if necessary, to get the degree of the Top less than the degree of the Bottom (the quotient is a polynomial which we already know about). The result, a proper rational fraction, is what we will investigate now. The Partial Fractions algorithm depends on factoring Bottom(x). I've tried to show with examples what to do if there are linear factors, like (x–3)²

What happens in general
Powers of linear factors need constants on top. Powers of irreducible quadratics need degree one "unknowns" on top.

The partial fraction algorithm
First there is a theorem which is not part of Math 152. The theorem states that every rational function has exactly one partial fraction description. What's below is an attempt to describe how to get that description.

Description	Feasibility
Step 0 Long division Take a rational function, and divide Top by Bottom until there is a Quotient (which will be a polynomial) and a remainder. Make the remainder the new top, and then pass that fraction (remainder/Bottom) to the next steps.	This is certainly easy to do by hand or by machine.
Step 1 Factoring Factor the bottom into powers of linear polynomials and powers of irreducible quadratics.	If the factorization is known, everything else is easy. This is the part that is authentically computationally difficult. There's a thereom which states that every polynomial can be written as a product of linear factors and irreducible quadratic factors. But actually finding these factors, even approximately, can be a difficult computational task.
Step 2 Write a symbolic sum (linear)power: if, say, (x–3)4 is a factor, write a sum of 4 terms: k1/(x–3)1+k2/(x–3)1+k3/(x–3)2+k4/(x–3)3. In general, there are as many terms as the degree of the power of the linear factor, and each has an unknown symbolic constant on top. (irred. quad.)power: if, say, (4x2+5x+11)3 appears, then write (k1+j1x)/(4x2+5x+11)1+(k2+j2x)/(4x2+5x+11)2+(k3+j3x)/(4x2+5x+11)3. In general, there are as many terms as the degree of the power of the irreducible quadratic factor, and each has two unknown symbolic constants on top, arranged as a linear polynomial.	This is easy to do, once you get used to it.
Step 3 Find the values of the constants Push together the fractions with the symbolic coefficients. Then if this is to be equal to the original rational function, the tops must be equal. Find the values of the symbolic coefficients.	This actually turns out not to be difficult in practice when you learn what you are doing. The equations for the symbolic coefficients are just a collection of linear (first power) equations in the unknowns. It is easy and practical to solve systems of linear equations, even very large systems of linear equations (people deal with systems of thousands of linear equations all the time!).

If you want to include Step 4, integrating the pieces, then you should already see that integrating the pieces which come from the linear factors is not difficult. Integrating the pieces which come from the irreducatible quadratic factors can certainly be tedious by hand, but not actually impossible. A complete-the-square step followed by a trig substitution (x=(constant)tan(θ) will do it). It won't be pleasant, but I could imagine doing low degree examples by hand.

The simplest example of an irreducible quadratic is 1+x². What sorts of pieces do we get if there is a factor of (1+x²)² and how can we antidifferentiate them? Well, we should get

 Ax+B        Cx+D
------  +  --------
(1+x2)      (1+x2)2

Checking an old formula
In lecture #2 I wrote the following formula (labeled clearly but remember that "clearly" is a codeword in a math course, almost always used for something which is not clear!):
    ∫(1/(1+x²)²dx=(1/2)(x/1+x²)+(1/2)(arctan x). At that time, I told people that the formula could be checked by differentiation. We can now see how this formula was created. If x=tan(θ) then 1+x²=1+(tan(θ))²=(sec(θ))² and dx=(sec(θ))²dθ. The integral becomes ∫[(sec(θ))²/(sec(θ))⁴] dθ. Some of the secant powers cancel, and since 1/sec is cos, we need to compute ∫(cos(θ))²dθ. We have done this a large number of times with double angle formulas, etc. What we got before is this:
    (1/2)cos(θ) sin(θ)+θ/2+C.
So ... if tan(θ)=x which is x/1 and tan(θ) is also OPP/ADJ, then I can think of OPP as x and ADJ as 1. See the picture to the right, please. Then HYP is sqrt(x²+1), and sin(θ)=x/sqrt(x²+1) and cos(θ)=1/sqrt(x²+1). Also, θ=arctan(x). So the whole darn formula in θ-land, (1/2)cos(θ)sin(θ)+θ/2+C, becomes in x-land this:
    (1/2)[1/sqrt(x²+1)][x/sqrt(x²+1)]+arctan(x)/2+C
which is, exactly (1/2)(x/1+x²)+(1/2)(arctan x), a formula which seemed so silly that there would be no simple way of finding it. Well, we now know enough to "discover" such formulas.

An example, not by hand The following computation took about half a second on my work computer, which is several years old and not especially fast. I am a bit surprised it took that long. (Ahhh ... I experimented further: just loading the appropriate software seems to take most of the time, not working on the specific example!) The first command defines the function. The second command asks to convert the rational function into its partial fraction form. I can't figure out the order the pieces are printed -- certainly not any order I can easily identify. All the pieces one might expect according to the general statement are there, however. Then the third command asks for the integral (antiderivative) of this. Take a quick look and see which pieces you can pick out, just for fun. > f:=1/(x*(x-2)^3*(3*x-7)^2*(x^2+4)^2*(x^2+2*x+3)); 1 f := ---------------------------------------------- 3 2 2 2 2 x (x - 2) (3 x - 7) (x + 4) (x + 2 x + 3) > convert(f,parfrac); 87 1 -54 - 11 x 59049 -------------- - ------- + ---------------- + ------------------ 2 18816 x 2 2 2 30976 (x - 2) 924800 (x + 4) 5967850 (3 x - 7) -11708 - 1727 x 3774530178 1 + ------------------ - ---------------------- + ------------- 2 104750687125 (3 x - 7) 3 314432000 (x + 4) 1408 (x - 2) -52145 - 44488 x 8229 + ------------------------- + -------------- 2 681472 (x - 2) 8033987874 (x + 2 x + 3) > int(%,x); 87 -108 x + 88 14003 - ------------- - 1/18816 ln(x) + ----------------- - --------- arctan(x/2) 30976 (x - 2) 2 628864000 14796800 (x + 4) 19683 1727 2 1258176726 - ----------------- - --------- ln(x + 4) - ------------ ln(3 x - 7) 5967850 (3 x - 7) 628864000 104750687125 1 11122 2 - ------------- - ---------- ln(x + 2 x + 3) 2 4016993937 2816 (x - 2) 1/2 7657 1/2 (2 x + 2) 2 8229 - ----------- 2 arctan(--------------) + ------ ln(x - 2) 16067975748 4 681472

Rationalizing substitutions
I showed you how the partial fraction algorithm gets used directly in integration. But lots of other antiderivatives use partial fractions because certain substitutions change the integration problem into the integration of a rational function. Such substitutions are called rationalizing substitutions. Let's look at a few examples.

∫1/[(2+sqrt(x))(1+sqrt(x))] dx
For this integral, try the substitution u=sqrt(x) because sqrt(x) is the center of all the trouble or ugliness. Then du=[1/(2sqrt(x))]dx and so 2sqrt(x)du=dx and 2u du=dx. The integral ∫1/[(2+sqrt(x))(1+sqrt(x))] dx becomes ∫(2u)/[(2+u)(1+u)] du. This we can easily handle:
    (2u)/[(2+u)(1+u)]=A/(2+u)+B/(1+u)=[A(2+u)+B(1+u)]/[(2+u)(1+u)]
So just looking at the tops gives 2u=A(2+u)+B(1+u). u=–1 says A=–2 and u=–2 says –B=–4 so B=4. Therefore
    ∫(2u)/[(2+u)(1+u)] du=∫[–2/(2+u)]+[4/(1+u)]du=–2ln(2+u)+4ln(1+u)+C.
We go back to x's and get –2ln(2+sqrt(x))+4ln(1+sqrt(x))+C.

Here's the Maple solution which is what we just got:

> int(1/((2+sqrt(x))*(1+sqrt(x))),x);
                                                   1/2              1/2
                                        -2 ln(1 + x   ) + 4 ln(2 + x   )

Just one more ...
What about ∫[1/(2+sqrt(1+x))] dx? Here I would probably try u=sqrt(1+x), so that du=[1/2sqrt(1+x)]dx and 2sqrt(1+x) du=dx and 2u du=dx. The integral becomes ∫[(2u)/(2+u)]du. Now we need Step 0 of the partial fractions algorithm since the degree of 2u is 1 and the degree of 2+u is 1. So I must divide 2u by 2+u:

       2
    -------
u+2 | 2u
      2u+4
      ----
        -4

> int(1/(2+sqrt(1+x)),x);
                                            1/2                    1/2                    1/2
                   -2 ln(-3 + x) + 2 (1 + x)    + 2 ln(-2 + (1 + x)   ) - 2 ln(2 + (1 + x)   )

QotD
Problem 50 of the partial fractions section of the book asks for ∫(x dx)/[x^1/2+x^1/3]. The question is: find a substitution and carry it out so that the result is an antiderivative of a rational function (you don't need to perform that antidifferentiation, though!).

I think that the substitution u=x^1/6. Well, then u²=(x^1/6)²=x^2/6=x^1/3. And u³=(x^1/6)³=x^3/6=x^1/2. Also u⁶=x. How about du? One way of getting this is to begin with u=x^1/6 and differentiate: du=(1/6)x^–5/6dx, so that 6x^5/6du=dx and 6u⁵du=dx. Or we could begin with u⁶=x and differentiate to get 6u⁵du=dx. Maybe that way is easier. In any case, we can go from x-land where we have ∫(x dx)/[x^1/2+x^1/3] to u-land, where we get ∫(u⁶6u⁵du/(u³+u²)=∫6u¹¹du/(u³+u²) and we can divide by u² top and bottom to get ∫6u⁹du/(u+1). Now we would need to divide through (Step 0) and then go on, but it can be done.

∫₀¹[sqrt(x²–1)/x³]dx
Because I see sqrt(x²–1) and the toolbox contains the equation (tan(θ))²=(sec(θ))²–1 I will "guess" at the substitution x=sec(θ), which gives dx=sec(θ)tan(θ)dθ and sqrt(x²–1)=tan(θ). With this know, I can rewrite the integral.

∫[sqrt(x²–1)/x³]dx becomes ∫[tan(θ)/sec(θ)³]sec(θ)tan(θ)dθ. Then cancel everything you can. The result is ∫[cos(θ)]²dθ. 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 by doubling the frequency.
so ∫[cos(θ)]²dθ= ∫(1/2)[1+cos(2θ)]dθ=(1/2)[θ+(1/2)sin(2θ)}+C=(1/2)[θ+sin(θ)cos(θ)]+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(θ)=x and we want to know what sine and cosine of θ are in terms of x. One way some people use is drawing a right triangle with one acute angle equal to θ, and with sides selected so that sec(θ) 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(θ)=sqrt(x²–1)/x and cos(θ)=1/x. Then (1/2)[θ+sin(θ)cos(θ)]+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 judgment 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 in terms of 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

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. The picture to the right is supposed to symbolize this method. Heh, heh, heh ... 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(θ))²=(sec(θ))²–1 and multiply by 2 to get 2(tan(θ))²=2(sec(θ))²–2. I "guess" that I would like to try t=sqrt(2)sec(θ). Then t²=2(sec(θ))², so t²–2 is 2(tan(θ))². Also dt=sqrt(2)sec(θ)tan(θ) dθ. The integral in θ–l and is ∫sqrt[2(tan(θ))²]sqrt(2)sec(θ)tan(θ) dθ which is ∫2sec(θ)[tan(θ)]²dθ. 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

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 (the standard form of a polynomial). 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. (Don't try this on a Rutgers computer!)

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. For example, the repelling force between two classical charged objects (electrons?) is rational (it is inverse square where the variable is the distance between the object). 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.

An example
I noted that the derivative of ln(x²–2x–3) is not [(1/(x²–2x–3)] and so therefore the antiderivative of the latter function is not just the log of the bottom. Then we wrote [(1/(x²–2x–3)] as a sum of constants multiplying (x–3)^–1 and (x+1)^–1 and then got specific values of the constants. This allows antidifferentiation. 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

 x3+4x–1              11x+7
---------- = x+2 +  --------
 x2–2x–3             x2–2x–3

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

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)

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 get 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!)

QotD
Write 1/(3x²+2x) as a sum of two terms, and then antidifferentiate this sum.

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

Imagine this ...
Some maniac has written an infinite list, actually a doubly infinite list, of functions. The beginning looks like what follows:

                sec(x)             (sec(x))2           (sec(x))3               (sec(x))4  

 tan(x)        tan(x)sec(x)     tan(x)(sec(x))2      tan(x)(sec(x))3          tan(x)(sec(x))4  

(tan(x))2   (tan(x))2sec(x)    (tan(x))2(sec(x))2    (tan(x))2(sec(x))3       (tan(x))2(sec(x))4  

(tan(x))3    (tan(x))3sec(x)   (tan(x))3(sec(x))2    (tan(x))3(sec(x))3       (tan(x))3 (sec(x))4  

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. Much 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:
What is the derivative of ln(sec(x)+tan(x))? We must use the Chain Rule. The derivative of the "outside" gives us 1/(sec(x)+tan(x)), and then we must multiply by the derivative of the inside. But sec(x)+tan(x) has derivative (look at the secant/tangent tools) sec(x)tan(x)+(sec(x))².
Therefore the derivative of ln(sec(x)+tan(x)) is 1/(sec(x)+tan(x)) multiplied by sec(x)tan(x)+(sec(x))². If you realize that the top (the second factor) is sec(x) times sec(x)+tan(x), you will then know that ∫sec(x)dx=ln(sec(x)+tan(x))+C.

After you know this, there are lots of ways of "justifying" this ludicrous formula. In fact, the formula was not discovered solely by math people. There is a historical account in the article An Application of Geography to Mathematics: History of the Integral of the Secant by Rickey and Tuchinsky appearing in Mathematics Magazine (volume 53, 1980). Here is a link which will take you directly to the article if you are on a Rutgers machine.

This "computation" is probably the single most irritating (because of the lack of motivation) in the whole darn course. The history 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.

∫(tan(x))⁴dx
Clearly we can write (tan(x))⁴ as (tan(x))²(tan(x))² and then since (tan(x))²=(sec(x))²–1, we have (tan(x))⁴=(tan(x))²(sec(x))²–(tan(x))². So let's compute two integrals:
∫(tan(x))²(sec(x))²dx: if we put w=tan(x), then dw=(sec(x))²dx, so the integral becomes ∫w²dw=(1/3)w³+C=(1/3)(tan(x))³+C. We need to subtract ∫(tan(x))²dx, but we just did that. It was tan(x)–x+C, so the combined result is this:
∫(tan(x))⁴dx=(1/3)(tan(x))³–tan(x)+x+C. Neat.

∫(sec(x))³dx
This is also tricky. The first thing is to write it in a rather strange way:
∫(sec(x))³dx=∫sec(x)·(sec(x))²dx
While you may not think that this helps, it does. Sigh. In fact, "one" could recognize that (sec(x))² is the derivative of tan(x). We can use integration by parts on this creature. So:

int;(sec(x))3dx=∫sec(x)·(sec(x))2dx=sec(x)tan(x)–∫tan(x)sec(x)tan(x)dx
                   ∫u dv        =    uv    –   ∫v du
where  
      u=sec(x)        du=sec(x)tan(x)dx     
     dv=(sec(x))2dx    v=tan(x)

∫(sec(x))³dx=(1/2)(sec(x)tan(x)+ln(sec(x)+tan(x))).

As I mentioned in class, the tricks just shown are enough to allow us to find antiderivatives of any function on the list we started with. And with sufficient energy and cleverness, I think a computer program could be written which would actually find the antiderivatives.

Today's tools
We will use the tools we have, with one more equation whose usefulness will be demonstrated shortly. I will exchange the previous variable, x, for a new variable, θ, which your text prefers here.

Derivative formulas	Functional equations
d --- sin(θ) = cos(θ)      dθ       d --- cos(θ) = –sin(θ)      dθ  d --- sec(θ) = sec(θ)tan(θ) dθ   d --- tan(θ) = (sec(θ))2   dθ	1=(cos(θ))2+(sin(θ))2   cos(2θ)=(cos(θ))2–(sin(θ))2   sin(2θ)=2sin(θ)cos(θ)   (sec(θ))2=(tan(θ))2+1   (tan(θ))2=(sec(θ))2–1

We will use these tools and the ideas we've already seen to investigate antiderivatives of functions which arise naturally in physical and geometric problems. The functions involve squares and square roots.

A circular area
What is the value of ∫₀¹sqrt(1–x²)dx? I truly expected people to recognize this as a quarter of the unit circle, and therefore report the value as Π/4. But if we change the upper limit, then maybe things aren't as simple. So what is ∫₀^1/2sqrt(1–x²)dx?
If I want to use FTC to compute this, I should find an antiderivative for sqrt(1–x²). I tried to motivate the following substitution: x=sin(θ) because that substitution will make both dx and sqrt(1–x²) come out "nice". Then dx=cos(θ)dθ and sqrt(1–x²)=sqrt(1–sin(θ)²)=sqrt(cos(θ)²)=cos(θ). Therefore:
∫sqrt(1–x²)dx=∫[cos(θ)]²dθ.
We use the double angle formula from the toolbox:
∫[cos(θ)]²dθ=∫1/2)[1+cos(2θ)]dθ=(1/2)[θ+(1/2)sin(2θ)]+C
The extra 1/2 comes from the Chain Rule worked backwards on the "inside" part of cos(2θ). Now I'd like to return to x-land by writing everything in terms of x instead of θ. Since sin(θ)=x, we know that arcsin(x)=θ. That allows us to translate the first part of the answer back to x's. But what about sin(2θ)? Here is where I need the new formula in the toolbox: sin(2θ)=2sin(θ)cos(θ). Each of these I already know in terms of x's because x=sin(θ) and cos(θ) 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 Π/6, and also "compute" that sqrt(1–(1/2)²)=sqrt(3)/2. Therefore the definite integral is (1/2)(Π/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 Π/3, so the circular sector has inside angle Π/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)(Π/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

∫₀¹1/sqrt(1+x²)dx
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 estimate of things to be computed. That gives us at least a rough way of checking on the methods and the result. For example, the value of the integral can't be –17 and it can't be 438. One is negative, and the other is positive but too large.

How to do it
Look in the toolbox and see (sec(θ))²=(tan(θ))²+1. If I look at sqrt(x²+1) then I think of trying x=tan(θ), so dx=[sec(θ)]² and sec(θ)=srqt(x²+1). Then ∫1/sqrt(1+x²)dx becomes ∫[1/sec(θ)][sec(θ)]²dθ which is ∫sec(θ) dθ. We officially know this integral because of the ludicrous computation done earlier. Its value is ln(sec(θ)+tan(θ))+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

APOLOGY I made a serious mistake in the first lecture. I computed ∫sqrt(1+x2)dx. This was done correctly (it is somewhat more complicated than the integral I did in the second lecture, though). And then I asserted that this antiderivative was arcsinh. This was just wrong wrong wrong and I apologize. I am sorry. What is written above in this diary is correct.

Another integral (QotD!)
We looked at ∫[x³/sqrt(1–x²]dx. Again, if x=sin(θ) this gets converted into a "power of trig functions" integral. Actually there was little that was new about this integral, but I thought it would be cute to assign finishing the integral as the QotD. I am too scared to look at what was handed in. Here is what I hope that I will read (done by me in green and done by students in blue. The exercise was yet another effort to encourage students to work together. Sigh.

If x=sin(θ) then dx=cos(θ)dθ and sqrt(1–x²)=sqrt(1–θ²)=cos(θ) and x³=sin(θ)³ so that the integral becomes ∫sin(θ)³dθ. In turn we can "split off" one of the sines, and the result is ∫sin(θ)²·sin(θ)dθ. This makes us think of the substitution w=cos(θ) with dw=–sin(θ)d(θ). So the integral changes into ∫(1–w²)–dw because again sine squared is one minus cosine squared. Now we {integrate|antidifferentiate} and the result is –w+w³+C (I pushed through the minus sign) and now back in θ-land we get –cos(θ)+[cos(θ)³/3]+C and, finally, back in x-land (we are not done officially until we are back in x-land!) we get –sqrt(1–x²)+[{sqrt(1–x²)}³/3]+C.

Just to complete your joy here is what Maple says, and I don't see the process from the answer:

> int(x^3/sqrt(1–x^2),x);
                                                  2
                           (–1 + x) (x + 1) (2 + x )
                           -------------------------
                                         2 1/2
                                 3 (1 – x )

∫ln(x)dx
In the early lecture, the volunteers were Mr.Gerstein and Mr. San Giacomo and, the later lecture, the participants were Mr. Boemo and Mr. Saporta. They kindly computed this straightforward antiderivative. If ∫udv=uv–∫vdu, then u=ln(x) implies dv=dx so du=(1/x)dx and v=x. Then:
∫ln(x)dx=x[ln(x)]–∫x(1/x)dx=x[ln(x)]–∫1dx=x[ln(x)]–x+C.

∫sin(x)sin(3x)dx
Here the vital (!) formulas were obtained by Ms. Gendelberg and Ms. Kanikicherla (first lecture) and by Mr. Bell and Mr. Poppy (second lecture). We want to apply ∫udv=uv–∫vdu to ∫sin(x)sin(3x)dx. This antiderivative is certainly not as simple as the previous one.

I think the volunteers took parts which were different from the ones I will use. Here I will try u=sin(x) and dv=sin(3x)dx. Then du=cos(x)dx and v=–(1/3)cos(3x). Just the computation of v needs some care, because there is a sign to worry about and then the factor (1/3) is needed to "fix up" what happens when the Chain Rule differentiates cos(3x). So ∫udv=uv–∫vdu becomes:
∫sin(x)sin(3x)dx=–(1/3)sin(x)cos(3x)+∫(1/3)cos(3x)cos(x)dx

The plus sign is really two minus signs packaged together. Now things look bad, because it seems like we have "exchanged" the original integral for something just as bad. Well, in this case we can try again. Use integration by parts on ∫cos(3x)cos(x)dx. (I'll pull out the multiplicative factor 1/3 but remember it later!). Take u=cos(x) and dv=cos(3x)dx. The resulting du is –sin(x)dx and the resulting v is (1/3)sin(3x). The signs and the multipliers are quite important. The integration by parts formula becomes:
∫cos(3x)cos(x)dx=(1/3)cos(x)sin(3x)+(1/3)∫sin(x)sin(3x)dx.

We get the original integral back. This plus sign is also two minus signs. All of the constants are important. Suppose we plug this into our first Integration by Parts:
∫sin(x)sin(3x)dx=–(1/3)sin(x)cos(3x)+(1/3)((1/3)cos(x)sin(3x)+(1/3)∫sin(x)sin(3x)dx)

Let's call ♠ the antiderivative we want. Then the equation above is:
♠=–(1/3)sin(x)cos(3x)+(1/3)((1/3)cos(x)sin(3x)+(1/3)♠)
so we can solve for ♠.
♠=–(1/3)sin(x)cos(3x)+(1/3)((1/3)cos(x)sin(3x)+(1/3) ♠) is
♠=–(1/3)sin(x)cos(3x)+(1/9)cos(x)sin(3x)+(1/9)♠ so
(1–{1/9})♠=–(1/3)sin(x)cos(3x)+(1/9)cos(x)sin(3x) and since 1–{1/9} is 8/9, we can divide by 8/9 to get
♠=(9/8)( –(1/3)sin(x)cos(3x)+(1/9)cos(x)sin(3x))
It is not at all clear to me that
∫sin(x)sin(3x)dx=(9/8)( –(1/3)sin(x)cos(3x)+(1/9)cos(x)sin(3x))+C

Comments I just did this antiderivative with Maple and the result was wonderfully weirdly different. But more about this later!
The parts chosen in class were different, but the result was the same.

A definite integral
Please notice that sin(x) and sin(3x) are both 0 when x=0 and x=Π (sines are always 0 at integer multiples of Π). This means (and this is almost clear, if you assume the antiderivative result above is correct!) ∫₀^Πsin(x)sin(3x)dx=0. A picture of sin(x)sin(3x) on the interval [0,Π] is shown to the right. Several people helped me with this in class. I thank them here. Notice that, even with the picture, it is not totally clear to me that the positive areas (above the x-axis) exactly balance out the negative area below the axis. They do, though. It turns out that integrals of this type come up very often when looking at vibrations of things, and that this integral is 0 (and other similar integrals, also) is quite useful.

The workshop problem I discussed ways of thinking about the workshop problem. First, I knew the answer to part a) was74 900Π≈235 305.29 ft-lbs (not Newtons, not Joules!) because this was the answer to problem 41 of the Chapter 6 review problems, and it was in the back of the book. Therefore I, as the grader, could check. If you, as the student, got a different answer, then you would know either that the text was incorrect (possible, but unlikely) or that you were incorrect. I also mentioned that some students ran into real difficulty even in a) due to bad specification of the coordinate system's origin. To me there seem to be three "natural" places: the center of the sphere, the bottom of the sphere, and the water source 2 feet down from the bottom of the sphere. I saw successful solutions using all three of these. But sometimes people put the origin in one place for one part of the formula, and in another place for another part. This inconsistency does not lead to correct results.

Moving to part b), I mentioned that the tank in part b) had the same setup, the same configuration, as what was considered in a). There was no change in the "geometry" of the tank. If W(h) is the work done in filling the tank to a depth of h feet, I knew several things. First, the natural domain of W(h) in this problem was 0≤h≤10. Also, we know two values of W(h). If h=0, there is no water in the tank, so no work has been done. Therefore W(0)=0. If h=10, then the tank is full, and we get the result of a). Therefore W(10)=74 900Π. So we have a function which "connects the dots" (0,0) and (10,74 900Π). What sort of function is it? Well, some thought should convince you that the function should be increasing. And considering the size of the cross-sections will show you that the function will increase faster (per unit of height or depth) around the middle of the tank, where the tank is widest. A bit more thought might convince you that W(h) is at least continuous (no breaks) and even smooth -- differentiable, with no corner. In fact, a number of people got the formula W(h)=64.2Π(–h4/4+(8/3)h3+10h2). A graph of this function is shown to the right (the horizontal line is at the height of total work done for filling the whole tank).

Exam warning!
The first exam for these sections will be in two weeks, on Monday, February 23 (usual class time and place). Further information, including extensive review material, will be available soon. Right now you can see the formula sheet which will be handed out along with the exam. Please: see both what is included and what is not included.

Integer powers of trig functions
The goal today is to show you some algebraic tricks which can be used to compute (in principal!) the antiderivative of any integer powers of trig functions. I sincerely hope that you won't have to use these tricks too often once you're out of Math 152, and that you will have "machinery" available to compute these integrals. But, as you will see, even with machinery, you will need to cope with the results, and sometimes they can be irritating to understand and use.

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

∫sin(x)³cos(x)³dx
This is (relatively) easy, as you will see. What we can do is "borrow" a cosine, say. What do I mean? Well:
∫(sin(x)³cos(x)³dx=∫sin(x)³cos(x)²[cos(x)]dx
Maybe this suggests the substitution w=sin(x). Then dw=cos(x)dx and we can write sin(x)³ as w³. What about the last chunk of the integrand (the function we're integrating)? Well, that's cos(x)², and since cos(x)²=1–sin(x)², we see that cos(x)²=1–w². Now:
∫sin(x)³cos(x)²[cos(x)]dx=∫w³[1–w²]dw=∫w³–w⁵dw=(1/4)w⁴–(1/6)w⁶+C
If we return from w-land to x-land, the answer turns out to be (1/4)sin(x)⁴–(1/6)sin(x)⁶+C.

And another way
An inquisitive person might notice the following: we could have borrowed a sine. Here:
∫(sin(x)³cos(x)³dx=∫sin(x)²cos(x)³[sin(x)]dx
and take w=cos(x), so dw=–sin(x)dx and –dw=sin(x)dx. Now it is sin(x)² which we should think of as 1–cos(x)²=1–w² and replace cos(x)³ by w³, so that:
∫sin(x)²cos(x)³[sin(x)]dx=–∫[1–w²]w³dw=∫w⁵–w³dw=(1/6)w⁶–(1/4)w⁴+C
Things may get a bit tricky. I pulled out a minus first, and then reinserted it when I multiplied the powers of w inside the integral. Then I antidifferentiated. So the answer is (substituting back cos(x) for w here) (1/6)cos(x)⁶–(1/4)cos(x)⁴+C

What's going on?
Let's see: we seem to have two answers for ∫sin(x)³cos(x)³dx. They are
(1/4)sin(x)⁴–(1/6)sin(x)⁶+C and (1/6)cos(x)⁶–(1/4)cos(x)⁴+C.
If you love (?) trig functions, you know that there are many more trig identities besides the basic ones listed above in the Sine & Cosine Tools. Maybe the functions (1/4)sin(x)⁴–(1/6)sin(x)⁶ and (1/6)cos(x)⁶–(1/4)cos(x)⁴ are actually the same.
Plug in x=0. Then (1/4)sin(x)⁴–(1/6)sin(x)⁶ becomes 0 (hey: sin(0)=0) and (1/6)cos(x)⁶–(1/4)cos(x)⁴ becomes 1/6–1/4 because cos(0)=1. And 1/6–1/4 is not 0. What is going on here? Let me in fact make things more difficult before explaining more.

And Maple's answer ...
Let's see what we get:

> int(sin(x)^3*cos(x)^3,x);
                                 2       4              4
                      –1/6 sin(x)  cos(x)  – 1/12 cos(x)

∫x(x+1)dx
So let me try to integrate something really really simple and get two "different" answers. It will be an artificial 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 (or a computer trying to follow a badly implemented algorithm) 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, algebra shows exactly this:
(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)

Huh? (and a picture) The two graphs are parallel to each other. To the right is a graph of both (1/3)x3+(1/2)x2 and (1/2)x(x+1)2–(1/6)(x+1)3. 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.

Back to explaining ∫sin(x)3cos(x)3dx To the right is a graph of the two different antiderivatives of sin(x)3cos(x)3 on the interval [0,2Π]. I think that (1/4)sin(x)4–(1/6)sin(x)6 is in green and (1/6)cos(x)6–(1/4)cos(x)4 is in red and if you believe in pictures, you can see that these curves are parallel also: they differ by a constant. So you and a friend who algebraically work on antiderivatives of functions may well come up with superficially different answers, and both of them may be correct! This is weird and wonderful.

What about definite integrals? Several students asked after class what happens to the values of definite integrals since we find antiderivatives in order to use FTC (Fundamental Theorem of Calculus). So let me try to explain, please. (I am copying most of an email message I sent in response to this question.) When we apply FTC to evaluate a definite integral, the setup is something like this: We have a function F with F´=f and want to know the integral of f from a to b. We compute F(b)-F(a). Indeed, EXACTLY as you suspect, an additive constant cancels out. So if we used a function G(x) defined by F(x)+37, then G´=f also and G(b)-G(a)={F(b)+37}-{F(a)+37}=F(b)-F(a). So the additive constants don't matter in this case. O.k., so how do they affect answers? Well, let's go to a simple physical example. Suppose f(t) is the velocity of a particle moving on the horizontal axis: positive means the particle is moving right and negative means the particle is moving left. That is probably not a difficult setup. What does a function F(t) with F´(t)=f(t) represent? Well, this now gets a bit delicate. I know (from the reasoning above) that F(b)-F(a) doesn't depend on the particular additive constant. That represents the displacement (the net change in position) of the particle during the time interval from t=a to t=b. But if you don't tell me more, then I can't tell you what F(t) represents. It may or may NOT represent the position of the particle at time t. IF you tell me an initial position, so that you specify, say, F(0)=-42, then I can evaluate the +C and then the data (initial position as well as velocity) will tell us in this model what the positions are at later times. And F(t), the specific F(t) mentioned, will be that position. I hope I have explained things a bit better than in class. We will return to specifying initial positions later (in fact, the whole idea is rather important in applications).

∫sin(x)³cos(x)²dx
This is easy. "Borrow" a sine, convert everything into cosines using a formula, etc. Things are easy when there is at least one odd power. But:

∫sin(x)²cos(x)²dx
Now everything is different. If I borrow, what remains can't be written very nicely in terms of the candidate substitution. A different trick is used. It converts squares of sines and cosines to first powers but it doubles the frequency. Here is what I mean:
Look at the tools. We have
          1=(cos(x))²+(sin(x))²
cos(2x)=(cos(x))²–(sin(x))²
Add the equations and divide by 2. The result is (1/2)[1+cos(2x)]=cos(x)².
Subtract the equations and divide by 2. The result is (1/2)[1–cos(2x)]=sin(x)².

Let's use these equations in ∫sin(x)²cos(x)²dx. So we get ∫(1/2)[1–cos(2x)](1/2)[1+cos(2x)]dx. We started out with degree 2+2 and now we have degree 1+1, but the argument in the cosine is 2x, not x. Next, multiply:
∫(1/2)[1–cos(2x)](1/2)[1+cos(2x)]dx= (1/4)∫1–cos(2x)²dx
and I pulled out the two 1/2 multiplicative factors to get the 1/4. But now we have cos(2x)² and what shall we do? A burst of genius (not really, but there is some inspiration!) turns (1/2)[1+cos(2x)]=cos(x)² into (1/2)[1+cos(4x)]=cos(2x)². (Halve the exponent and double the frequency!) So:
(1/4)∫1–cos(2x)²dx= (1/4)∫1–(1/2)[1+cos(4x)]dx.
I can "do" each of the pieces (although I got the constants confused in class!). The answer is (1/4)(x–(1/2)[x+(1/4)sin(4x)])+C.

Incidentally, Maple's answer is the same as this one, except (sigh) it is written differently: 3 –1/4 sin(x) cos(x) + 1/8 sin(x) cos(x) + x/8 I know it is the same since both answers have value 0 when x=0. The "constant" that separates them is therefore 0. Reduction formula Here is probably what Maple uses on powers of sine. Since ∫(sin(x))ndx= ∫(sin(x))n–1 sin(x)dx we can do this: ∫(sin(x))n–1 sin(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 for small powers of n, but it is lovely for a program to "know". You can see this reduction formula being used: > int(sin(x)^9,x); 8 6 16 4 –1/9 sin(x) cos(x) – 8/63 sin(x) cos(x) – --- sin(x) cos(x) 105 64 2 128 – --- sin(x) cos(x) – --- cos(x) 315 315

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!

QotD
What is ∫sin(x)³cos(x)²dx?
Hey, isn't this an antiderivative which was called "easy" earlier? Huh. Nasty, nasty, nasty lecturer.

Several people among you have already discussed students copying textbook homework solutions or writeups. The first time copying occurs (let me be charitable!) please give the student(s) involved 0's for the assignment. Inform them in writing that if this is repeated, the evidence will be brought to my attention and I will (darn it!) proceed against them formally. I am not lazy, but the punishments involved can be somewhat severe, so I would rather assume students are naive and may need some advice.

Numerical stuff ...
Numerical routines which can be used on some common graphing calculators are available. Also, I included in the diary for the last lecture details about computing approximations and error estimates for one specific definite integral. I didn't have time to do this in class, and this may be helpful for you when preparing homework problems or workshop solutions or studying for exams.

Now we'll begin a major part of the course dealing with the symbolic (not numerical) computation of antiderivatives. The word "antiderivative" is long, and so I'll use the shorter word "integral".

There are two methods more important than any others.

There are also some miscellaneous algebraic tricks which help find many integrals, and we'll see them later.

Meeting integration by parts
The derivative of u(x)v(x) is u´(x)v(x)+u(x)v´(x). So this has to be also an integration formula. Let's see:
∫(u´(x)v(x)+u(x)v´(x))dx=u(x)v(x)
but even that looks a little bit silly. The left-hand side is also ∫u´(x)v(x)dx+∫u(x)v´(x)dx. That doesn't seem neat. But here is another way of writing the same equation:
∫u(x)v´(x)dx=u(x)v(x)–∫u´(x)v(x)dx
where we took one of the integrals and put it "on the other side". This is the formula for Integration by Parts. The u and v letters are used by almost everyone. There is also some standard abbreviation which most people use. It's this: v´(x) is dv/dx, so v´(x)dx=dv. And du=u´(x)dx. Then the formula becomes more compact and easier to remember:
∫udv=uv–∫vdu
which is very telegraphic. The remainder of the lecture is a sequence of examples.

Example #0
∫e^xdx=e^x+C. Well, that's not too illuminating. Let me try something a bit more complicated.

Example #1
How about ∫xe^xdx. Notice, please, that integration is NOT multiplicative. While the integral of x is (1/2)x² (yeah, yeah, +C), the integral of x²=x·x, is not {(1/2)x²}·{(1/2)x²} (the degree is wrong, the constant is wrong, and the idea is wrong!). So we need a new trick, and this new trick is Integration by Parts. I tend to think of Integration by Parts whenever I see a product of functions that somehow seem unrelated to each other, "like" x and e^x. Your intuition will develop as you do more and more examples.

Here I will take u=x and dv=e^xdx. I will need du, which is dx, and v, which is e^x. I can do the derivative and the integral "on the fly" -- this is the first real example. Then ∫udv=uv–∫vdu becomes
∫xe^xdx=xe^x–∫e^xdx=(easy to finish!)xe^x–e^x+C.

Comments
Before jumping forward to other examples, let me analyze this "easiest" example first.
Comment A Is the result correct? Well, the method says that xe^x–e^x+C is an antiderivative of xe^x. Should we believe this, or, better, is there a way of checking the result. Well, certainly: differentiation is almost always easier than integration (antidifferentiation). So let us d/dx the result:
d/dx(xe^x–e^x+C)=d/dx(xe^x)–d/dx(e^x)+d/dx(C)=(e^x+xe^x)–e^x+0=xe^x.
This shouldn't be so amazing, after all, since integration by parts and the product rule are inverses of each other. Integration by parts just arranges things well.
Comment B This is a relatively easy example. There aren't many possibilities for the "parts". Certainly there will be more choices in other cases.
Comment C This is my feeling about Integration by Parts: I have udv, and then I trade it in for vdu, "paying" a penalty of uv. The idea is that the uv cost allows you to change the function you need to integrate. As soon as you select, say, u, you then have dv, and you can sometimes look forward and "see" vdu. You will be successful if that is somehow simpler or easier than the starting integral. Machines are fairly good at this, but, honestly, experienced human beings are even better!

Example #2
Let's do ∫x²e^xdx. (I am trying to build a hierarchy of examples, moving from easy to more difficult.) Here try (although there are other choices!) u=x² and dv=e^x. Then we need du=2xdx and v=e^x. I mentioned in class and I want to restate here: even when I am alone, and don't need to teach, I tend to write out all of the details, including explicit descriptions of u and dv and du and v and then the Integration by Parts formula. I have found that if I am lazy and don't do this, then disaster (errors!) occur far too often, and I just spend more time, because I need to do the computation again and again. So ∫udv=uv–∫vdu becomes
∫x²e^xdx=x²e^x–∫2xe^xdx.

What about ∫2xe^xdx? We can pull out the multiplicative constant 2 and just get 2∫xe^xdx. But we know what that integral is -- we just computed it! Therefore
∫x²e^xdx=x²e^x–∫2xe^xdx=x²e^x&ndash(xe^x–e^x+C)
The final result would usually be written
x²e^x–xe^x+e^x+C
and I should make several comments about this. First, notice that even in this still relatively simple computation several minus signs occur. There will be a plethora of minus signs as we use integration by parts. The word plethora is a great SAT vocabulary work, and one meaning is "A superabundance; an excess". The minus signs are very easy to get confused about. A second thing is the apparent silliness of –(+C) becoming just "+C". Everyone does that. This just means an additive constant, and its specific name doesn't matter. I think 3C and –5C etc. will all get renamed +C during this semester during antidifferentiation exercises.

Example #3
Let's do ∫x³e^xdx. As I began to address this problem, I was aware of some unrest in class. So I abandoned the problem entirely, and decide instead to do

Example #n
Suppose n is a positive integer. What can we say about ∫xⁿe^xdx? Well, if u=xⁿ and v=e^xdx, then du=nx^n–1dx and v=e^x. And ∫udv=uv–∫vdu becomes
∫xⁿe^xdx=xⁿe^x–n∫x^n–1e^xdx.
No, I have not told you a formula for the antiderivative, but what I have done is explore the whole collection of integrals, and I hope that I have convince you that I could compute any specific "instantiation" of this collection (so I could find an antiderivative of x⁵e^x or even x¹²³e^x).

This sort of setup is called a reduction formula. In its simplest manifestation, there are a sequence of integrals, and the reduction formula shows how to work your way down the sequence. Properly speaking, you ought to also know how to do the lowest member of the sequence, which we do: that would be ∫xe^xdx or ∫e^xdx, both of which we know how to do.

Example #7
Let's try ∫arctan(x)dx. If you are new at this game, this integral is quite puzzling. Hey: this is the Integration by Parts lecture, so we should compute this integral using integration by parts. So the integrand, what's inside the integral sign, must be udv. What? Well, there are not many choices. In fact, I can only see one choice, and maybe it won't be too good: u=arctan(x). Then dv=dx, so that v=x. And du=[1/{1+x²}]dx (you must know the basic facts totally well, or you won't be able to get started!). So ∫udv=uv–∫vdu becomes
∫arctan(x)dx=arctan(x)·x–∫x[1/{1+x²}]dx.
This example is slightly more realistic. We now need to compute ∫[x/{1+x²}]dx so the integral we get in exchange for the original one is not at all "clear".

What is ∫[x/{1+x²}]dx? Actually, this can be computed with a fairly straightforward substitution. I will use w for the substitution variable since u is used almost always in the Integration by Parts formula. Well, if w=1+x² (the worst part of the integrand!) then dw=2x dx. We don't have 2x dx, but we do have x dx. So (1/2)dw=x dx, so that:
∫[x/{1+x²}]dx=(1/2)∫[1/w]dw=(1/2)ln(w)+C
You must learn to be totally sure about simple antiderivatives, such as the very last step. Now substitute back to get (1/2)ln(1+x²)+C. The final result is
∫arctan(x)dx=arctan(x)·x–((1/2)ln(1+x²)+C)
so that the antiderivative of arctan(x) is actually x·arctan(x)–(1/2)ln(1+x²)+C.

Example #8
I tried a definite integral, just to change pace a bit. What is ∫₀¹x·sqrt(1–x)dx? First I'll get an antiderivative.

So we need ∫x·sqrt(1–x)dx. I don't think that choosing a "successful" u and dv pair is totally clear here. If u=sqrt(1–x), then dv=xdx, so v=(1/2)x², and (I can do this!) du would be some stuff like –(1/2)(1–x)^–1/2. The vdu we would get looks worse or more complicated than the original. So I don't like this choice. What if we try u=x and dv=sqrt(1–x)dx? Well, du=dx. If dv=sqrt(1–x)dx, then I need to make an auxiliary computation:
sqrt(1–x)=(1–x)^1/2, and ∫(1–x)^1/2dx=–∫w^1/2dw (if 1–x=w, so –dx=dw and dx=–dw). Now ∫w^1/2dw=(2/3)w^3/2 (yeah, yeah, +C). So (don't forget the minus sign!) ∫(1–x)^1/2dx=–(3/2)(1–x)^3/2. Wow.
Therefore v=–(3/2)(1–x)^3/2. Then ∫udv=uv–∫vdu becomes
∫x·sqrt(1–x)dx=–x(2/3)(1–x)^3/2+∫(2/3)(1–x)^3/2dx.
The last integral, ∫(2/3)(1–x)^3/2dx, we can do with the substitution w=1–x. The answer will be –(2/3)(2/5)(1–x)^5/2. Now put things back together. Here is the final result.

∫x·sqrt(1–x)dx=–(2/3)x·(1–x)^3/2–(4/15)(1–x)^5/2+C.
I wouldn't want to check the answer by differentiating! What a mess. And I made the minus sign larger, because that error is embarrassingly easy to make: it is three minus signs concatenated, and I have fouled up such signs too many times myself.

We wanted a definite integral. The antiderivative is so confusing that I wanted to be able to check the answer. What should the answer "look like"? Here is a bunch of pictures. If you think through them, I hope you will see that the answer should be a positive number, certainly less than 1. (The bump is not symmetric, since sqrt is steeper than x, certainly.)

The definite integral
∫₀¹x·sqrt(1–x)dx=–(2/3)x·(1–x)^3/2–(4/15)(1–x)^5/2|₀¹=0(from x=1)–(–4/15)(from x=0)=4/15.
There are many opportunities for sign errors in this computation.

Another way Suppose we want ∫x·sqrt(1–x)dx. Hey: just directly try the substitution w=1–x. Then dw=–dx, and, since w=1–x, solve for x to get x=1–w. The integral changes: ∫x·sqrt(1–x)dx=∫(1–w)·sqrt(w)(–dw)=–∫w1/2–w3/2dw=–((2/3)w3/2–(2/5)w5/2)+C=–((2/3)(1–x)3/2–(2/5)(1–x)5/2)+C. The two antiderivatives we got are here: –(2/3)x·(1–x)3/2–(4/15)(1–x)5/2 and –((2/3)(1–x)3/2–(2/5)(1–x)5/2) I think these are the same functions, but the algebra needed is irritating. I can do it with calculus, though, in a second. Both of these are antiderivatives of the same function. They must differ by a constant (+C). If we plug in x=1, both of them are 0. So the +C is 0, and they are the same! By the way, here is what Maple tells me: > int(x*(1–x)^(1/2),x); 3/2 2 (2 + 3 x) (1 - x) - ---------------------- 15 Funny, huh (again, the same function, but written differently again!).

Example #9
Look at ∫sin(ln(x))dx. The integrand here is an absurd function. It is NOT a product, but a composition (read carefully!). It really does occur in applications (really real applications). You'll see it again when you study certain kinds of differential equations. Since this is the Integration by Parts lecture, we should compute using Integration by Parts (startling). Again, not many choices, so take u=sin(ln(x)) and dv=dx, so v=x and du=cos(ln(x))(1/x)dx (use the Chain Rule correctly) and ∫udv=uv–∫vdu becomes
∫sin(ln(x))dx=sin(ln(x))·x–∫x·cos(ln(x))(1/x)dx=sin(ln(x))·x–∫cos(ln(x))dx
because the x and 1/x cancel. We don't seem to have made any progress, but in fact, be courageous.

Do Integration by Parts again, this time for ∫cos(ln(x))dx, with u=cos(ln(x)) and dv=dx. Then du=–sin(ln(x))(1/x)dx and v=x, so ∫udv=uv–∫vdu becomes (be careful!)
∫cos(ln(x))dx=cos(ln(x))·x–∫x·[–sin(ln(x))(1/x)]dx=cos(ln(x))·x+∫sin(ln(x))dx

Suppose we push the formulas together carefully.

∫sin(ln(x))dx=sin(ln(x))·x–∫cos(ln(x))dx= sin(ln(x))·x–(cos(ln(x))·x+∫sin(ln(x))dx)

This is so very very very tricky. The original integral appears on the other side with an opposite sign!

∫sin(ln(x))dx=sin(ln(x))·x–(cos(ln(x))·x+∫sin(ln(x))dx)=sin(ln(x))·x–cos(ln(x))·x–∫sin(ln(x))dx

Therefore we can solve for the original integral by taking the minus indefinite integral to the other side and dividing by 2. Here is the result:
∫sin(ln(x))dx=(1/2)[sin(ln(x))·x–cos(ln(x))·x]+C.
Check the result if you like by differentiating. This is a ludicrous and wonderful example.

QotD
Find ∫arcsin(x)dx. Remember that the derivative of the arcsine function is 1/sqrt(1–x²).

Here and now, in Math 152
On Thursday evening I met with Mr. Jin and Mr. Yin and we discussed the course so far and the students. It is interesting that they both said, independently, there were numerous diligent, intelligent, and talented students. They also remarked, however, that they had already noticed students trying to get through the course minimally. Our job isn't educational coercion, but rather helping you learn. Perhaps you might find discussing college with one of these people (Mr. Jin or Mr. Yin) interesting and informative. This experience might broaden your appreciation of the opportunity involved in higher education.

How to get through this course minimally

How to get through this course with adequate proficiency

Definite integrals
These topics are covered in the first chapters of the text:

What to do? Some alternatives ...
So someone comes up to you on the street and says, "What's ∫_a^b f(x)dx?" Here if you're lucky the integrand, f(x), is some easily recognized function. There are a bunch of algebraic tricks which may be helpful, and we will begin to study them next time. But if this is the real world, maybe all you need to know is some numerical approximation to the value of the integral.

Numerical computation
Scientific and technical computation is present everywhere. The algebraic tricks we'll see don't always work and sometimes all you can do is get an approximate value.

Computer processors and memories have been improving in speed and size respectively very quickly over the last 40 years, but studies made of standard engineering problems requiring lots of computation show that the speedups which have come about resulted equally from faster processors and from advances in the computational algorithms.

You should know something about the algorithms. Yeah, much of the time you will (as I do!) just essentially "push a button" to get an approximate value. But some of the time, perhaps when you least want to worry about it, your method of choice may not work too well. So you should have some background knowledge. That's what we will do today. The two most important considerations are certainly how easy the method is to use, and how accurate it is.

The very beginning ...
I'll show you three methods of approximating definite integrals. The first is one you sort of know already, but I want to discuss it more quantitatively. The second and third methods are quite practical, and if you wanted to, you could actually compute fairly well with them. Programming them is not too hard. I will admit, though, that what I show you today is actually not what is used in, say, most hand-held devices (such as graphing calculators). Somewhat more elaborate (sophisticated?) methods are used there.

I'm going to try to use the notation in the text. So we'll investigate the definite integral of f over the interval [a,b]. For all three methods, we will chop up the interval into equal subintervals, say N subintervals (you're supposed to think that N will be large). Since the length of the whole interval is b–a, the length of each subinterval will be (b–a)/N. We will call this Δx.

There are N subintervals. The function, f(x), will have N+1 values at all of the endpoints of all of the subintervals. (There's one more f value then there are subintervals.) The function values will be labeled y₀, y₁, y₂, .... up to y_N. The y-value before the last one will be y_N–1. And, if you desperately wanted to know it, y₂, for example, is f(a+2Δx)=f(a+2(b–a)/N).

Level 0: Riemann sums, the definition Just approximate each piece by a rectangle. This is just the Riemann sum, which appeared in the definition of definite integrals. The side that's used doesn't matter very much, but to be specific here I'll choose the left-hand endpoint and evaluate f there, and then multiply it by the length of the subinterval.

Error estimate
What turns out to be important in this business is, first, how much work is the computation, and second, what sort of error estimate is valid. Let me copy what I did in class. I looked at an interval from 0 to small, so you should think of this as a very narrow interval (yeah, but the picture ...).
The true value of the integral is ∫₀^smallf(x)dx. The rectangle approximating is has height f(0), the value on the left-hand endpoint, multiplied by the width of the subinterval, small. This is f(0)·small. But notice that this is also an integral: ∫₀^smallf(0)dx (there is no x in this integrand!). So here is the error:
∫₀^smallf(x)dx–∫₀^smallf(0)dx
We are integrating over the same interval so this is the same as
∫₀^small(f(x)–f(0))dx.
The integrand is f(x)–f(0). One of the two results I said you had to know from the first semester was MVT, the Mean Value Theorem. I will use it here, and so f(x)–f(0)=f´(blah)x. I don't much care (and I actually don't know much about) where blah is. In fact, let me just assume that somehow I know some estimate for the largest value of the first derivative of f: I will call this, just as the textbook does, K₁. (As I mentioned in class, you could think about f(x) as modeling motion on a line with x=time. If f(x) is position, then f´(x) is velocity, and you could think of K₁ as a sort of speed limit.) Then
|∫₀^small(f(x)–f(0))dx|≤∫₀^smallK₁x dx.
I can actually compute the integral of x, and the result is that the error is at most K₁[small²/2].

This is the error in one piece of the rectangular approximation. But there are N pieces, and we can't assume that the errors cancel: in general, they won't, and if you're using this for real, you'd better assume the worst. So we need to multiply by N, and the total error is at most N· K₁[small²/2]. Things don't look too good, since N is large, but then several students observed that small is actually (b–a)/N so there are two N's "downstairs". One will cancel but this is the result.

Level 0
We approximate by (Δx)y₀+(Δx)y₁+(Δx)y₂+...+(Δx)y_N–1 (we only go up to N–1, we don't use the last height on f because this is the left-hand endpoint rectangles!). We can factor, and the result is:

Formula (Δx)(y₀+y₁+y₂+...+y_N–1)
Error Less than K₁(b–a)²/N where K₁ is some overestimate of |f´(x)| on all of [a,b].

The next trick, straight lines
Let's put y=f(x) on the interval [0,1] so computations will be easier. A likely better approximation would be to use a straight line, maybe tilted, something like A+Bx. Here A and B will be selected so that the values at 0 and 1 will agree with f(x). So the integral is easy:
∫₀¹A+Bx dx=Ax+(B/2)x²|₀¹=A+B/2.

We know that f(0) should be A+Bx at 0, which is A, and f(1) should be A+Bx at 1, which is A+B. In applications we'll be given f(0) and f(1), and we will need to compute A+B/2. What should we do? Well, notice (?) that
Value at 0+value at 1=A+(A+B)=2A+B, and this is twice what we want, A+B/2.
Therefore we should just approximate the definite integral by [f(0)+f(1)]/2. Another way to see this is just to notice that the figure I've drawn is four-sided (a quadrilateral) with two of the sides parallel: this is a trapezoid (which also names the method!) and the formula comes from the formula for the area of a trapezoid.

Putting it together
The picture attempts to illustrate what happens when a bunch of trapezoids is used. Due to the scale of the picture and the thickness of the lines, it is hard to see the error, but it is there. The error is there because the graph is not made of line segments: it curves (in fact, it curves, and somehow the amount the graph changes from being a line is measured by the second derivative, which magically appears below).

What does the trapezoid formula look like? Here:
[(Δx)/2](y₀+y₁)+[(Δx)/2](y₁+y₂)+[(Δx)/2](y₂+y₃)+...+[(Δx)/2](y_N–1+y_N)
If you notice, this can be reorganized a bit.

Trapezoid Rule
Here is the reorganized formula, and an error estimate (getting this error estimate is not as easy as the one before). The formula is called the Trapezoid Rule.

Formula [(Δx)/2](y₀+2y₁+2y₂+...+2y_N–1+y_N)
Error Less than K₂(b–a)³/(12N²) where K₂ is some overestimate of |f´´(x)| on all of [a,b].

Here what is important is the N² on the bottom, since that gets larger much faster than N alone.

Ha: use three sample points, and parabolas!
Back to y=f(x) on [0,1], and let's use A+Bx+Cx² where A and B and C are selected so that at 0 and 1/2 and 1 the values will be f(0), f(1/2), and f(1). Because parabolas curve and maybe most functions have graphs that curve also, we might hope that the integral of the parabola is closer to the true value of the integral of the function. Now
∫₀¹A+Bx+Cx²dx=Ax+(B/2)x²+(C/3)x³|₀¹=A+B/2+C/3.

Well, f(0) is A and f(1/2) is A+(B/2)+(C/4) and f(1) is A+B+C. Now one needs some inspiration in order to write A+B/2+C/3 in terms of those three quantities. Well, here is the inspiration. Suppose you take 4 times the value at 1/2: that's 4A+2B+C. Add on the values at 0 and at 1. The result is A+(4A+2B+C)+(A+B+C)=6A+3B+2C. (There are other ways of guessing, ways which are more systematic, and allow generalization, but I don't have time to show them to you: I am sorry.) Now 6A+3B+2C happens to be 6 times the integral, so we should divide f(0)+4f(1/2)+f(1) by 6. Sigh, really, this is not at all clearly!

Simpson's Rule
I want to lift the formula from the [0,1] case and use it in general. Let's "transport" the parabola setup to the more general picture, piecewise. So what do we need? First, we need to assume that N is even. O.k.: fine. But the width of the subinterval is Δx, and this corresponds to ... what? In the interval case, this corresponds to 1/2. So we should just divide by 3 here. So this is what happens in the first few pairs of intervals:
1y₀+4y₁+1y₂ gets multiplied by (Δx)/3.
1y₂+4y₃+1y₄ gets multiplied by (Δx)/3.
1y₄+4y₅+1y₆ gets multiplied by (Δx)/3.
·  ·  ·
1y_N–2+4y_N–1+1y_N gets multiplied by (Δx)/3.

If you observe the "structure" of the formula, you can see what happens. The 1 4 1 patterns overlap, and the "inside" 1's become 2's. So this is Simpson's Rule:

Formula [(Δx)/3](1y₀+4y₁+2y₂+4y₃+2y₄+...2y_N–2+4y_N–1+1y_N)
Error Less than K₄(b–a)⁵/(180N⁴) where K₄ is some overestimate of |f⁴(x)| on all of [a,b].

The implementation of Simpson's Rule is not much more work than the Trapezoid Rule or the "Rectangular Rule": just keep a running total of the odd and even heights inside the interval, multiply the odd final total by 4 and the even final total by 2, add on the end heights, and finish by multipling by a third of the interval length. Seriously, that's not too much work, especially since multiplications by 2 and 4 on a binary computer are just some shifts.

What happened to the error? The fantastic thing is that the error is some number divided by the fourth power of N. That's unexpected. I have gone through a direct proof of the error formula. It is not pleasant, but it can be done (there are some very clever proofs -- I only went through an "elementary" proof). What is amazing is that there's an N4 and not a cubic power. Why? Well, accidentally (more or less!) the Simpson's Rule formula is valid for x3: we're just lucky. What do I mean? Well, ∫01x3dx=1/4, certainly (since ∫x3dx=[x4/4]+C). And the idea was that we should compute f(0)+4f(1/2)+f(1) divided by 6. Well, if f(x)=x3, then f(0)=0, f(1/2)=1/8 and f(1)=1. So f(0)+4f(1/2)+f(1)=0+4(1/8)+1=3/2. When we divide that by 6 we get (3/2)/6=1/4: the correct answer! So we gain another power of accuracy, and the error formula is even smaller.

Some numerical evidence
I computed (well, a machine did, at my direction!) approximations to two examples and wrote the results on the board. 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(Π/4)–4(0)=Π, something else I presumably know. 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 (ln 2)
n=	Left Hand Rule	Trapezoid Rule	Simpson's Rule
10   100   1,000   10,000   100,000	0.7187714032 0.6956534302 0.6933972431 0.6931721811 0.6931496806	0.6937714035 0.6931534305 0.6931472430 0.6931471810	0.6931473748 0.6931471808 0.6931471803
∫01(4/{1+x2}) dx, approximately 3.141592654 (Π)
n=	Left Hand Rule	Trapezoid Rule	Simpson's Rule
10   100   1,000   10,000   100,000	3.239925989 3.151575986 3.142592487 3.141692652 3.141602654	3.139925989 3.141575988 3.142592487 3.141592652	3.141592652 3.141592653

Things to think about
Although in theory any one of these computational schemes can give you any accuracy desired, in practice there are other considerations. First, you can worry about the amount of computational time. Function evaluations take time on real machines. The lowest row (n=100,000) took about half a minute. Second, numbers are represented on real machines using floating point techniques, and arithmetic using floating point numbers passes along and increases errors. That is, sort if, there is "dirt" in the lowest digit or two, and the more things you do, the more the dirt is spread around. Therefore 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.

What you should know ...
The textbook discusses the Midpoint Rule in addition to the Trapezoid Rule and Simpson's Rule. You should know about these three, and have some feeling for how the error estimates work. Please note that there are calculator programs available for all three of them which work on some of the more common graphing calculators. This may help you with some of the textbook homework problems.

A specific example, with details I will try to give more details with this example. I should have done something like this in class. My function will be f(x)=sin(x2+5) and the interval will be [1,2]: those x's with 1≤x≤2. A graph of the function on this interval is to the right. What do you think is the value of &int12sin(x2+5)dx? Well, there's a little chunk of negative on the left-hand side, and the rest of the graph sort of is almost a triangle. It is a bit bigger than the triangle on the right, but the left negative chunk will bring the number down. If I had to "guess", I would say that the integral is about (1/2)(1)(1) (area of a triangle with base 1 and height 1). A sophisticated numerical integration program reports that the value is 0.5651369238 (indeed!). I want to try to estimate the value using the three methods mentioned in this lecture. I will also try to estimate the accuracy using the stated error bounds. I will use N=10. This means there will be N+1=11 y-values: y0=f(1), y1=f(1.1), y2=f(1.2), y3=f(1.3), y4=f(1.4), y5=f(1.5), y6=f(1.6), y7=f(1.7), y8=f(1.8), y9=f(1.9), and y10=f(2). For example, y7=f(1.7)=0.9993514088 (look at the picture and see that the value at x=1.7 is near the top!). Also, (b–a)/N=(2–1)/10=.1 and this is Δx. Level 0, rectangles sampled on the left Approximate value So here we want Δx(y0+y1+y2+y3+y4+y5+y6+y7+y8+y9). If "you" do all the arithmetic (I had a silicon friend do it) 0.5259155121 is the result. Error analysis Since f(x)=sin(x2+5), we get f´(x)=cos(x2+5)2x. How big can this get on [1,2]? Well, laziness is important. Look, the biggest that |cos(x2+5)| can be is 1. The biggest that |2x| can be on [1,2] is 4. So I think that one good overestimate for K1 is 4. Certainly b–a=1, and therefore we know K1(b–a)2/N=(4)(12)/10=.4 is an upper bound for the error. Comparison of theory and the approximation we got Well, 0.5651369238–0.5259155121 is about .04, so, in fact, the result is considerably more accurate than the theoretical bound. If you look at the picture, you may be able to see why. To the left of the max, the rectangles are below the curve, and to the right of the max, they are above the curve. So there is nice cancellation. We are fortunate. Level 1, Trapezoid Rule Approximate value So here we want (Δx/2)(y0+2[y1+y2+y3+y4+y5+y6+y7+y8+y9]+y10). My silicon pal reports that the result is 0.5604922111 which is nice, I guess. Error analysis We need the second derivative. Since f´(x)=cos(x2+5)2x, we get f´´(x)=–sin(x2+5)(2x)(2x)+cos(x2+5)2. Now I will be even lazier, and just analyze each part separately. Also, I am interested in absolute values, so I will discard any minus signs. I just want a lazy estimate of K2. What about |–sin(x2+5)|? Overestimate by 1. And |(2x)(2x)| on [1,2]: overestimate by 16. And |cos(x2+5)2| is overestimated by 2. Put these all together and I think that K2 is overestimated by 18. Then K2(b–a)3/(12N2) becomes (18)(13)/[12(10)2] and this is .015, our error estimate. Comparison of theory and the approximation we got True value minus this approximation is 0.5651369238–0.5604922111 and this is about .005, which is again less than our estimate. The estimate we got, by the way, is how people actually do this in practice. Generally no one wants to spend time getting the absolutely best possible error estimate -- we just want an approximation. But one caution: your approximation should always be an overestimate of the error since you want your result (the result you report, with "error bars") to be reliable. Level 1, Simpson's Rule Approximate value The weird 1-4-1 pattern needs to appear, with the inside 1's doubled: very strange. Therefore we want (Δx/3)(y0+4[y1+y3+y5+y7+y9]+2[y2+y4+y6+y8]+y10). Electrons again move and compute and 0.5651387559 is the reported result. Error analysis This will certainly be painful. We need the fourth derivative. Start with f´´(x)=–sin(x2+5)(2x)(2x)+cos(x2+5)2 and then get f(3)(x)=–cos(x2+5)(2x)3–sin(x2+5)(8x)–sin(x2+5)(2x)2. Let me clean this up a bit, and instead write f(3)(x)=–cos(x2+5)(8x3)–sin(x2+5)(12x). Now another derivative: f(4)(x)=sin(x2+5)(2x)(8x3)–cos(x2+5)(24x2)–cos(x2+5)(2x)(12x)–sin(x2+5)(12). I am trying here not to cheat and use any shortcuts. There is a bit of "simplification" which can be done: f(4)(x)=sin(x2+5)(16x4)–cos(x2+5)(48x2)–sin(x2+5)(12). Now estimate. Well, all of the sines and cosines with + or – will be estimated by 1. And 16x4 on [1,2] is at most 16·24=162=256, while 48x2 is overestimated by 48·22=192 and K4 is overestimated by 256+192+12=460. The error overestimate will be K4(b–a)5/(180N4)=460(15)/[180(104)]), wow. In decimals, this is .000256. So the Simpson's Rule approximation is at least this accurate. Comparison of theory and the approximation we got True value minus the approximation is 0.5651369238–0.5651387559, about .000002, so the Simpson's Rule approximation is actually closer by a factor of 100 times to the true value. This is o.k. to me. But I don't know why it is so much more accurate (that is, I can't give a heuristic reason): sorry. So I went to my faithful companion, Maple, and had it compute the fourth derivative, which I really did "by hand" above. And my/our computation was correct. Then I asked for a graph on [0,1]. Not so oddly there is some cancellation. The analysis above to get some value for K4 assumes there is no cancellation. And the K4 suggested by the graph is about 47.2 compared to 460 above. So one order of magnitude difference in the error observed is explained. But I still can't understand why the result is 10 times better than expected: sorry.

QotD

Time	Rate, in gal/hr
1:00	50
2:00	70
3:00	35
4:00	60
5:00	40

I gave some data ("flow rates" measured at various times) shown in the table to the right. Of course this had to be for toxic waste or something like that. Sigh. Then the definite integral would measure the total flow: flow rate multiplied by time interval, so the units work out o.k. (hours times gallons per hour). I wanted approximations for the definite integral.

• A (left) "Rectangular Rule" approximation for the flow (in gallons, which is what you get by integrating gals/hr over time: the flow rate integrated gives you the net flow) is, with Δx=1:
  (1)(50+70+35+60).
• A Trapezoid Rule estimate is, with Δx=1:
  (1/2)(50+2·70+2·35+2·60+40).
• A Simpson's Rule estimate is, with Δx=1:
  (1/3)(50+4·70+2·35+4·60+40).

If you've gotta have the numbers (?) I think they are 215 and 210 and 226.6666...

Problem from the textbook
Here is problem 48 (must be very difficult!) from section 6.4.

Use the Shell or Disk Method (whichever is easier) to compute the volume of the solid obtained by rotating the region shown (y=x–x¹²) about
(a) the x-axis (b) the y-axis.

Let's solve (a): we do it dx and get
∫_Left^RightΠ(Radius)²dx=∫_x=0^x=1Π(x–x¹²)²dx=Π∫₀¹x²–2x¹³+x²⁴dx=Π([x³/3]–2[x¹⁴/14]+[x²⁵/25])|₀¹=Π([1/3]–[2/14]+[1/25]).

Now for (b). If I wanted to do this dy, I would need to solve y=x–x¹² for x. I don't know a formula for this, and NO ONE ELSE IN THE WORLD DOES. This would not be a feasible method. But dx things are easy (thin shells). Here we go:
∫_Left^Right2Π(Radius)(Height)dx=∫₀¹2Π(x)(x–x¹²)dx=2Π∫₀¹(x²–x¹³)dx=2Π([x³/3]–[x¹⁴/14])|₀¹=2Π([1/3]–[1/14]).
Why is this problem 48? For many people the "natural" way to compute the volume for part b) is dy, and get the rotated dy strip as an object which is dy thick and which has cross-sectional area equal to the area between two concentric circles. This is all fine, but getting useful formulas for the two radiuses (radii?) involved is not possible. So this problem is difficult for psychological reasons!

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.

Hooke's Law
I gave a fantastic demonstration of Hooke's Law, using specially obtained titanium (?) wire hangers and specially obtained elasticium (?) bands on a specially crafted teak rod. Maybe now I can take a $10,000 tax deduction (doubtful). Anyway, an approximate demonstration of Hooke's Law was given. Hooke was a strange person.
It is somewhat interesting that validity of Hooke's Law can be shown even with a rather crude setup.

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, I sometimes make an "easy mistake". I 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. Generally, of course, stuff near the bottom compresses, but it turns out that water has 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).

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 the chain problem, the force was constant (3dx) and the distance varied.
• In the spring 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 the problem solution was organized the problem so that we took advantage of the geometry.

QotD
I asked people to compute ∫_θ=0^θ=Π/3e^2cos(θ)sin(θ)dθ. This integral is unrelated to what the lecture is about, but it continues my effort to learn more about what students know and to inform them that if they don't know something, they should learn about it. In this case, the problem is again one which could have been asked on a Calc 1 exam. It is a relatively simple substitution. I can also see if people are familiar with the exponential function and with sine and cosine, and even if they can find values of these functions. These are all things which should be known by students at the start of the course.
I hope student solutions look something like what follows.

To compute ∫_θ=0^θ=Π/3e^2cos(θ)sin(θ)dθ, maybe take u=2cos(θ). Then du=–2sin(θ)dθ, so that –(1/2)du=sin(θ)dθ, and we see
∫e^2cos(θ)sin(θ)dθ=∫e²(–1/2)du=–(1/2)∫e^udu=–(1/2)e^u+C=–(1/2)e^2cos(θ)+C. The value of the definite integral can now be computed:
∫_θ=0^θ=Π/3e^2cos(θ)sin(θ)dθ=–(1/2)e^2cos(θ)|₀^Pi/3=–(1/2)e^2cos(Π/3)–[–(1/2)e^2cos(0)]=(1/2)e²–(1/2)e¹.

How to do the first workshop
Some helpful comments are here.

A strategy to improve your learning of course material
It turns out that education, both teaching and learning, is complicated, and there are many confusing ideas about how to increase successful outcomes (that is, students learning well). Size of classes and specific teachers matter, but not so much. Style of instruction matters, which is one reason we have workshops. There is one learning strategy that leads to great improvements. That's students getting together in groups outside of class, say groups of 4 to 6 people, and studying together, doing homework and preparing for exams. Students working together are even better than visiting an instructor, because in such visits students tend to be too passive, and defer to the "authority" of the instructor. Study groups of students which stay on task can be your very best method leading to success in this course.
I recommend that you look at the published list of students and send messages to some, suggesting that a group be formed and meet for, say 2 hours at a time, maybe once or twice a week. Be serious about this, please, because the course material will get more difficult and there are always many distractions.

Another volume by slicing
Suppose we have a solid whose base is the same region we discussed last time (bounded by y=4x and y=x³ in the first quadrant), so that the cross-sections or slices of the solid which are parallel to the y-axis and perpendicular to the xy-plane are half-discs (semicircles) with diameters in the xy-plane. What does the solid look like, and what is its volume? Here of course we will slice the solid dx. The cross-sectional area is the area of half of a circle whose diameter is L. Well, the area of a whole circle with radius Π is Π·r². Here r is (1/2)L, so that the cross-sectional area is (1/2)Π[(1/2)L]². Repeating: the first 1/2 is because we have half of the area of a circle, and the second 1/2, inside the parentheses, is because we are changing diameter into radius. Here the setup is dx. The limits of integration look back to the first computation of the area, so that x=0 is the lower limit and x=2 is the upper limit. L is what I called Height then, so L=4x–x³. What is the volume of the solid? Here we go. Notice that the (Π/8) factor comes from Π(1/2)(1/2)² pulled out of the integral (we can do that with multiplicative factors!).
(Π/8)∫₀²(4x–x³)²dx= (Π/8)∫₀²16x²–8x⁴+x⁶dx=(Π/8)([16/3]x³–[8/5]x⁵+[1/7]x⁷|₀²=(Π/8)([16/3]2³–[8/5]2⁵+[1/7]2⁷–0
Again, this is the way I would leave this answer, please please please!!!. If you are a weirdo, and need "precision" then the exact value is Π([128]/[105]). I don't know why there is a coincidence with the previous computation. I didn't plan on it. Sigh.

Average value
Suppose a function f(x) is defined on an interval [a,b]. Then the average or mean value of f on [a,b] is defined to be
    (1/[b–a])∫_a^bf(x) dx.
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.

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. Then 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. This is known as the Monte Carlo method for estimating the value of the definite integral (really!) because it seems to use gambling ideas (random numbers).

Another 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 (!). The value of the integral doesn't change, but the approximation does: very strange. # 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            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 operations which are needed.) Such a technique is called a Monte Carlo method because of its dependence on randomness (Monte Carlo↔gambling/casino↔random). I'd 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. Here is further information about the Monte Carlo method. Getting efficient ways of generating random numbers (really, pseudo-random because they are obtained in various deterministic ways) is very important when using this method. Heck, this is also important in other applications, such as cryptography.

Solids of revolution
A collection of solid objects with central symmetry are known as solids of revolution. They occur often enough in applications that we should spend some time discussing them. Below are pictures of some of these solids. Revolving a rectangle around one of its sides gives a cylinder. Revolving a right triangle around one of its legs gives a cone. Revolving a semicircle around its diameter gives a sphere. Revolving a circle around an external line gives a torus (a doughnut).

Volume of a right circular cone
Formulas for the volumes of all of these solids can be gotten using the methods of today's class. I'll just do one, the cone. Really this is called a right circular cone, "circular" because the slices of the solid perpendicular to its axis of symmetry are circles, and "right" because the axis of symmetry is perpendicular to the base. The textbook formula for the volume of a cone with base radius R and height H is (Π/3)R²H. We can get this cone by revolving the area under a line segment going from (0,0) to (H,R) around the x-axis. The slope of this line is R/H, and its equation is y=(R/H)x.

If we slice perpendicular to the x-axis, by slices which are perpendicular to the x-axis, we get (approximately!) a bunch of circular discs which are dx thick and which have radius (R/H)x. So dV, a piece of the volume, is Π[(R/H)x]²dx. The total volume is the sum of the dV's added up from x=0 to x=R. So the volume is
∫_x=0^x=RΠ[(R/H)x]²dx=Π(R²/H²)∫_x=0^x=Rx²dx=Π(R²/H²)x³/3]_x=0^x=R=Π(R²/H²)R³/3=(Π/3)R²H
The final result is the formula which everyone "knows". The first equality is true because we can pull multiplicative constants out of integrals, and then the next is a simple use of FTC.

Computing the volume another way
The method about is called circular discs or coins. Let me now show you a different way. Let's slice the region dy: by many horizontal lines very close together (dy apart). Now take one of the rectangles which is dy high, and revolve it around the x-axis. The result is what's called a thin shell. We can cut it with magic scissors (?) and flatten it out. Yes, there will be some distortion of volume, but some volume is expanded and some is compressed, and it turns out that the effects cancel. The result is just about a rectangular parellelopiped. Its volume with be the product of the lengths of the edges. One edge is dy, and another is 2Πy, the circumference of the circle which the thin rectangle describes. What about the last edge? It is the distance between y=(R/H)x (or x=(H/R)y) and x=H. So this distance is R–(H/R)y. This chunk of volume, dV, is 2Πy[H–(H/R)y]dy. We need to add this up from bottom (y=0) to top, (y=R). Here we go:
∫_y=0^y=R2Πy[H–(H/R)y]dy= 2Π∫_y=0^y=R[Hy–(H/R)y²]dy= 2Π∫_y=0^y=R [Hy²/2–(H/R)y³/3]]_y=0^y=R=
2Π[H·R²/2–(H/R)R³/3]= 2ΠHR²[(1/2)–(1/3)}= (Π/3)HR²
So we get the same result (thank goodness!).

The Cauchy distribution
To the right is a graph of f(x)=1/(1+x²). This function is alway positive, and it is even (f(x)=f(–x)). When |x| is large (either positive or negative) the values of the function →0. The curve is sometimes called the Cauchy distribution. It is a very important object in statistics and physics, and is what's known as a "fat-tailed" distribution (really!). General information about it can be found here and here, and there is a neat applet involving the Cauchy distribution here.

Revolving a chunk around the x-axis
Suppose we want to revolve f(x)=1/(1+x²) from 0 to 1 around the x-axis. If we integrate dx, this is again a collection of thin discs or coins, and the radius at x of the coin is 1/(1+x²). The thickness is dx, and so the dV is approximately (Cross-sectional area)dx, which is Π(1/(1+x²)²dx. Now the total volume is ∫_Left^RightΠ(1/(1+x²)²dx, and Left is x=0 while Right is x=1. We can finish the computation with FTC if we know an antiderivative of (1/(1+x²)².

But clearly ∫(1/(1+x²)²dx=(1/2)(x/1+x²)+(1/2)(arctan x).

The word "clearly" is one of the worst to hear in a math course. It usually doesn't mean anything good. It could mean that the instructor doesn't understand what's going on, or at least doesn't understand it well enough to explain it. It could mean that the instructor is lazy. Etc. There are lots of things it could mean, and very few of them are good. What about here?

I do not think that the antiderivative given is at all clear. Well, but is it correct? How could we check the claimed result? The simplest way is to differentiate the claimed antiderivative and see if we get the original function. Differentiation, you see, is usually easy, and antidifferentiation is frequently much harder. Let's try. The derivative of x/(1+x2) using the quotient rule is [(1)(1+x2)–(2x)(x)]/(1+x2)2 and this simplifies on top to 1+x2–2x2=1–x2 while the bottom is (1+x2)2. The derivative of arctan(x) is 1/(1+x2), and if we multiply top and bottom by (1+x2) we get (1+x2)/(1+x2)2. Now add the two previous results. The bottom of both of them is (1+x2)2 and the tops work like this: 1–x2+1+x2=2. Remember, finally, that there were 1/2's in front of both terms which finishes our derivative check. So actually, yeah, we have just verified the clear fact, and we could now compute the integral with FTC. This doesn't really tell much, since we still don't understand where this strange formula came from. Sigh. I will show you in a few weeks. This is a magic trick.

Revolving a chunk around the y-axis
Suppose we want to revolve f(x)=1/(1+x²) from 0 to 1 around the y-axis. Again, let me integrate dx. We can use the thin shell method. The height of the shell is 1/(1+x²), and it is a distance of x from the y-axis making the circumference 2Πx, and its thickness is dx. The definite integral we must compute is this:
∫_x=0^x=12Π[x/(1+x²)]dx.

The QotD
Compute this definite integral. There were some hints offered: u and substitute and ...
Indeed, if u=1+x², then du=2x dx, so (1/2)du=x dx, and [x/(1+x²)]dx=(1/2)du/u. This antidifferentiates to (1/2)ln(u) or (1/2)ln(1+x²). Now don't forget the 2Π! So we need (2Π)(1/2)ln(1+x²)]_x=0^x=1=Πln(2)–Πln(1)=Πln(2). Again, I think this would be an allowable Math 151 question.

Comment
Several students computed the volume of this solid by integrating with respect to y. This is not totally straightforward, since the profile in y is a piecewise function, with one formula in [0,1/2] and another in [1/2,1]. Although they did compute the volume correctly, please note that on an exam they would have gotten no credit: they invented their own problem, and didn't do "my" problem which was to evaluate the given definite integral. Things are sometimes unfair.

What do you need to know?
Math 152 is the second semester of the three semester calculus sequence which is given for students who really need to know calculus (or who are quite curious!). In many respects it is the most technical, most computational of the three semesters. I personally think it has many fewer big ideas and lots of computational tricks. So what should you know to take this course?

1. You need to know quite a bit about specific functions. You should know about the derivatives and integrals and domains and ranges and graphs of powers (x^b), exponential (e^x), logarithm (ln x), trig functions (certainly sine, cosine, tangent, sometimes secant), inverse trig functions (certainly arcsine and arctangent, and sometimes arcsecant).

   Some poor engineering and physics majors may need to learn about the hyperbolic functions, so I will describe them later this semester.

2. The most important results to remember from earlier calculus are the Fundamental Theorem of Calculus (FTC) and the Mean Value Theorem (MVT). By FTC I mean a bunch of ideas, something like:
   • The definite integral is defined to be a limit of a sum, and the sum is gotten in the following strange way: slice up the domain interval into many subintervals; multiply function values in each subinterval by the length of that subinterval; add up the products. Amazingly, that process gets close to one specific number (converges in more precise language) as the largest length of the subintervals gets shorter. The specific limiting value is the definite integral.
   • One way to compute the definite integral is to find an antiderivative of the integrand (the function whose values you're looking at), and then taking the difference of that function's values.
   • The definite integral itself, when considered with a variable upper bound, has derivative equal to the integrand function.
   MVT is a slightly more slippery idea, because it is presented as a fact about secant lines and tangent lines (the average value [difference of the function values at the endpoints divided by the length of the interval] is equal to at least one instantaneous rate of change [derivative] in the interval). In fact, the way MVT is used computationally is as part of an equation, so f(a+h)=f(a)+f´(c)h etc.. We will relearn MVT and extend it enormously later this semester.
3. You should also know about how to use simple substitution to find antiderivatives. This semester we'll discuss more complicated strategies.

What this course is about
Math 152 is quite technical. The principle excuse for the technicality is that if you will have a career in science or engineering, the methods of the course are assumed to be part of the background of everyone. The course asks students to do some hand computation which can be very elaborate. In most workplaces, this computation will likely be strongly assisted by Maple or Mathematica or Matlab. (I use Maple myself but the others have similar qualities.) The excuse for having students learn to compute by hand is that they should then have some feeling for what strategies are useful and for what the "shape" of the answers is like.

More precisely (about as precise as I can be at this stage) here is what the course is about:

Part 1: some uses of the definite integral
Definite integrals are used to compute lots and lots of quantities: areas, volumes, masses, moments, pressures, ... hundreds of other things you will learn about.

I'll use this background color for material I didn't have time to do in class. A friend of mine wrote in a recently published book that "This method of finding areas is paradigmatic for an entire class of problems in which one is multiplying two quantities such as area = height x width, volume = cross-sectional area x width, moment = mass x distance, work = force x distance, distance = speed x time, or velocity = acceleration x time, where the value of the first quantity can vary as the second quantity increases." He is a great scholar, and uses words precisely. The word "paradigm" means "an example or pattern." The definite integral can be used to compute or approximate many quantities (many more than are in the list just given). That's the real power of calculus.

Very few of you will compute many areas professionally (or, actually, any areas professionally!). But I am fairly sure that almost all of you will need to investigate and compute many definite integrals. So initially we will consider some simple uses of definite integrals. All of the quantities will be computed by the sequence of ideas:
    Chop up → approximate → add → take limits
which turns out to be wonderfully useful.

Part 2: computing definite integrals
Once you have the creature (a definite integral), what is the vale of the integral? 38.5? Π/sqrt(2)? We need some methods.

• Some simple numerical methods
  In many cases, all that can be obtained is some approximation to the true value. And that is frequently good enough. Although, again, it is likely most students will just use "canned" programs or routines created by other people, students should have some basic idea of what's going on. Numerical routines can easily give incorrect or misleading answers.
• "Well-known" symbolic tricks
  For example, in 151 we showed some substitutions were found which allow antidifferentiation and then evaluation of the definite integral via FTC. In fact, human beings, using substitution, can find antiderivatives that TI-89's and even Maple and Mathematica can't! There are some other "standard" methods which students should know about. While they are implemented by computers, knowing these methods also gives insight into the behavior of the integrals (how big, how small, what to expect). There's lots of algebraic computation needed here.
• Geometric methods
  Here we learn how to use pictures (!) to get information about integrals. The context in which this is presented is solutions of differential equations. The pictures can sometimes allow predictions of asymptotic behavior (what happens to competing species over a long time, or what happens in a solution of various chemicals with several reactions going on) which would be difficult or impossible to predict numerically or algebraically. These methods are much more useful than students initially guess.

That's the plan of the course. There are a few minor additional topics, but please take a look at the syllabus. It will give you a good idea of where we are headed.

How do we teach it?
There are two lectures each week. The lecturer is correct 72.6% of the time, and students should point out the almost inevitable exxors. The diary may help. There is an additional meeting each week. Most of those meetings will have most of their time devoted to workshops. Small groups of students will investigate calculus problems. One of these problems will later be designated for a writeup. Attendance will be taken at every meeting of the course, and quizzes may or may not be given without further announcement. The workshops and writeups are an extremely important and useful part of your technical education.

Last semester I taught 251, the successor to this course. Every student with poor attendance did badly on exams and failed the course. This does not always happen, and certainly there were students who came to class faithfully and did not succeed, but there is very good correlation between class attendance and learning the material so that mastery can be displayed on exams.

    Do all of the homework! Work with other students! Read the textbook!

Let's begin
We will compute the area in the plane enclosed by y=x³ and the line segment which joins (0,0) and (2,8). The region is shown to the right. The points (0,0) and (2,8) are on y=x³, and the region is bounded by the line y=4x and the cubic curve. (This is not supposed to be difficult to see!).

First method
We slice the area by lines parallel to the y-axis. I think of the lines as dx apart. The region is chopped up into things which are almost rectangles, and the area of each piece is going to be HEIGHT·dx. We will get a good approximation to the total area by taking the Sum of the areas of these almost rectangular pieces. Of course what I mean by "Sum" is a definite integral, because that includes a limiting process which makes increases the accuracy of the approximations. So in this setup, we will have ∫_Left^RightHeight dx. We add up the slices from the left to the right of the region. Here we have slices which are dx thick, and we should try to write everything in terms of x. Let's see: the Left and Right are easy enough, because the left-most point of the region occurs when x=0, and the right-most point of the region occurs when x=2. What about the height of the rectangularish (that's a word?) slice? Well, the height will be the difference between the top measurement, that is, the curve which forms the boundary of the region with larger y values, and the bottom, the curve which forms the boundary of the region with lower y values. This is a rather simple region, and it isn't too difficult (I hope!) to see that Top(x)=4x and Bottom(x)=x³. In more complicated cases, the curves can cross, and to compute the area we may need to break up the definite integral so cancellation doesn't occur (if you don't know what I mean, please reread section 6.1, and learn again that the definite integral computes the net signed area). Please remember also that the Height will always be gifven by Top(x)-Bottom(x): this formula is valid no matter what the signs of the functions are, as long as Top(x) is geometrically higher than Bottom(x) (well, of course, since then Top(x)>Bottom(x)).

O.k., now we have ∫_x=0^x=2(4x–x³)dx. This definite integral is sufficiently simple that it can be computed directly using FTC. We "guess" an antiderivative, and here's the computation:
∫₀²(4x–x³)dx=2x²–x⁴/4|₀²=(2(2²)–2⁴/4)–0=8–4=4.

And again
Let's compute the area of the region in the plane enclosed by y=x³ and the line segment which joins (0,0) and (2,8). Huh? Yeah, well, I want to do it again, but I want to do it a different way. The more techniques you have available, the more likely you will be able to analyze a model successfully or complete a computation, even if your favorite method fails. In this case, I want to "dissect" or take apart the region dy.

Draw many horizontal lines in the region, and space the horizontal lines dy apart. The region is broken up into pieces, each of which is sort of a rectangle. The rectangle has some Width and a height of dy, so its area is just about Width·dy. Again we will Sum these strips, now from the bottom (the smallest value of y) to the top, the largest value of y. So the area of the region is ∫_Bottom^TopWidth dy. To carry out the computation, we need to write everything in terms of y. Well, the bottom value of y is 0 and the top value of y is 8. That's easy. What about the width? Each strip has a right and left border, and the distance of these borders to the y-axis varies with y. So we need these distances, which I called in the accompanying diagram Left(y) and Right(y). For example, the Left(y) border corresponds to the line segment, which has equation y=4x. Here x is the distance to the y-axis, so we need x in terms of y. This is a toy problem! We take y=4x and solve for x to get x=(1/4)y. Similarly, for Right(y), the equation y=x³ becomes x=y^1/3, and Right(y)=y^1/3. Now let's compute the area again.
∫_y=0^y=8(y^1/3–(1/4)y)dy=(3/4)y^4/3–(1/8)y²|₀⁸=((3/4)8^4/3–(1/8)8²)–(0)=(3/4)2⁴–8=12–8=4.

Comments
I'm really happy that the two results were the same. I mentioned in class that the second computation took me three tries in "rehearsal" before I got the answers to coincide. Embarrassment is a great motivator. It turns out (as you will see in Math 251) that these dual computations are really quite interesting. We got 12–8=4 and 8–4=4. Some much more complicated computations are really similar.

Yes, these is a toy computation. Let's see: I'll work with toys because I have limited time in class. I'll work with toys because the computational details would otherwise obscure the ideas. I'll work with toys, well, because I can do them. In order to finish the second computation, I needed to find the inverses of the functions x→4x and x→x³. I can do this. Many other functions do not have inverses which are easy to compute. I hope you recognize this problem.

The pictures I have been drawing are "stylized". That is, I tried to draw what I thought were the essential features (straight and curved lines where they should be straight and curved, for example). But I neglected some proportions. People who use computers and calculators sometimes are not sensitive to this feature, which even in this case leads to some distortion. To the right is a picture (produced with Maple) showing the true proportions. You can compare it to the other illustrations.

Now a volume
Suppose a solid object has as its base the region we've been describing: enclosed by y=x³ and the line segment which joins (0,0) and (2,8). Suppose the cross-sections or slices of the solid by planes perpendicular to the xy-plane and parallel to the x-axis are isoceles right triangles, with one leg of the triangle in the xy-plane. The right angle of the triangle is on the straight line. Sketch this solid, and compute its volume.

If you are new to this subject, much information is "encoded" in the paragraph above. I hope you know the base, the bottom of the solid. It is a flat region in the plane. The solid itself is above this region. The important sentence in the paragraph is

Suppose the cross-sections or slices of the solid by planes perpendicular to the xy-plane and parallel to the x-axis are isoceles right triangles, with one leg of the triangle in the xy-plane. ...

What can the solid object look like? Here I'll just make an attempt. If you are new to this stuff, your attempts may not be terrific the first few times, but they will improve if you practice. I think my version of the solid looks like what is shown to the right. It has a flat bottom and a flat side, and the third side is curved, but is made up of lots of hypotenuses (is that the plural? I also worry about hippopotamuses, or musi ...). Of course, such solids can't exist in real life, and, for example, engineers therefore should not worry about them. This is emphatically NOT CORRECT!!!: all sorts of machine parts, buildings, and molecules "look" like this. Even if your first attempt at drawing the solid is not agreeable, I hope that you will eventually feel you could build one by carving it out of wood or shaping clay.

Another word ... I don't think in the previous discussion that I emphasized enough that the cross-sections were to be perpendicular to the base. There are lots of ways to slice solids, and then to describe the solids. I could take a loaf of bread and slice it perpendicular to the base, the bottom of the loaf. Or I could be more artistic (?) and slice it in some way that was angled to the base ("obliquely"). Below is an attempt to show you what I mean. We are analyzing a solid which is similar to the left-hand loaf of bread, where the slices are perpendicular to the base.

How to compute the volume
We could get the total amount of bread by adding up the bread in all the slices. Similarly, we will get the total volume of our solid by imagining that it is dissected by numerous appropriate slices. To take advantage of the description, the slices will be parallel to the x-axis and perpendicular to the xy-plane. We also will think that the slices are about dy thick. Each individual slice is almost a nice solid, dy thick, with sides that are almost equal, having an area of (1/2)L² where L is the indicated length in the figure. So dV, a piece of the volume, will be about (1/2)L²dy. And the total volume is obtained by adding up the pieces of volume, V=∫dV.

Let's be more precise. We will integrate with respect to y. The limits of integration will be as in the second computation of volume: y=0 to y=8. The length labeled L here is exactly what I called Width there, so L=Width=Right(y)–Left(y)=y^1/3–(1/4)y, and we need to compute ∫_y=0^y=8(1/2)(y^1/3–(1/4)y)²dy. We can "pull out" the (1/2) but the best thing to do to the rest is "expand" it (multiply out). Here we go:
∫_y=0^y=8(1/2)(y^1/3–(1/4)y)²dy=(1/2)∫_y=0^y=8y^2/3–(2/4)y^4/3+(1/[16])y² dy=(1/2)((3/5)y^5/3–(3/[14])y^7/3+(1/[48])y³(1/2))|₀⁸=([128]/[105]) clearly

No, not clearly!!!. In fact, if you had to do this on an exam or for homework, please report the answer as (1/2)((3/5)8^5/3–(3/[14])8^7/3+(1/[48])8³(1/2)). That's a very very good way to report the answer. I got the number given because a silicon friend told me.

> int((1/2)*(y^(1/3)–(1/4)*y)^2,y=0..8);
                                                128
                                                ---
                                                105

QotD
I'll try to give a simple computation, usually towards the end of each lecture, which I'll call the Question of the Day. Students get full credit for handing in any answer. They may work with other students, use a calculator, look at their notes or textbooks, but they must hand in some answer. Yes, this a way to take attendance. It is also a way for me to see if people understand what I am doing. This is also a way for students, if they wish, to see if they can do (correctly!) what the lecturer considers to be a "simple computation".

And another volume Suppose we have a solid with the same base, so that the cross-sections or slices of the solid which are parallel to the y-axis and perpendicular to the xy-plane are half-discs (semicircles) with diameters in the xy-plane. What does the solid look like, and what is its volume? Hew of course we will slice the solid dx. The cross-sectional area is the area of half of a circle whose diameter is L. Well, the area of a whole circle with radius Π is Π·r2. Here r is (1/2)L, so that the cross-sectional area is (1/2)Π[(1/2)L]2. Repeating: the first 1/2 is because we have half of the area of a circle, and the second 1/2, inside the parentheses, is because we are changing diameter into radius. Here the setup is dx. The limits of integration look back to the first computation of the area, so that x=0 is the lower limit and x=2 is the upper limit. L is what I called Height then, so L=4x–x3. What is the volume of the solid? Here we go. Notice that the (Π/8) factor comes from Π(1/2)(1/2)2 pulled out of the integral (we can do that with multiplicative factors!). (Π/8)∫02(4x–x3)2dx= (Π/8)∫0216x2–8x4+x6dx=(Π/8)([16/3]x3–[8/5]x5+[1/7]x7|02=(Π/8)([16/3]23–[8/5]25+[1/7]27–0 Again, this is the way I would leave this answer, please please please!!!. If you are a weirdo, and need "precision" then the exact value is Π([128]/[105]). I don't know why there is a coincidence with the previous computation. I didn't plan on it. Sigh.

Maintained by greenfie@math.rutgers.edu and last modified 1/21/2009.