Course diary for Math 135, section F2, summer 2006

Tuesday, August 15
The instructor wrote the following list of topics:

Computation of limits ... Definition of derivative ... Computation of derivatives ... Log/exp problems ... Continuity & differentiability ... Linear approximation ... Max/min problems ... Optimization ... Related Rates ... Intermediate Value Theorem ... Mean Value Theorem ... Curve sketching ... Riemann sums ... Fundamental Theorem of Calculus ... Area ... Definite integral ... Substitution ... Initial value problems
All Math 135 final exams seen by the instructor have questions covering all or almost all of these topics. Sometimes, of course, the questions cover more than one of them, etc.

We did a few problems of various types. In particular we found the area under y=2x+x4 as x ranges from 0 to 2. That was almost easy. But then I asked people to sketch the graphs of y=x2 and y=1-x2, and to compute the area of the region enclosed between the curves. For this we needed to think: to translate the statements into pictures, do some algebra, and then compute some definite integrals.

We also discussed a number of problems from the exam handed out last time.

Monday, August 14
We wrote out solutions to all ten of the homework problems and then discussed some further examples of substitution. You had to be there to enjoy it!

If u=5x4, then du/dx=20x3, du=20x3dx, and (1/20)du=x3dx. Then sin(5x4)x3dx= sin(u)(1/20)du=-cos(u)(1/20)+C=-cos(5x4)(1/20)+C.

Here I would try u=17x3 so that du/dx=17(3x2) and du=17(3x2)dx. But look at the original integral. We "need" only x3dx, so du=17(3x2)dx becomes (1/[17·3])du=x2dx. The original indefinite integral in x-land, sin(5x4)x3dx, becomes the integral (1/[17·3])eudu. This is easy, and the result is (1/[17·3])eu+C. Now back to x-land: (1/[17·3])e17x3+C. You can always check by differentiating.

As I mentioned in class, students by now (we've worked through 10 or 15 examples!) should see how the examples are constructed:

  • e8x11 WHAT? dx: the WHAT? should be some constant multiple of x10. Then everything will work out nicely.
  • sin(ex) WHAT? dx: here the WHAT? should be some constant multiple of ex.

    This is, to me, a more complicated integral. I guess my "eye" focuses on the bottom of the fraction. I wish that were not there. So I will try u=2x3+6x. This yields du/dx=6x2+6 and then du=(6x2+6)dx. But the top of the fraction in the original integral is x2+1 (and there is a dx also, of course). Well, 6x2+6=6(x2+1). So (1/6)du=(x2+1)dx, and the original integral becomes (1/u)(1/6)du. The (1/6) is a multiplicative constant and "comes out" of the indefinite integral. The remainder is (1/u)du, which is ln(u)+C. So together we get (1/6)ln(u)+C=(1/6)ln(2x3+6x)+C.

    Here the psychological (?) effect was to make students "see" the likelihood of success when the substitution u=cos(x) is used. Then du=-sin(x)dx and -du=sin(x)dx. The original integral becomes [1/u](-du)=-ln(u)+C. So, back to x's, we have -ln(cos(x))+C.

    In fact, we have deduced the last entry in the Math 135 antiderivative table, since sine divided by cosine is tangent.

    f(x)f(x) dx=F(x)+C
    tan(x) -ln(cos(x))+C or ln(sec(x))+C

    -ln(cos(x)) is the same as ln(sec(x)) since -ln(A)=ln(1/A) and 1/cos is sec.

    Why the "+C"?
    The "+C" helps to remind people that there is a whole family of antiderivatives. Historically calculus started with problems from physics and geometry, and the "+C" is used in physics to pick out preferred antiderivatives. Let me show you an example.

    Throwing a ball
    Most people who have taken a high school physics course have seen problems similar to what follows:
    I stand on a cliff which is 10 feet high, and then I throw the ball straight up in the air at 20 feet per second. I will assume that (earth) gravity pulls the ball down, and the figure I remember for earth's gravity is that it pulls the ball down at 32 feet per second2. Suppose y(t) is the height in feet from the ground level and up is positive. t is the time in seconds after the ball is thrown.
    What is the position of the ball t seconds after it is thrown?

    How do the statements in the previous paragraph translate to calcspeak?

    EnglishcalcspeakPhysics words
    I stand on a cliff 10 feet highy(0)=10This is position.
    I throw the ball straight up in the air at 20 ft/sec.y´(0)=20This is velocity.
    Gravity pulls a ball down at 32 ft/sec2.y´´(t)=-32
    (the minus sign
    means down)
    This is acceleration.

    Since y´´(t)=-32, we get y´(t)=-32 dt=-32t+C. Now we plug in t=0, -32·0+C=20, so C=20. Therefore y´(t)=-32t+20.

    The next step backwards is y(t)=-32t+10 dt=-16t2+20t+C. This C will be identified using y(0)=10: Therefore -16(0)2+20(0)+C=10 and this C is 10.

    Therefore y(t)=-16t2+20t+10 is a formula giving the height above ground at time t (valid for t≥0).

    Some standard questions

    1. At what time is the ball highest? What is its height at that time?
      The ball is highest when y´(t)=0 (critical number!). Since y´(t)=-32t+20, -32t+20=0 implies that t=20/32. To get the height at that time, plug into y(t). The height is y(20/32)= -16(20/32)2+20(20/32)+10. This turns out to be (65)/4 feet.
    2. At what time does the ball hit the ground? What is the velocity of the ball when it hits the ground?
      The ball hits the ground when y(t)=0. (We assume [something I never really understood!] that the cliff got pulled away and the ball, coming down, doesn't hit the cliff.) So we need -16t2+20t+10=0 and (quadratic formula) t={-20+/-sqrt([20]2-4(-16)(10))}/(2[-16]). What's under the square root is 400+640=1040 and this is positive, so the quadratic has two real roots. Amusingly, since the bottom is negative, the root we want is tied to the minus sign (in the +/- before the square root). The two roots are numerically about -.383 and 1.633: the answer here is 1.633. The velocity at time 1.633 is about -32.256, which means the ball is moving down.

    As I mentioned in class, equations such as y(0)=10 and y´(0)=20 are called initial conditions. In practice, initial conditions are used to specify one out of a whole bunch of candidate antiderivatives. The simplest and first models in ecology and epidemiology and business are frequently differential equations with initial conditions: these models are called initial value problems. The initial condition could represent something like the lemming population at a certain time or the amount of money available to invest at a certain time. The differential equation is then a way of describing how the quantity changes with time.

    Example #1
    Suppose dy/dx=-5ex/4+2e3x and y(0)=7. Find a formula for y(x).
    We find the possible formulas by computing the indefinite integral:
    Each part of this integral can be computed with a substitution (but a different substitution for each part!):
    So for -5ex/4dx please use u=x/4 so that du/dx=1/4, du=(1/4)dx, and dx=4du. The integral then becomes -20eudu=-20eu+C and back in x-land this is -20ex/4+C.
    And for 2e3xdx please use u=3x so that du/dx=3, du=3dx, and dx=(1/3)du, The integral then becomes (2/3)eudu=(2/3)eu+C=(2/3)e3x+C.
    The formula for y(x) is y(x)=-20ex/4+(2/3)e3x+C and the initial condition y(0)=7 becomes -20+(2/3)+C=7, so that C=27-(2/3)(79/3) and y(x)=-20ex/4+(2/3)e3x+[79/3].

    Example #2
    Suppose y´´(x)=-3cos(x)+4x2, and we know that y´(0)=5 and y(0)=-2. Find a formula for y(x). Here we need too antidifferentiations, one going from y´´(x) to y´(x) and the next, going from y´(x) to y(x). After each transition, we will use an appropriate initial condition to pick out the specific function of interest.
    Since y´´(x)=-3cos(x)+4x2, y´(x)=-3cos(x)+4x2dx=-3sin(x)+(4/3)x3+C. Since we know y´(0)=5, if we "plug in" x=0, we get -3sin(0)+(4/3)03+C=5. This C is 5, and the specific formula for y´(x) we get is -3sin(x)+(4/3)x3+5.
    Since y´(x)=-3sin(x)+(4/3)x3+5, y(x)=-3sin(x)+(4/3)x3+5 dx=3cos(x)+(1/3)x4+5x+C. Plug in x=0 and use y(0)=-2. The result is 3cos(0)+(1/3)04+5·0+C=-2. Please be careful about this equation, since the result is 3+C=-2, and therefore C=-5. The desired formula is then y(x)=3cos(x)+(1/3)x4+5x-5.

    Thursday, August 10
    Students volunteered to write solutions of the homework problems. The word "volunteer" is used ... in the sense of impressment (this means "enforced service in the army or navy").

    Rewriting FTC
    FTC stands for Fundamental Theorem of Calculus.

    I remarked that most of what is important in a first semester calculus course is included in FTC and MVT. The latter is the Mean Value Theorem which I rewrote.
    Rewriting MVT
    [f(b)-f(a)]/[b-a]=f´(c) for some c between a and b.

    Functions which have zero derivative
    If the derivative of a function is always 0, then the MVT tells us something interesting. We may not know what much about the c which appears, but "always 0" means f´(c)=0. Then the top of the fraction on the left-hand side of the equation is 0, so that f(b) must always equal f(a). This means the function's values are constant -- they can't change! So any function whose derivative is always 0 is a constant function.
    We can restate this using the physical model, where f(t) is the position of a particle, then f´(t) is the velocity of the particle. Well, if the particle's velocity is always 0, then "clearly" (?) the particle never changes position. In the history of math, this "clearly" took over a century to understand. This is because the velocity is the instantaneous rate of change of position (the limit!) and it wasn't clear, really, how to relate the instantaneous rate of change with the position at two times. The Mean Value Theorem became an accepted part of calculus quite a while after the subject started.

    Functions which have the same derivative
    Suppose we know two functions, F(x) and G(x), and that these functions have the same derivative for all x: F´(x)=h(x) and G´(x)=h(x) for all x. Well, the difference F(x)-G(x) will have derivative w(x)-w(x) and so it is 0 for all x. Therefore F(x)-G(x)=C (some constant) for all x. So F(x)=G(x)+C. The geometric consequence is shown to the right. I hope that the picture reflects the following implication:
    if two functions have the same derivative for all x, then the graphs of the functions are up/down (don't know which) translations of each other. The curves must be parallel.

    All antiderivatives ...
    If you know one antiderivative, then you know them all. What I mean is that if F(x) is an antiderivative of f(x), then all antiderviatives of F(x) have the form F(x)+C. Example (stupid)
    Well, sqrt(x3+17)-9 has derivative equal to (1/2)(x3+17)-1/2. Therefore a complete description of all antiderivatives of (1/2)(x3+17)-1/2 is sqrt(x3+17)-9+C where C is any constant.
    People didn't like this because the darn -9 looked silly. Sometimes, though, you can't detect the "silly" part. We will see further examples.

    F(x) is called an antiderivative of f(x) if F´(x)=f(x).
    F(x) is called an primitive of f(x) if F´(x)=f(x).
    F(x) is called an indefinite integral of f(x) if F´(x)=f(x).
    Yes, all of these words and phrases are commonly used for the same thing: I'm sorry. And there is notation. If F´(x)=f(x) people frequently write f(x) dx=F(x)+C.
    There are no numbers at the top and bottom of the long S, the integral sign. So the integral does not refer to a definite value or area or region ... it is indefinite. And most people write "+C" to remind themselves that "the" antiderivative can refer to any one of a large family of functions. This turns out to be important in some other computations. In definite integral computations, most people omit the +C because evaluating the result at the top minus the result at the bottom cancels +C's.

    Antiderivative information
    Any information about derivatives is information about antiderivatives: just reverse the connection. So our table of antiderivatives on the formula sheet can be turned into a table of antiderivatives. Of course, some folks like the table to look better. For example, the derivative of xn is nxn-1. "Thus" the antiderivative of nxn-1 is xn+C. But people like the table to be more user friendly. They would prefer to be able to find the antiderivative of xn. Well, this likely would result from differentiating xn+1. But that function's derivative is (n+1)xn: so we need to fix up xn+1 by 1/(n+1). Therefore the antiderivative of xn is [1/(n+1)]xn+1. But when division cccurs, check the bottom to make sure embarrassing things aren't happening: hey, this will only work if n is not equal to -1. So we need to think about the antiderivative of x-1=1/x: but that's ln(x) (how wonderful). Anyway, the child's table of antiderivatives using the information we have accumulated so far might look like this:

    f(x)f(x) dx=F(x)+C
    0 C (a constant)
    if n not -1
    x-1=1/x ln(x)+C
    k a constant
    if F(x) is one antider. of f(x)
    f(x)+g(x) F(x)+G(X)+C
    if F(x), respectively G(x), is one antider. of f(x), respectively G(x).
    sin(x) -cos(x)+C
    cos(x) sin(x)+C
    ex ex+C
    ax [1/ln(a)]ax+C

    Example #1
    Here please notice that 3 is 3x0 and 5/x8 is 5x-8. If you have this in mind, then the antiderivative can just be written out, glancing up at the table for help:

    Example #2

    Example #3
    Here you need to "decode" sqrt(x): it is x1/2 and then this can be handled with the xn entry in the table. Notice that the antiderivative of sine gives a minus cosine!

    Example #4
    Here the problem is that ... we may not know what to do! The best thing is to "square out" the integrand (that's what the function to be integrated is called). The result:
    Therefore we need to find the antiderivative of x4+2/x+x-6. Each piece of this can be handled by referring to our table and simple arithmetic. The result is (1/5)x5+2ln(x)-(1/5)x-5+C. I think this is a bit tricky.

    Another wrinkle: #1
    Let's consider (x3+17)83x2dx. If you look long enough at this, maybe you can see the function the integrand should come from must be something like (x3+17)9. When this function is differentiated, the Chain Rule will "spit out" the derivative of the inside multiplying 9(x3+17)8. The derivative of the inside is 3x2, so the anbswer is (1/9)(x3+17)9+C.

    Another wrinkle: #2
    How about (3x+2)100dx? I could imagine, barely, computing this integral by "expanding" the 100th power. The result would be an enormous polynomial, and then I could antidifferentiate each piece of the polynomial, one by one. Or I could think a bit: how could we get a function of the form (3x+2)100 as the output of differentiation? Well, maybe one higher power could be considered. Thus we could guess at (3x+2)101. But when we differentiate this, the 101 comes out in front, and a 3 comes out in back (Chain Rule). I can fix up these little problems, and I bet that (3x+2)100dx=[1/{101·3}](3x+2)101+C. You can check this guess by differentiating. The guess is correct.

    Another wrinkle: #3
    Now this: 4e5x+6dx. Here again we could guess and correct. The process will yield (if we make no mistakes!) the following result: (4/5)e5x+6+C. Again, you can check this guess by differentiating. The guess is correct.

    This is sort of the chain rule backwards and is a way of keeping track of the constants which arise in computations such as those "guesses" I just wrote. The idea of this method is to prevent mistakes that guesses (especially those guesses which are too elaborate!) sometimes cause. Let me go over the three examples that I just did, with a slightly different language.

    Wrinkle #1 rewrinkled
    (x3+17)83x2dx. Take u=x3+17. Then du/dx=3x2 and (if you appreciate the symbolism!) du=3x2dx. Now let me Translate the integral I started with from x-land to u-land. Each piece involving x needs to be changed to a piece in terms of u. So:
    Now I will integrate the indefinite integral in u, and the result is (1/9)u9+C. And now use our dictionary to return to x-land from u-land: (1/9)(x3+17)9+C. This is the answer we guessed at and can confirm by differentiation. I invented this example to show off the substitution method, and the success therefore should not be too surprizing.

    Wrinkle #2 rewrinkled
    (3x+2)100dx: here u=3x+2 so du/dx=3, du=3dx, and (1/3)du=dx. The integral goes from x-land to u-land:
    Now u100(1/3)du=(1/3)u100du=(1/3)(1/{101})u101+C. And back to x-land, using the dictionary for this problem: (1/3)(1/{101})(3x+2)101+C. I hope that you see the substitution method is an effort to run the Chain Rule "backwards".

    Wrinkle #3 rewrinkled
    4e5x+6dx. Here I'll try u=5x+6 so that du/dx=5 and du=5dx. I notice that I actually have 4dx "left over" so I will modify the equation du=5dx. It becomes (1/5)du=dx and then (4/5)du=4dx. The result for translation from u-land to x-land follows: 4 e5x+6dx=eu(4/5)du
    Then the antiderivative is (4/5)eu+C and, back in x-land, this is (4/5)e5x+6+C.

    All of these examples are arranged to work well with this substitution "trick". Essentially we are working with f(u(x))u´(x)dx and rewriting it as f(u)du. If we then "happen" to know an antiderivative of f is F, we can "see" that f(u)du=F(u)+C, and the original indefinite integral gets solved with f(u(x))u´(x)dx=F(u)+C. The method is successful sufficiently often that it is the first general antidifferentiation "trick" shown to calculus students. There are several questions which occur almost immediately. How should u be "chosen"? Well, I usually first try some "mess" whose disappearance will make the integral "easier". And I look, at the same time, for other pieces which will combine to be close to the resulting du. And, no, it doesn't always work. There are integrals which can't be done using this sort of trick. Then other tricks are tried. And sometimes they don't work.

    Today's last try: sin(7x3+17)x2dx
    If u=7x3+17, then du/dx=21x2, and du=21x2dx. But we have x2dx in our integral, so we need to play a bit: du=21x2dx turns into (1/{21})du=x2dx, and sin(7x3+17)x2dx in x-land becomes sin(u)(1/{21})du in u-land. This indefinite integral I can do by using the table. The result is -cos(u)(1/{21})+C and back to x-land: -cos(7x3+17)(1/{21})+C.

    You can always check the answer. Differentiation is basically easy to do, and so if someone suggests an antiderivative, verifying to suggestion is usually not too hard.

    Wednesday, August 9
    I urged students to view much of what was discussed today as entertainment of the most magical kind. The aim is to show how at least three different civilizations, mostly independently, computed definite integrals (their integrals were area and volume numbers). Then I want to show how analyzing a harder problem amazingly gives what is an easier method (in many examples) for computing definite integrals.

    Back in the good old days (China/Greece/India ...)
    This is the approach many very clever people worked on for centuries. I will start with a problem whose answer we already know, and hope that we will get the answer we anticipate. I hope then that the approach will be successful when we try a problem whose answer is not generally known.

    Area of a triangle
    Let's find "the area under y=x when x is between 0 and 1". This is the language used for the area indicated in the picture to the right. It is a triangle whose vertices are (0,0) and (1,0) and (1,1). The base and height are both 1 unit long, and the area is half the product of the base and height, so the area should be 1/2.

    I'll split up the interval [0,1] into n equal-lengthed pieces. So this means the intervals will be determined by
    0 1/n 2/n 3/n ... (i-i)/n i/n ... (n-1)/n n/n (yes, that's equal to 1).
    The length of each subinterval is 1/n. The ith subinterval is determined by its endpoints: [(i-1)/n,i/n]. I'll approximate the area of the triangle by "constructing" a bunch of approximating rectangles and computing the sum of their areas. The ith approximating rectangle will have the height of the formula y=x at the right-hand endpoint of the subinterval [(i-1)/n,i/n], so its height will be i/n. The area of the ith rectangle will therefore be height·base, and this is (i/n)·(1/n), which is i(1/n2). There are n rectangles, and the sum of their areas is
    Trying to describe this sum in a culture which does not know algebra is an intriguing and difficult task: how many words and sentences and ... would be necessary to describe what we wrote so compactly above? And, since we do have algebra, I hope that you can see that 1/n2 can be "factored out" and the expression becomes what follows:

    For example, 1+2+3+4+5+6, as I computed in class, is equal to 21. But I am not really interested in what happens for n=6. I want to examine the "asymptotic behavior" as n gets large. The most efficient way to do that is to get some neat formula for 1+2+3+...+i+...+n. This is what's called an arithmetic progression. You may remember a neat formula, or you might look at the following:
    1 + 2 + 3 + 4 + 5 + 6
    6 + 5 + 4 + 3 + 2 + 1
    If we add up the "columns" -- that is, add things vertically, then the result is
    7 + 7 + 7 + 7 + 7 + 7
    When the entry on the first line goes up, the entry on the second line goes down. So the column sums are all the same. There are six of them, so the sum is 6(7). But this is the sum of two repetitions of 1+2+3+4+5+6, so that sum should be the same as 6(7)/2. Well, this is correct. And, indeed, 6(7)/2=42/2=21, the same answer as we got by direct computation. Therefore I can suggest the following formula:
    (1+2+3+...+i+...+n=n(n+1)/2: a magic formula!

    Now I can write the sum of the areas of the approximating rectangles, (1+2+3+...+i+...+n)·(1/n2), using this formula. The result is:
    We know algebra, and much of the whole course has been an effort to coax you through the next few equalities.
    The last step is accomplished by multiplying the top and bottom of the expression by 1/n2. But I want to know what happens as n-->infinity. Well, clearly [1+{1/n}]/2-->1/2 since 1/n-->0. The word clearly is used here not in truth, but to signal, almost, hey, something significant has happened, and the whole process may not (probably is not!) "clear".

    We computed an expression for the sum of the areas of the approximating rectangles. Then we computed the limit of this expression. The area of the triangle is the limit of the area of the approximating regions, so the area of the triangle should be 1/2. We know this already. I did not use the language of the past few class meetings, but I hope you "see" the appearance of regular partitions, right-hand endpoints as sample points, Riemann sums approximating a definite integral, etc.

    Area of a parabolic region
    I'm going to try the same approach with "the area under y=x2 when x is between 0 and 1". Here the region whose area I want to compute shares two straight line segment boundaries with the right triangle (the x-axes and part of the line x=1). The hypotenuse is replaced by a part of the parabolic arc y=x2 connecting (0,0) and (1,1). My approach to computing this area will be to imitate the previous work as much as possible. Alll of this could have been done (with no algebra!!!) by apppropriate representatives from the three civilizations mention. There will be an interesting increase in difficulty at one stage, but let me begin:

    I'll split up the interval [0,1] into n equal-lengthed pieces. So this means the intervals will be determined by
    0 1/n 2/n 3/n ... (i-i)/n i/n ... (n-1)/n n/n (yes, that's equal to 1).
    [This paragraph copied exactly.]

    The length of each subinterval is 1/n. The ith subinterval is determined by its endpoints: [(i-1)/n,i/n]. I'll approximate the area of the parabolic region by "constructing" a bunch of approximating rectangles and computing the sum of their areas. The ith approximating rectangle will have the height of the formula y=x2 at the right-hand endpoint of the subinterval [(i-1)/n,i/n], so its height will be (i/n)2. The area of the ith rectangle will therefore be height·base, and this is (i/n)2·(1/n), which is i2(1/n3). There are n rectangles, and the sum of their areas is
    ... I hope that you can see that 1/n3 can be "factored out" and the expression becomes what follows:
    [Here some changes needed to be made, and I tried to indicate them with a change in type font and color.]

    Now let me abandon copying for a while. I'd like to try to find a compact formula for the sum 12+22+32+...+i2+...+n2. I computed 12+22+32+42+52+62 in class. I think the result was 1+4+9+16+25+36=91. I can't just reverse the sum and add things up, since the gaps between the successive numbers change. There is a magical formula for this sum. I found this link to a web page which discusses how the formula is verified. I want to use the formula, and am willing to accept it. So:
    12+22+32+...+i2+...+n2 is equal to [n(n+1)(2n+1)]/6
    I did verify that if we plug in n=6 into [n(n+1)(2n+1)]/6 the result is [6(7)(13)]/6 which is 7(13) which is 91, and I had added up 12+22+32+42+52+62 and also got 91 as the answer. No, this is not a "proof" but at least the formula agrees with the data we have. Anyway, we go back to the old folks:

    Now I can write the sum of the areas of the approximating rectangles, 12·(1/n3)+22·(1/n3)+32·(1/n3)+...+i2·(1/n3)+...+n2·(1/n3), using this formula. The result is:
    We know algebra, and much of the whole course has been an effort to coax you through the next few equalities.

     n (n+1) (2n+1)    (1+{1/n})(2+{1/n}) 
    ---------------- = ------------------
         6 n·n·n               6
    (This was done by dividing each of the factors of the top by one of the n's in the bottom.)
    ... But I want to know what happens as n-->infinity.

    I'll stop copying. What happens as n-->infinity? Well, clearly since 1/n-->0, our approximating area amount --> 2/6=1/3. So 1/3 "is" the area (I guess 1/3 square units, as mentioned in class).

    The analysis of the area problem is very clever. But if other, similar problems are thought about, then it seems like we will need a magical formula in every case. Imagine doing this with trig functions or exponential functions (the old-time wizards did handle these cases). But I will show another approach to this problem.

    Another way: do a harder problem
    The approach is almost paradoxical. You study a more difficult problem, and solve that problem. As a result, the problem you started with is solved easily! The whole process is mechanized, almost. Here in this case, we want to compute the area of the parabolic region, which had boundary goiven by the horizontal axis and the parabolic arc and the vertical line segment. Let me temporarily relabel the horizontal axis with t (time?). Then the curve becomes y=t2. I will define a new function, A(x), in the following way:

    A(x) is the area under the parabola from 0 to x. So A(x) is the area bounded by the curve y=t2, y=0, and t=x.
    We want to compute A(1).

    Still photography versus moving pictures?
    Maybe you can think about the difference between computing A(1) and investigating A(x) as a difference between the static/still and kinetic/moving viewpoints. There are more pictures, and maybe the insight of how the pictures "connect" will allow us to understand just one of the pictures better.

    What can we say about this function?
    I want to learn about A(x). This is what I will learn.

    One value of A(x)
    I know A(0). This is an almost trivial observation. If I move the x-value, the right size of the picture, back towards 0, then the region just becomes a dot. Its area will be 0. So A(0)=0.

    Differentiate it
    One thing that is learned in a calculus course is to differentiate everything in sight. The derivative of A(x) uses A(x+h)-A(x) divided by h. We usually think of h as small. Suppose that h is a small positive number (the same logic works with some changes of inequalities if h is negative). Then A(x+h) is the area of a region which is almost the same as the region whose area is A(x). We just move the boundary line segment on the right somewhat more to the right. That is, we move the line segment to x+h. What about A(x+h)-A(x)? This is the area of the region which is below the parabola, above the horizontal axis, and between x and x+h. What can we say about it? Since "squaring" is particularly simple function (well, for example, it is increasing) I know that the height in the slice between x and x+h will be between x2 and (x+h)2. The width of the slice is h, so that
    Suppose we divide this inequality by the positive number h. The result is:

          A(x+h) - A(x)
    x2 < --------------- < (x+h)2
    What happens as h-->0? Certainly the x2 doesn't change. On the farthest right, consider the term (x+h)2: as h-->0, this term must -->x2. Huh. Well, the term is the middle is squeezed between two formulas which both -->x2 as h-->0. So it should also -->x2. Therefore we have "computed" A´(x): it is x2.

    So we can solve the problem!
    We now know that:

  • A(0)=0.
  • A´(x)=x2. Well, if A´(x)=x2, we can guess what A(x) might be. It might be (1/3)x3+98. It might be (1/3)x3-22. It might be (1/3)x3+Pi. Etc. There are many candidates for A(x). Which one should we pick? The condition A(0)=0 forces us to choose A(x)=(1/3)x3 as the only answer. But then A(1)=(1/3)13=1/3. This is the answer. Certainly it coincides with the answer given by the wonderful magicians of long ago. But this answer is an indication of a systematic approach to such problems. Here is the very important statement of that approach.

    The Fundamental Theorem of Calculus (FTC)

    First part Suppose f(t) is a continuous function, and F(x) is the definite integral of f from t=a to t=x.
    Then F(x) is differentiable, and its derivative is f(x).
          In algebra d/dxaxf(t) dt=f(x)

    Second part If F(x) is any antiderivative of f(x), then the definite integral of f(x) from x=a to x=b is F(b)-F(a).
          In algebra abf(t) dt=F(b)-F(a) if F´(x)=f(x).


    There is even special notation for "F(b)-F(a)" because it occurs so often. Usually people will write F(x)]ab or F(x)|ab. The text uses the second version, without the little hooks at the end. The text used when I was taught calculus had the hooks, so surely I will slip sometimes and write them.

    Any antiderivative?
    We decided in the previously discussed area problem that the area, which equals 01x2 dx, must be 1/3. We actually found A(x)=(1/3)x3 and got 1/3 because it was A(1). But what if I wanted (?) to use F(x)=(1/3)x3+73 as my antiderviative of x2. Then the FTC says that the area should be F(1)-F(0). But this is ((1/3)13+73)-((1/3)03+73) and the 73's just cancel each other, and we again get 1/3 as the answer. So in computing the definite integral, the constants involved in specifying the antiderivative cancel. The constants are important in connection with other computations, as you will see.

    Example #1
    x5 on [2,5]

    Example #2: how big is the hump?
    Look at the curve y=sin(x).
    What is the area under the first bump of this sine curve?
    The accompanying picture is supposed to be very sloppy. I wanted to emphasize the need to have some size of the answer. The answer should not be very large (456 is too darn large). The answer should not be very small (1/456 is too darn small).

    Although I sincerely hope that in most situations in the future you will have computational help (machines!) you should have some idea of the size of your answer, so that when the machine reports something really far from the correct answer, you'll know enough to worry a bit. So I try to imagine the answer. Look at the (better!) picture to the right. The sine bump is inside a box whose height is 1 and whose width is Pi, so the area, when we compute it, should be less than Pi. Also the area under the bump contains a triangle (as shown) whose height is 1 which has a base of width Pi, so the area of the bump should be more than Pi/2. Now maybe I will compute the area.

    So I need 0Pisin(x) dx. I know a function whose derivative is sin(x). Such a function is -cos(x) (keep track of minus signs!). So the area is -cos(x)|0Pi. This is -cos(Pi)-[-cos(0)]. This is -(-1)-(-1)=2 (keep track of minus signs!). You can check that 2 is between Pi and Pi/2 as we thought it should be.

    Example #3
    I computed this definite integral:
    -11x3-x dx. According to FTC, I can evaluate this by first finding an antiderivative, and I guess the antiderivative:(1/4)x4-(1/2)x2. So I must compute (1/4)x4-(1/2)x2|-11=((1/4)14-(1/2)12)-((1/4)(-1)4-(1/2)(-1)2)=(-1/4)-(-1/4)=0. I don't think errors were made. Is there no area "under" this curve?

    The answer is that we computed a definite integral. Remember that the definite integral counts area below the horizontal axis negatively. The function f(x)=x3-x is antisymmetric or odd. Its graph is shown to the right. The geometric area inside the two bumps is the same, but the signed area in the left-hand bump is positive, and the signed area in the right-hand bump is negative. The result when they are combined (in the definite integral from -1 to 1) is 0.

    The instructor is not too smart ...
    When I discussed the problem involving f(x)=x3-x in class, I tried to sketch a graph rapidly. So I wrote x3-x=0 and then x(x2-1)=0 and then x(x-1)(x+1)=0. There are roots at 0 and 1 and -1. There aren't any other roots (a polynomial of degree 3 can only have 3 roots) and the polynomial changes sign as the x "crosses" each of the roots. So I drew bumps in the wrong order (minus the truth). Oh well, I do apologize but the idea is still "correct": the geometric areas in the picture cancel and the signed or oriented areas reported by the definite integral are, together, equal to 0.

    Example #4
    Let look at y=x(x-1)(x-3), slightly changed from the example I actually did in class. I wrote this already factored, and the roots are 0 and 1 and 3. A picture is shown to the right. Can we find the geometric area (not the signed area!) enclosed by this curve and the x-axis? I look at the picture and see that the definite integral from 0 to 1 will be positive and will equal the geometric area. The definite integral from 1 to 4 will be negative, and if the sign is "corrected" (take away the minus!) the result will be the geometric area. If we computed the integral over the interval [0,3] some of the positive and negative values will cancel and the result will not be what we wnat. So I'll compute two integrals. In both cases we'll need an antiderivative of x(x-1)(x-3). But when the polynomial is written in factored form, I can't readily guess an antiderivative. Let me multiply out:
    Now I "guess" an antiderivative: (1/4)x4-(4/3)x3+(3/2)x2.
    Note There isn't that much guessing going on. I hope you already suspect this. An organized method for obtaining some antiderivatives will be presented in the next class.

    We compute:

    01x(x-1)(x-3)dx=01x3-4x2+3x dx= (1/4)x4-(4/3)x3+(3/2)x2|01=

    13x(x-1)(x-3)dx=13x3-4x2+3x dx= (1/4)x4-(4/3)x3+(3/2)x2|13=

    Well, maybe not doing this example in class was a good idea: the numbers are not pleasant. But please not that the first definite integral has value 5/12, positive as it should be. The second definite integral is -32/12, and this is a negative "area", below the horizontal axis. The geometric area bounded by the line and curve must then be (5/12)+(32/12)=37/12. I made several errors with the arithmetic (signs and fractions!) in finishing this example. Oh well.

    Tuesday, August 8
    We went over some homework problems. Specifically, we looked at problems 1 and 6 and 7 and 8. In each of these problems, we were given the function f(x)=x2+x and the interval [0,3]. We were then given various partitions and sample points, and asked to write (actually, to compute!) the Riemann sum associated with these choices.

    I'll discuss problem 6, because I asked some additional questions which were relevant to an important assertion I made later in the day's presentation.

    Problem 6 of chapter 8
    In problem 6, we were given the function f(x)=x2+x and the interval [0,3]. The problem asked about the Riemann sum which resulted from using a regular partition with six subintervals, and the choice of same points was determined by using the right-hand endpoints of each subinterval.

    A regular partition has equal subintervals. since the length of [0,3] is 3, the length of the subintervals of a regular partition with six subintervals is 3/6: that's one-half. The partition points will be {0,1/2, 1, 3/2, 2, 5/2, 6}: there are seven of these, one more than the number of subintervals requested. What about the sample points? They will be 1/2 and 1 and 3/2 and 2 and 5/2 and 3 respectively: these are the right-hand endpoints of each subinterval. So the Riemann sum will be this:
    We were also requested to sketch y=x2+x on the interval [0,3], and draw areas corresponding to the Riemann sum. This is what I've attempted to do in the picture shown. I mentioned in class that the curve y=x2+x goes through (0,0) and (3,12). I also mentioned (hey, y´=2x+1, positive in [0,3]) that the curve is increasing between these points. I also mentioned (hey, y´´=2, also positive in [0,3]) that the curve is concave up between these points. So the graph drawn is qualitatively correct. I should mention that the darn curve does go through the correct points, so it is more or less quantitatively correct, also (note that the vertical and horizontal axes have different scales, however).

    A question which is not in the book
    Suppose we wanted to compute the area bounded by y=x2+x, x=0, x=3, and the x-axis. This area is usually called the area "under y=x2+x between 0 and 3". How close is this specific Riemann sum to the actual area? Let me ask two questions:

    Answer Well I might first look at the picture and see that the boxes contain the curvy area, so the Riemann sum is an overestimate of the area. But if I wanted to be algebraically certain, I could mention that in each little subinterval, the function is increasing, and therefore the function values in that subinterval are all less than or equal to the function value at the right-hand endpoint. Therefore I am "algebraically" certain that the sum written above is an overestimate. Answer Well, one comment is that the darn Riemann sum is certainly with 10,000 of the true area. That's sort of silly, because all of the "action" in this picture is taking place inside a rectangle which is 3 by 12, and has an area of 36. So the Riemann sum is certainly within 36 of the true area. While that statement is correct, it is certainly not useful. Can we get a neater answer, taking advantage of the special situation we have?
    Yes, we can. Look at the "errors" in each rectangle. Now move them, say, to the right. Since the function is increasing, the errors can all be made to line up, one on top of another. The width of the subintervals in this regular partition is always the same, 1/2. The error areas all line up inside a rectangle whose width is 1/2 and whose height is 12. So the total error will be less than 12·(1/2), which is 6. I agree with some students who commented that I'm "losing" some precision because the areas involved are almost triangular, and their total is actually maybe about half of 6. But I don't want to work too hard. I'm lazy.

    And then problem 7 ...
    The same function, and the same interval, and a regular partition with 100 subdivisions, and the sample points at the right-hand endpoints of the subintervals. Well, what happens? The subinterval width is now .03, because we are dividing the interval into 100 equal parts. The Riemann sum itself is something like:
    SUM as i goes from 1 to 100 of (({3/100}i)2+({3/100}i))·(3/100)
    In the ith subinterval, the right-hand endpoint is (3/100) (the width) multiplied by the number of the subinterval, and the result is {3/100}i. Since we're "starting" from 0, this is the correct "formula".

    What about the error?
    Now if I try to draw a picture similar to what's above, well, the details are too darn difficult. But the idea is the same. There are 100 error pieces. They all have width 3/100 and they all can be moved to the right to stack up inside a rectangle of width 3/100 and height, still, 12. Now we can see again that the Riemann sum is an overestimate that the overestimate is no bigger than 12·(3/100), which is 36/100, about one-third. This is certainly less than the error of the approximation in problem 6, which we estimated as 6.

    Huh: a mystery function!
    The instructor now asked students: what if f(x)=sqrt(x5+x2+x+1)? The correct response of the students might have been, well, what if? But we tried to graph this function on the interval [0,1]. The strangeness of the function was partially explained by the "nice" facts: f(0)=1 and f(1)=2. I defined the function so that it had nice values at the endpoints. Then I remarked that this function was increasing (calculus tells me this, since f´(x)=(1/2)(STUFF)-1/2(5x4+2x+1), and this is certainly positive in [0,1]). The graph is also concave up, but I don't care since I really don't want to find another derivative. In fact, a machine-generated graph is shown to the right.

    Such a simple graph. What if, for some really silly reason, we want to find the area "under" y=f(x) on the interval [0,1]. Again, this phrase refers to the area bounded by the graph y=f(x) and the lines x=0, x=1, and the x-axis (y=0). I do not believe that any application I know of would require computation of this area. But there are much more complicated "areas" (not really areas, but see below, please!) which people very much want to compute. Well, I turn to the program which handles much of the symbolic computation that I want to do, since my brain is small. I ask the program (I'll show you later, really) to compute this area. The program, Maple, is one of the standard programs (similar to Mathematica) used all over the world to do lots of computations. The program is like a very powerful version of the TI-89. Its design was begun in 1980, and it has been developed by literally tens of thousands of academic and industrial people since then. Well, this wonderful, wonderful program, in its latest version (less than a year old), can't compute this area: really. Let me ask something more modest.

    Can we get a Riemann sum for this f(x) on the interval [0,1] which is within .01 of the true value? More precisely, can we describe such a sum, which I'll then have a machine compute?
    Abstraction (but just a bit!)
    Well, let's try to approximate this area by a Riemann sum. And maybe we can find a Riemann sum which is within .01 if we are a little bit careful. We will begin with a regular partition with n subintervals. We will choose the n later. Since the whole interval, [0,1], has length 1-0=1, the width of each subinterval is 1/n. Let's also, just to be definite and simple, choose the right-hand endpoint in each subinterval as our sample points. So, since we are "starting" at 0, the first right-hand endpoint of a subinterval will be 1/n, and the second will be (1/n)·2, the third will be (1/n)·3, etc. The ith right-hand endpoint will be (1/n)·i. The i will go from 1 to n. Each piece of the Riemann sum will look like f((1/n)·i)·(1/n). We are supposed to add up these pieces as i goes from 1 to n.

    And the error here?
    I did not try to draw a picture with lots and lots (n large!) of thin rectangles. I did try to draw a graph which has the area shown and shows what I think is one "typical" rectangle whose area contributes to the Riemann sum. The error is again pushed over to the right. As Mr. Acosta observed, these little errors will line up nicely inside a rectangle which is itself fairly thin (the width is 1/n) and which is 2-1 high. So the total error will be less than (1/n)(2-1) and this is 1/n.

    If I want to have this error less than .01: well, .01 is 1/100 and so the error will be less than 1/100 if I choose n to be, say, 100. (I am sorry almost that there are so many coincidences although the numbers work out very simply.) Then the Riemann sum for the function f(x) on the interval [0,1] associated to the regular partition with 100 subdivisions and with the choice of the right-hand endpoints as sample points must be within .01 of the "true area".

    Here is the actual command and the response of the Maple program which computes this Riemann sum:

    I asked the program to time this command. Errr ... the response I got to that request was the disconcerting time = 0.00 which doesn't mean the command took no time (!) but that the time needed was less than one-hundredth of a second. Sigh. If I wanted more accuracy, using this idea of approximating area by Riemann sums, of less than .001, I could alter the command above by changing the ".01" to ".001" and the "100" to "1000". The result is 1.387097725 and took the program .07 seconds. With 10,000 terms, the result was 1.386647601 (.67 seconds!). The true answer, using other techniques, is 1.386597600 (to ten digit accuracy)

    What you should know ...
    You should know that fairly simple ideas can be used to compute these Riemann sums and that, using widely available programs, the results don't take long. I am more interested in the conceptual background of these sums (not only area, but also blood flow, distance, future value of an income stream, etc.).

    The definite integral
    I hope you are willing to believe, after the rather involved discussion about error, that if the partition of the interval is chosen to be "fine enough" (so the lengths of the subintervals are all small) then the Riemann sums will all be close to something. This is true, and the "something" has a name.

    As the number of points in the partition increases, as the length of the subintervals approaches 0, if the function involved is fairly nice (let's say continuous, although the function can have some jumps and breaks) then the Riemann sums, for any possible selection of sample points, approach a unique limit, which is called the definite integral of f(x) from a to b. The "any possible" is italicized because in practice when data is gathered it may be difficult to get exactly the sample points desired, so the "any possible" says that, more or less, the specific choice of sample points doesn't matter too much!

    Here is the rather wild traditional notation. Riemann sums themselves are frequently abbreviated with capital Greek sigmas as I mentioned last time. The notation for the definite integral is the following: ab f(x) dx
    The long "S" is an abbreviation for Sum. So you can look at this notation and see that it is an abbreviation for summing up, as x goes from a to b, the "rectangles" of height f(x) and width dx. Here "dx" stands for some really really really small width. Yes, we will relate it to the other uses of dx in this course.

    You can think that the definite integral stands for "area" but that isn't quite correct. Let me look at some examples.

    Some definite integrals
    I drew the graph shown to the right, and stated that this was a graph of y=f(x). I then asked for the following definite integrals:
    13f(x) dx    24f(x) dx    35f(x) dx    46f(x) dx
    Since I "identified" pieces of the curve as simple shapes, probably the integrals can be computed.
    I deliberately did not write a formula or formulas (since this function would need a piecewise definition!) for f(x), since I wanted people to think about the concepts involved.

    13f(x) dx
    Here the region involved is just a rectangle. The definite integral's approximating Riemann sums divide this region into many thin rectangles with height 1 and tiny width. They "add up" to the whole rectangle.
    The base is 2 units long, and the height is 1 unit, so that the area is 2. This is the value of the definite integral.

    24f(x) dx
    The region here is a square and a triangle joined on a side. The thin Riemann sum rectangles exactly fill up the square, and in the triangular part I hope you can see the "error" will be some fuzzy stuff around the diagonal line segment, and this error-->0. So we will just get the area of the square and the triangle as the limits.
    The square is 1 by 1 unit and has area 1. The triangle, with base and height both 1, has area 1/2. Therefore the definite integral, in this case measuring the total area, has value 1.5.

    35f(x) dx
    The geometry is clear, but what the definite integral computes is sometimes not exactly what the geometry seems to say. The Riemann sum definition of the definite integral is a sum of values of f(x) multiplied by litte changes in x, dx. This makes area above the horizontal axis count positively and area below that axis, negatively. The picture below shows the signed area which a definite integral would compute. The sign of the definite integral in applications can have meaning which is central to the computation: are those dollar amounts profit or loss? Is there a net amount of fluid being pumped around the body, or is there a backflow?
    In this case with this f(x) and this interval, the definite integral counts the triangle with positive area and the quarter circle with negative area. So the value of the definite integral is 1/2 -Pi(1)2/4. This is a negative number.

    46f(x) dx
    Here all of the curve is below the axis, and we will "count" all of the area negatively. This is half a circle of radius 1. The value of the definite integral is -Pi(1)2/2.

    So when the definite integral is interpreted as an area, it actually computes what's called an oriented area or a signed area. Regions above the x-axis count positively and regions below the x-axis count negatively. Profit and loss are different quantities.

    What's next?
    Tomorrow I hope to show you one method that Chinese and Greek and Indian people used to compute the actual values of definite integrals. To them the definite integrals represented the values of areas or, sometimes, volumes. The methods they used were extremely clever but perhaps very idosyncratic. By this I mean that the methods were special to almost every computation. (Look, please, in chapter 9 of the text.) It turns out that there's a more systematic method. This is the Fundamental Theorem of Calculus. (Look, please, in chapter 10 of the text.)

    Maintained by and last modified 7/25/2006.