Math 151 diary, fall 2007
In reverse order: the most recent material is first.

Earlier material
Much earlier material

Tuesday, December 11 (Lecture #28)
The final exam and related items
Please see this for information about the final, the TV review, and the review for our lecture. Also have questions ready for Mr. Yin tomorrow, please!

Another attempt to foul up ...
I choose, more or less at random, y=x(x-1)(x-2). I asked people to sketch this curve, and then attempted to find the (geometric) area enclosed between the curve and the x-axis. A picture of this is shown to the right. Sigh. And then we computed two definite integrals, 01x(x-1)(x-2)dx and 12x(x-1)(x-2)dx. This problem was my attempt to show that the definite integrals would not balance out. And, of course, since I made the choice at random, they did balance. Sigh.

The key to the computation of 01x(x-1)(x-2)dx is the first step, which is multiply out the factors, and get the standard representation of a polynomial. Integrating that representation is easy. So x(x-1)(x-2)=(x2-x)(x-2)=x3-2x2-x2+2x=x3-3x2+2x. Now we easily can get an antiderivative: (1/4)x4-x3+x2 and use FTC 1 to evaluate the definite integrals. So:
The total geometric area inside the two bumps is 1/2.
Mr. Louides Ferdinand triumphed again! There is symmetry here also, and I should have known the two bumps had the same size.

Maybe we should have done ...
How much geometric area is enclosed by y=x(x-1)(x-3) and the x-axis? Again, a graph is shown to the right, and these bumps definitely have different sizes. But my patience (?) is running out. In a spare tenth of a second, we get:

> int(x*(x-1)*(x-3),x=0..1);

> int(x*(x-1)*(x-3),x=1..3);
So the geometric area is 5/12+8/3, which is, I can tell you, 37/12. So this is 03|x(x-1)(x-3)|dx (the absolute value signs make both bumps "sit" on top of the x-axis.

I continued my attempt to reach a new level ineptitude by miswriting the next problem I wanted to analyze. That won't happen here, because I have the opportunity to think and correct. So I ask, what does the curve defined by the equation (y-1)(y-2)=x look like? With a bit of thought, we decided that it was a parabola with axis of symmetry parallel to the x-axis, and that it was opening to the right. A graph is shown to the right.

Notice, please, that the point (2,0) is on the graph, as well as the point (0,2). Therefore there's a (exactly one!) straight line through these two points. Its equation is x+y=2 (Admission: I sort of guessed at the equation, since the sum of each pair of coordinates is 2, that must be the equation!). What's the area of the region bounded by the line and the parabola?

The dy method
If we have a region in the plane bounded by two curves, x=Left(y) and x=Right(y) (where Left(y)<Right(y) for the y's of interest here) and also bounded by lines y=Bottom and y=Top, then we could imagine slicing this area by lots of horizontal lines dy apart. The region would be divided into lots of pieces, and each piece would approximately be retangular. The height of the (almost) rectangle would be dy, and the width would be Right(y)-Left(y). So the area would be (Right(y)-Left(y))dy. Since this is the area of a slice, we would get the total area by taking the Sum of the areas of the slices, from the Bottom to the Top: y=Bottomy=Top(Right(y)-Left(y))dy.

In our case ...
Top is 2 and Bottom is 0. Left(y) is (y-1)(y-2)=y2-3y+2 (you should multiply/distribute/expand/foil whatever! before integrating) and Right(y) is 2-y. Thus the area can be computed by
y=0y=2([2-y]-[y2-3y+2])dy= y=0y=2([2y-y2])dy=y2-(1/3)y3|02=4-8/3=4/3.

Certainly this is not the only method of getting this area. You could possibly imagine dissecting the area dx, and chopping it up somehow so the computation could be done. That would be much more work, I think. Or maybe you just interchange x and y, and then redraw the graph, and do it dx. That's certainly possible. I just report, though, that for many students in the class (physics students, engineering students) many other geometric and physical quantities will occur (moment of inertia, center of gravity, etc.) which can be computed more naturally dy, and therefore "chopping up" regions in this way will be useful.

What was done very wrong ...
We had a bit more time before the official end of the class meeting. This was the last lecture of the course, and I want to help students prepare for the final. I believe that students should prepare for what they want to avoid in the following sense: assume that whatever type of problem was most difficult will appear on the final exam. So I looked at the student performance on the two exams we've had, and decided to discuss the problems which students had done the worst (in terms of grades).

Exam #1
Problem 6 had the worst results, followed perhaps by problems 4 and 9. So let's look at a sample problem asking for information similar to the what was needed for problem 6.

A problem
Suppose that f(x)=x6-5x3+2. Verify that f has a root, and locate this root in an interval of length 1/2. Explain why your assertion is correct.

It isn't even immediately clear to me that this f has a root since the highest order term is x6, and I know that as x--> and as x-->-, x6 will get large positive. So I will compute some values of f. Certainly, f(0)=2 is the easiest value to check. Now if I were doing this on an exam, I would probably check either f(1) or f(-1). Notice that if we could assume that students had calculators, the function to be investigated could be considerably more complicated! Well, f(-1)=(-1)6-5(-1)3+2=1+5+2=8. This is positive, just like f(0), so this is no help. But f(1)=16-5·13+2=-2<0. I bet there's a root inside the interval [0,1]. I need finer information, so I compute f(.5), which is 1/26-5/23+2: I think that the +2 determines the sign, and f(.5) is positive. So here is my answer to the problem:

Answer f(1)=-2 is negative and f(1/2)=1/26-5/23+2 is positive. The function is a polynomial and is continuous on the interval [.5,1]. The Intermediate Value Theorem applies, and asserts that for at least one x in the interval, f(x)=0.
If I would grade the answers to such a problem, I would look for certain indicators: "Intermediate Value Theorem" and "continuous" as some abstractions, surely, but applied to the specific function under consideration. I would also look and assure myself that the student had supplied specific evidence showing that these abstractions were correctly applied in this case. I would like all of these to be present.

Problem 9 asked for computations of limits. Problem 4 asked for the location of what we've since defined to be a critical point. In both problems, students made serious mistakes in algebra. I urge people to check their answers and their work. There will be sufficient time on the final exam for this.

Exam #2
Certainly problems 2 and 8 had the worst results, followed perhaps by problems 6 (a geometrical question about Newton's method) and 7 (more limit computations!). Problem 8 asked students to use (and cite!) a result in the course connecting information about the distance traveled if some restrictions on velocity are given. This is a problem about the Mean Value Theorem, which, as I wrote on the guide to preparing for the final, is one of the two major results in the course (the other being FTC). Problem 2 was an optimization problem copied directly from the textbook. So I thought I would do another textbook problem. This is problem #23 in section 4.6; the test problem was problem #27.

A problem
Find the angle that maximizes the area of the trapezoid with a base of 4 and sides of length 2, as in the figure shown.

The steps we should follow, as told to me:
    Function: area in terms of --> differentiate --> set equal to 0 --> solve for
I remarked, I pleaded, I requested ... please find out what the "eligible" 's are first (the domain that's appropriate for this independent variable in this problem). Then go on and start the steps listed above. You need to know what the likely answers should be, and you'll need to know why you have found a max. All this will be helped greatly if you have the appropriate domain when you start the steps indicated above.

Diary entry in progress!

Friday, December 7 (Lecture #27)
The final exam and related items
Please see this.

Exponential growth: a textbook problem
This is problem #16 of section 5.8: An insect population triples in size after 5 months. Assuming exponential growth, when will it quadruple in size?
I added the following question:
    What is the doubling time of this type of insect?

The phrase "Assuming exponential growth" means that, if B(t) gives the number of bugs at time t, then we should assume that B(t)=Cekt. So (units!) B(t) will represent the population of the insects at time t measured in months since the start. We know that B(0)=Ck·0=C and B(5)=3C. So 3C=Cek·5, and if we divide by C and ln both sides and divide by 5, we get k=ln(3)/5.
So B(t)=Ce[ln(3)/5]t and we want to know when the population quadruples, that is, reaches 4C. (By the way, it is a good idea to have some estimate in mind just so we can check the answer, or, if the answer is correct and way off the estimate, we can improve estimating skills. In this case, I would guess that about another month or so is needed to get to four times the initial population: so my guess is 6, maybe a little bit more.) So let's solve 4C=Ce[ln(3)/5]t: we divide by C, ln both sides, and divide by [ln(3)/5]. The result is t=[5ln(4)/ln(3)]6.31 months.

I had also asked "What is the doubling time?" Doubling time is a standard measure of growth, and is the time needed for the population to double (indeed!). If the formula governing growth rate is exponential, then doubling time is a constant (hey, if the population was given by t2+1, then the population doubles from 0 to 1, and it also doubles from 1 to sqrt(3): the time intervals change!). In this case, we are told that population is given by an exponential formula and that a population quadruples in 6.31 months. I think the doubling time is 3.15 months.

Exponential decay: a textbook problem
This is problem #16 of section 5.8: A 10-kg quantity of a radioactive isotope decays to 3 kg after 17 years. Find the decay constant of the isotope.
I added the following questions:
    Also, what's the half-life? What is half-life?

Well, suppose R(t) is the quantity in kg of the radioactive substance at time t in years. We assume (and mostly this is true) that amount of the substance is given by R(t)=Cekt. We know that R(0)=10, so that C=10 (always try to start the "clock" in these problems at 0 so that C will be the initial amount). Also since R(17)=3, 10ek·17=3. Now divide by 10, ln both sides and divide by 17: k=ln(3/10)/17. I think this is the decay constant. Please notice that since 3/10 is less than 1, ln(3/10) is negative so the constant in the exponential formula is negative, and this is, indeed, decay.

The half-life of a radioactive substance is the time needed for an initial quantity to reduce to half. Since we are told that in 17 years, 10 kg reduces to 3 kg, I am sure that a half-life of this substance will be less than 17. In fact, a half-life will be more than, say, 8, since a half-life of 8 would result in 2.5 kg at 16. So a casual estimate of half-life is somewhere between 8 and 17.
Here we want t so that 10e[ln(3/10)/17]t=5. Divide by 10, take lns, and divide by [ln(3/10)/17]: the result is t=ln(1/2)/[ln(3/10)/17]. This is about 9.79.

Radioactivity: a very short discussion
There are lots of radioactive isotopes. OA web page I found mentions these "widely used industrial isotopes ... 192Iridium with a half-life of 74 days ... and 60Cobalt with a half-life of 5.3 years". If radioactive substances have the same activities (there are three principal types: alpha and beta particle emissions and gamma rays) then it is likely that a substance with a shorter half-life will be more dangerous than one with a longer half-life. But, as Ms. Kravitz reminded me, things can get more complicated. Please don't think that I know more than a superficial amount about radioactivity. I also mentioned tritium, which is an isotope of hydrogen, 3H, widely used to make paint glow (signs, rifle sights, sometimes as a tracer in the body. It has a half-life of 12.3 years, although one source I found declared that when used as part of a biological sample (administered as water) then the "biological half-life" (a phrase I had not seen before) in humans was about 10 days (this refers not to the decay of the tritium, which takes quite a while, but to how a normal human body processes water). Tritium is not good in the body.

Radiocarbon dating is one of the most widely used methods of dating human remains and cultural objects. Please follow the link supplied to read about it a bit. The basic assumption is that an organic (living) object has a certain percentage of 14C ("Carbon 14") and this percent is not replenished which the object is dead. So as the 14C decreases, a reasonable guess about the age of the object can be made. 14C is convenient because there's a good amount of carbon in most organic objects, and because the 14C half-life is commensurate (appropriate amount) to measure much of human history.

One source I found said that the half-life of 14C is about 5,730 years. The text asserts that the decay constant for 14C is about -.000121. Are these two numbers consistent? Is there some way to check? Well, suppose 14C follows a Cekt formula. If C=1 and k=-.000121 and t=5,730, then Cekt should be about a half. In fact, we computed this (or rather, calculators did) and the result is .499905, which is quite close to a half. I was happy.

Area between two curves
Here's the final topic of the semester, which is a simple introduction to the uses of the definite integral. The definite integral has hundreds of important applications in science and engineering.

Suppose we are given two functions defined on an interval, Top(x) and Bot(x) ("Bot" is an abbreviation for "Bottom") and we know on that interval that Top(x)>Bot(x). How can we compute the area enclosed by the two curves and by vertical lines on the sides of the interval?

I always imagine that the interval is chopped into lots of little pieces, each of length dx. Then these pieces chop up the area, as shown in the picture to the right. Each little slice of area is almost a rectangle (if we ignore the possible tilts at the top and bottom). The area of the approximate rectangle is [Top(x)-Bot(x)] (the length of the vertical side, the difference in the heights of the graphs) multiplied by dx, the width. Now I need to take the Sum of these approximating slices, and add them up from a to b, or, as I think, from Left to Right. Therefore I believe that the area between these curves on this interval is

Random example #1
This example is not random, but it is constructed so I could do it easily in class. I would like to find the area of the region enclosed by the line y=x+2 and the parabola y=x2. I would probably try to begin almost any problem of this type by sketching the region first. The picture to the left is such a sketch. Where do these curves intersect? Well, we need to find x's which give the same y, so we solve x2=x+2 so that x2-x-2=0 and this is (wow! Oh, hold on, he made it so it would work out) x2-x-2=0 so that (x-2)(x+1)=0. The roots are x=2 and x=-1.
I honestly think that the phrase "the region enclosed by the line y=x+2 and the parabola y=x2" specifies exactly one piece of the plane. You can argue with me -- I discuss this a bit more in the example below. Here I think we have Left=-1 and Right=2 and Top(x)=x+2 and Bot(x)=x2. So let's compute:
This turns out to be 9/2. I hope. And, by the way, if you look at the picture, I hope you can see that the region fits inside a box with height 4 and width 3, so the area should be less than 3·4=12, which it is. Getting some approximate idea of the answer is useful, if you, like me, make sign errors and ... well, other kinds of errors.

Random example #2
Again, this is an arranged example and everything will work out neatly. Real problems are rarely like this. I'd like to compute the area of the part of the plane which is between the parabolas y=x2-5 and y=3-x2. The first curve opens up and has its bottom at (5,0). The second opens down, with its top at (3,0). A picture of the two curves is shown to the right.

I mentioned in class that it is certainly possible to "misunderstand" the statement of this problem, if you work at it. The plane is actually divided into five different regions by these two curves. And maybe someone could maybe declare that there's more than one candidate for the region between the parabolas (maybe all five are candidates, or maybe only three of them are?). I guess maybe I might agree if you really argue with me about it, but if the discussion is just a way to get out of computing the area of the only region with finite area then I would not agree. Please compute the area of that region, and then, later, argue about it.

The curves intersect where x2-5=3-x2. Since this is a problem in a math class, let's see: that's 2x2=8 or x2=4 so that x=+/-2. I think that Left=-2 and Right=+2.

Something new!?
There is a little bit that's new here. Notice that if Top(x)=3-x2 and Bot(x)=x2-5, then actually Bot(x) is always negative in the interval [-2,2]. Even so, the quantity Top(x)-Bot(x) gives the geometric length of the vertical side of a (dx) thin slice of area. Top(x)-Bot(x) will be positive. In fact, if you look really carefully at the picture, you'll see that there are two pieces of [-2,2] where both Top(x) and Bot(x) are negative. But Top(x) is still bigger than Bot(x), so Top(x)-Bot(x) will be positive, even though both of them are negative. The geometric length is still Top(x)-Bot(x).

The computation
This is less than a box which is 8 units high and 4 units wide (such a box would enclose the entire area). Also, several students remarked that we could have computed the integral from 0 to 2 and doubled the result, to take advantage of symmetry.

Random example #3
I would like to find the geometric area between the sine and cosine curves over one period of these functions. So I drew a picture similar to what's shown to the right. Again, it may be possible to misunderstand the question since this situation is more complicated than previous ones. But "one period" is 2Pi here, and the word "between" in this situation refers, in fact, to three specific regions in the picture.

So the area which I wanted to compute is shown to the right. Here is what I thought that I needed to do (clear evidence that the brain was not totally functional!). I thought that I would need to split the area computation into three pieces, and compute three different integrals, and then add them up. The idea would be as shown to the left where first one formula then the other, and then the other again, is on top. Well, this could be done but it would actually be much more work than is needed.

Mr. Louides Ferdinand kindly pointed out that the left-most region and the right-most region when put together are congruent to the central region: so the sum of the areas of these two regions must be equal to the area of the region in the middle. Thank you, Mr. Ferdinand! (Next time I'll get a non-symmetric example, darn it!)

The curves intersect where sin(x)=cos(x). This is at x=Pi/4 and x=5Pi/4. So Left=Pi/4 and Right=Pi/4 and between these numbers, the sine curve will be Top(x) and the cosine curve will be Bot(x). So I need to compute twice the definite integral over this interval to get the total area requested. Here we go (and be careful with minus signs!):
There were only SEVEN minus signs in the next-to-last expression. Mistakes are easy!

Tuesday, December 4 (Lecture #26)

Antiderivatives of complicated functions can now (maybe!) be done by machine. But people can sometimes still do things by hand which complicated programs may not be able to handle (see below!). People have millions of years of pattern recognition in their brains. In all of the examples we did in class, I kept repeating that attention to details was important, that inserting parentheses was important to help yourself, that ... well, just be careful!

I always try to substitute for the center of the difficulty first (this does not always work, but it helps in lots of examples). In this example I would try u=5x+4. So then du=5dx so (1/5)du=dx, and, as an antiderivative (let me leave out the limits right now):
sqrt(5x+4)dx=sqrt(u)(1/5)du=(1/5)u1/2du=(1/5)(2/3)u3/2+C=(2/15)u3/2+C =(2/15)(5x+4)3/2+C
The outline of this computation is: from x-land to u-land, antidifferentiate, back to x-land. Now if we really want to compute the definite integral, look:
Please notice that the x=0 limit does contribute to this answer -- you can't just ignore it because, golly, x=0. This antiderivative has a non-zero value at x=0.

My silicon pal does a great job on this (it even cleans up the fractions and powers!):

> int(sqrt(5*x+4),x=0..1);

As I remarked in class, I think that most people could have made a good guess at the previous antiderivative. I can't do this one without a substitution and some subsequent manipulation. So again let us try the center of the difficulty, u=5x+4, so du=5dx and (1/5)du=dx. But now to get to u-land we need to know how to translate x. Since u=5x+4, we know that u-4=5x and (1/5)(u-4)=x. Therefore:
x·sqrt(5x+4)dx=(1/5)(u-4)u1/2(1/5)du [to u-land]
(1/25)(u-1)u1/2du=(1/25)u3/2-u1/2du [pull out constant multiplier, distribute u powers]
(1/25)((2/5)u5/2-(2/3)u3/2)+C=(1/25)((2/5)(5x+4)5/2-(2/3)(5x+4)3/2)+C [antidifferentiate, and back to x-land. Notice that 1/(5/2)=2/5. etc.].
Although it may be possible to check an antiderivative by differentiating, you may not want to!

Here I pushed up a power of the x. If you use u=5x+4 again (the center of the difficulty) then as before, (1/5)du=dx and (1/5)(u-4)=x. So the integral becomes:
This is not so pleasant but we can do it. Now [(1/5)(u-4)]2=(1/25)(u2-8u+16) and so:
So what did I do there? Well, I squared 1/5 to get 1/25, and then multiplied by another 1/5 to get 1/125. Multiplicative constants can be pulled "out" of the indefinite integral. Other kinds of constants can't! Then I distributed the sqrt(u)=u1/2 over the other terms. Powers of u can be antidifferentiated easily, so we get:
Now back to x-land:
I surely would not want to check this by differentiating! My "pal", in about two-hundredths of a second (.02 seconds) gives this:
> int(x^2*sqrt(5*x+4),x);
                                3/2                     2
                     2 (5 x + 4)    (128 - 240 x + 375 x )
I hope (I guess!) this is what we have. It sort of looks the same!

Here u=5x2+4 (the most "horrible" thing in the integral), du=10x dx, (1/10)du=x dx (because notice that there is an x inside the integral -- not an accident!). From x-land to u-land, then, we get:
The x did not just disappear! It was needed in this case to be part of du. And finally we change (1/10)(2/3)u3/2+C to (1/10)(2/3)(5x2+4)3/2+C. You can check this by differentiating, if you use the Chain Rule correctly. The 2/3 cancels the 3/2 coming from the power outside, and the 1/10 cancels the 10 in the 10x coming from what is inside.

I remarked that I could show people how to compute an antiderivative that commonly used programs would not do. This antiderivative is the beginning. Here we tried the substitution u=1+sqrt(x), so du=(1/2)x-1/2dx. If we solve for dx, the result is dx=2sqrt(x)du, which is 2(u-1)du. So:
O.k.: this is good. It is sort of ugly, but it is certainly not the most complicated computation in the world.

So let's try the same u as before. Then:
If we make another substitution: v=1+sqrt(u), then dv=(1/2)u-1/2du, so 2u1/2dv=du and 2(v-1)dv=du. So go from u-land to v-land. We also need to know u in terms of v, but v=1+sqrt(u) leads to (v-1)2=u.
O.k. Let's see what the mess in v-land actually looks like if we multiply out things (sorry: "expand expressions"):
v1/2)2((v-1)2-1)2(v-1)=4v1/2((v2-2v)(v-1)=4v1/2(v3-3v2+2v)= 4v7/2-12v5/2+8v3/2
The last expression is just a bunch of powers of v, so its antiderivative is easy. Here it is:
Now back to u-land with v=1+sqrt(u):
Now back to x-land with u=1+sqrt(x):
There we go!
We've computed an antiderivative that the
calculators students had in class couldn't do!

Cheers for the human beings among us, who have both pattern recognition and patience.

O.k., I bet if you paid me enough (I'm cheap!) I could find an antiderivative for this function. The big and powerful program Maple, available on most Rutgers systems gives:

> int(sqrt(1+sqrt(1+sqrt(1+sqrt(x)))),x);
                     |                  1/2 1/2 1/2 1/2
                     |  (1 + (1 + (1 + x   )   )   )    dx
This response indicates that the program could not find an antiderivative in terms of familiar functions.

Mathematica, probably the most widely known Maple competitor (and also available on some Rutgers systems) has an integration webpage on line. When asked to find the antiderivative of sqrt(1+sqrt(1+sqrt(1+sqrt(x)))), the response is

Mathematica could not find a formula for your integral. Most likely
this means that no formula exists.
I don't agree. I could do it, just ask (but politely and persuasively!). I believe people and computers can work together usefully, each with their strengths.

Pattern: the worst thing is certainly inside sine. So I will try u=5ex. Then du=5exdx, and (1/5)du=exdx. From x-land to u-land, to antidifferentiate, and back to x-land;

Three ln integrals
I asked groups of students to go to the board and compute three increasingly difficult (at least to me!) integrals involving ln which are problems in section 5.7. Here they are:

  1. [(ln x)2/x]dx
    Solution Take u=ln x (since ln is the center of the difficulty), so du=(1/x)dx, and recognize please that we have dx/x in the integral! So from x-land to u-land:
    [(ln x)2/x]dx=(ln x)2[dx/x]=u2du
    Now integrate and get u3/3+C, which in x-land is (ln x)3/3+C. Is this good?
  2. [1/{x·ln x)]dx
    Solution Take u=ln x and again du=(1/x)dx. The integral changes:
    [1/{x·ln x)]dx={1/u]du=ln(u)+C=ln(ln(x))+C
    This really is not too horrible. People use functions like this one, really.
  3. [{ln(ln(x))}/{x·ln(x)}]dx
    O.k., this example is more of a textbook example. So, again, I will try u=ln x and again du=(1/x)dx. We will go from x-land to u-land:
    Please notice that the substitution absorbed (?) one of the things in the bottom of the integrand. Now what should we do? If you were "properly conditioned" by the previous examples, you would try a substitution and so I will. What I choose to try (it works, that's why!) is v=ln(u) and dv=[1/u]du. So from u-land to v-land:
    [{ln(u)/u]du=v dv=(1/2)v2+C
    Now back and back again:

A differential equation and initial condition again
I'm following the text here. We are next supposed to consider a new sort of differential equation. We have studied such problems as these:
    y´=3x4+{2/x4} and y(1)=2.
This is an initial value problem, with a differential equation and an initial condition. We find what's called the general solution by computing an antiderivative:
3x4+{2/x4} dx=3x4+2x-4dx=(3/5)x5-(2/3)x-3+C
The perhaps interesting aspects of what was just computed are these: I first changed {2/x3} to 2x-3, which is a more standard way to write powers of x. I regard things like sqrt(x) (just x1/2) and 1/sqrt(x) (surely x-1/2) as invitations to error, directed personally at me. Unfortunately I accept these invitations sometimes. Then I antidifferentiated. Making sure that a minus sign appears in the answer (a consequence of the transition from -3 to -2) is also something I foul up sometimes. Oh well. Now onward.
The initial condition y(1)=2 allows us to pick out a particular solution by finding a value of C in the general solution:
Since y(1)=2, (3/5)15-(2/3)1-3+C=2. This means (3/5)-(2/3)+C=2 or C=61/15.

The solution is f(x)=(3/5)x5-(2/3)x-3+(61/15).

A different kind of differential equation
We look at dy/dt=ky where k is a constant. The independent variable is usually called t here instead of x. The reason for t (time) will become apparent, I hope. Notice that this is a very different kind of equation, because what's on the right-hand side is a constant multiplied by y. We can't solve it by just antidifferentiating ky because we don't know y. So a different approach must be used. There will be more about such equations in Math 152. This is a simple example designed to help you work with a number of useful applications.

Translation into English; what is this?
I asked people to translate the equation dy/dt=ky in English. This is difficult. Here is a possible candidate for such a translation:

The rate of change of y over time (dy/dt) is directly proportional to y itself. ("Directly proportional" means that when the amount of y changes, that the rate of change of y is changed by k mutliplied by the change.)
This is a terrible "translation". I am sorry. Maybe I'd better stick with adding fractions. English is too difficult!
There are many real phenomena which satisfy this sort of growth (or decay) rule. Examples include:

What are the solutions?
I would hope that dy/dt=ky would have a family of solutions (the general solution) and we would use an initial condition to pick out one of these (the particular solution).
Well, first let's guess one solution of dy/dt=ky. For example, I know a wonderful function which is its own derivative: the exponential function (so that's a solution when k=1). After some further consideration, the function ekt was suggested. Indeed, if we differentiate this, the Chain Rule "spits out" a multiplicative factor of k, so the derivative of ekt is ektk, and this is ky.

Are there other solutions? Here is a trick to learn about other possible solutions. It is a fairly clever trick, and is used in other computations, which is why I'm showing it to you. Suppose y is another solution of dy/dt=ky. I want to compare it to the solution we know, ekt. The trick is to compare this way:
Look at y/ekt (the unknown solution y divided by the known solution, ekt). Let's differentiate this. The Quotient Rule gives uys:
Look carefully at the top of the fraction. We are assuming that y´ is ky. Then the top becomes:
because things exactly cancel. This means (MVT tells us: a function with 0 derivative is constant) y/ekt is a constant, C. So y=Cekt.

I admit totally truthfully that I don't go through any process remotely like what I just showed you in practice. In fact, if I think that the differential equation dy/dt=kt is a good description of a situation, then I immediately jump to Cekt. If you work with such situations for a while, I think you will do this also.

Mr. Patel's bacteria
Mr. Nitesh Patel graciously suggested the following (hypothetical) observations of a bacterial colony:

Initially, there are 50 bacteria. In 3 hours, 100 bacteria are observed (3 hours is called the doubling time). In 6 hours, 200 bacteria are observed.
People believe that dy/dt=kt is a differential equation which approproiately describes growth in such a situation (some comments on this assumption are below!) because bacteria (generally) reproduce asexually by division, and new bacteria are created by a fraction of the current bacteria splitting. I would like to make a mathematical model (in this case, get a simple formula) for the bacteria at time t.

I will measure t in hours from the start of the observation. B(t) will be the number of bacteria at time t. So B(0)=50 and B(3)=100 and B(6)=200. If we suppose that B(t)=Cekt (that follows from the differential equation's applicability to this situation) then we need to identify numbers for C and k.

Since B(0)=50 and B(0)=Cek·0=Ce0=C·1=C. So C is 50. Now we know that B(t)=50kt. We need to identify k, and we can use B(3)=100 for this. So: 100=50ek(3) which is 2=e3k which is (take ln's!) ln(2)=3k so that k=[ln(2)/3]. (People who work with these equations a great deal develop lots of computational shortcuts!).

So the number of Mr. Patel's bacteria is given by B(t)=50e[ln(2)/3]t. We could check this formula by plugging in t=6, since we know the answer should be 200. Here:
B(6)=50e[ln(2)/3]6=50e2ln(2)=50eln(4)=50·4=200 (because exp and ln are inverse functions).

Defects of the model?
Maybe bacteria don't reproduce the way we think (indeed, when I was younger, mostly people did believe bacteria could only reproduce asexually, but this is not always true). There's a more subtle assumption hidden in the model. The exponential function grows very fast. For example, there will be 1010 bacteria in about 82 hours (I did these computations secretly). That isn't so big, biologically, since there may be about 1014 cells in the human body. But if you continue to believe this model is valid, there will be 10100 bacteria after about five and a half weeks. Now to understand 10100 is difficult. Uh ... there are about, say, 6 billion people in the world, and so there are about 6·109·1014=6·1023 human cells in the world. Multiplied exponentials add "upstairs". This means we'd need lots and lots of worlds (more than 1075) to have maybe one bacteria per cell.

In fact, the bacteria will grow exponentially as long as conditions allow. That is, there needs to be adequate food, places to excrete poisons, etc. I think, in reality, these limits to growth impose themselves in a realistic way fairly soon in the case of bacterial growth. So the exponential model for bacterial growth is valid, but really only for relatively brief intervals. Exponential decay (with k<0) can be applied more appropriately over long periods of time to radioactivity. I'll try to discuss this next time.

Friday, November 30 (Lecture #25)

If F´=f, then abf(x) dx=F(x)|ab=F(b)-F(a).

Simple examples

  • 01x1/3dx=(3/4)x4/3|01=(4/3).
    Magic? I think it is very easy to not think about even "simple" computations like evaluating this integral. If we used Riemann sums, I believe getting an exact answer would require quite a bit of manipulation. Here all we need to do is guess (?) an antiderivative of x1/3 (hey, it comes from a power one unit higher, well, then we need to correct for the number that comes down in front, so the answer should be ...) and then the number practically presents itself. This is amazing technology to me.
  • 49(3x2-1/sqrt(x))2dx. There are various approaches which are possible. The simplest which occurs to me is just to expand the square. That is:
    So now we compute 499x4-6x3/2+1/x dx= (9/5)x5-6(2/5)x5/2+ln(x)|49=(9/5)95-6(2/5)95/2+ln(9)-(=(9/5)45-6(2/5)45/2+ln(4)). I would be content with this answer on an exam. Also, please note the big parentheses. I tell you with true humility that I have messed up such parentheses too many times. They are easy to forget.
  • 015ex+2e-xdx. Here the challenge in applying FTC 1 is getting an antiderivative. Well, ex is its own antiderivative. What about an antiderivative of e-x? First, I would "guess" e-x. But then I would quickly check and correct. That is, if we d/dx e-x, the result would be -e-x, so a suitable antiderivative would be -e-x. Now look:
    015ex+2e-xdx=5ex-4e-x|01= 5e1-4e-1 -(5e0-4e-0)=5e-4/e-1.

    FTC 2
    If F(x)=axf(t) dt, then F´(x)=f(x).

    An alphabet lesson
    What is 01x2dx? In the last lecture, we saw with several different techniques that this is 1/3. Then let's play a little logical game.

    The labels d(something) inside the definite integral do not get "communicated" to the outside. This is just like the index of summation in one of those 's. This is the analogue of what is called a local variable in computer science, defined inside a for loop or a subprogram or subroutine. In math, one phrase used for such a thing is local variable. The letter is there to help the computation, but only has a meaning within the scope of the computation of the integral.

    Simple example
    Describe a function whose derivative is cos(x+ex).
    This is either very very easy, or, for reasons that will be discussed more in Math 152, very very hard. I will give the very very easy answer here.

    One function is F(x)=-3xcos(t+et) dt. This is a function whose derivative is cos(sqrt(x)+ex) by using FTC 2.

    Another function is G(x)=5xcos(w+ew) dw.

    Let me compare the two answers. One has inside variable t and the other has inside variable w. This doesn't matter. They both are "satisfactory" answers to the question asked because FTC 2 implies that the derivative of a definite integral with a variable upper parameter is the integrand's value at that upper parameter (hey, I'm using all of the high-priced words!) One answer is a definite integral from -3 to x and the other is a definite integral from 5 to x. Using property (*) from last time, the two answers are different by this: -35cos(s+es) ds. What is this? It is a constant (I'm just using another variable, s, to again bring up the idea of a "dummy variable"). Two functions which differ by a constant have the same derivative. So that's o.k.

    By the way, it can be proved that no one can do much better in this problem than write the answer as a definite integral. There isn't any much simpler answer. Software can then plot this function, because people have spent a great deal of time and effort learning how to compute good numerical approximations to definite integrals very rapidly. The picture of F(x), the antiderivative of cos(x+ex) which is defined above, on the interval [0,2] was produced by Maple in less than three-tenths of a second. I am sure other software would do the job as well.

    More examples (a textbook problem)
    This is problem 23 in section 5.4. The graph of f is given to the right. The problem asks us to sketch a graph of A(x)=0xf(t) dt. I presume, since f's graph is given on the interval [0,4], that we are supposed to sketch a graph of A(x) on the same interval.

    There are various strategies for analyzing this sort of problem. Since the information is furnished graphically, I'd probably try to do the problem graphically, or at least begin it that way. I would think of the limits, 0 and x, and try to see how the definite integral (which I'd think of as signed area) varied as x varied.

    Let's consider first x's between 0 and 1. Here I think of a vertical line somewhere over the [0,1] interval, moving to the right. The area between that line and the y-axis, under y=2, is 0xf(t) dt. Since the line has height 2, the area is 2x, and the graph of A is a straight line segment starting from (0,0) with slope 2.
    When the vertical line passes x=1, we have accumulated 2 units of area, and A(1)=2. But now the "profile curve", f, changes height. It has height 1. We must add to the 2 units of accumulated area the new area we are getting between x and 1. Since the height of f is 1, and the base is x-1, we are add on 1(x-1) units of area.

    The graph of A in this interval will be linear with a slope of 1, and it will start from (2,2).

    At x=2, we have accumulated 3 units of area, 2 over the interval [0,1], and 1 more over the interval [1,2]. Now x moves to the right in the interval [2,3]. The height of f is -1, so the definite integral will decrease: the geometric area is below the x-axis. We know A(2)=3 becuase of the accumulated area.

    But if x is between 2 and 3, the change in A is (-1)(x-2). So we need to show the graph of A decreasing, and it will be decreasing linearly in x, with a slope of -1.

    By the time the vertical line has gotten to x=3, it has accumulated 3-1 units of area. Yes, I know that 3-1=2, so actually A(3)=2, but somehow writing and thinking 3-1 helps me recall that + is assigned to areas above the x-axis and is assigned to areas below the x-axis.

    Between 3 and 4, the height of f is 0 (zero: nothing). There is therefore no change in the area as we move the upper limit on the definite intergral to the right. And therefore there is no change in A, and the graph of A is a horizontal line segment beginning at (3,2).

    From this we get a very good idea of the graph of A(x). It is 4 line segments with slopes of 2, 1, -1, and 0. It is a continuous function. The slopes of A correspond to the heights of f, and actually A is an antiderivative of f (the antiderivative with the initial condition A(0)=0), so f is the derivative of A. I think the graph looks like what's shown to the right.

    Part B
    Well, there's actually a part B) to this problem. The graph to the right is given, and, again, we are asked to graph A(x)=0xf(t) dt. Here the graph is a bit more complicated (well, it is more complicated to me: the other function had exactly 4 values, and this function has lots and lots of values!).
    I'll use a strategy different from what we did in A) to understand and solve this problem.
    Probably I would look at the graph and learn that A(0)=0 (there is no area from 0 to 0!). I would look a bit more and find that A(2)=1, because 02f(t) dt is the area of a triangle with base 2 and height 1, and (1/2)(2)(1)=1. Finally, I would see that A(4)=2, because there are two of those triangles. So I have the three dots shown to the right on the graph of A.

    Notice also that the graph of A is increasing because as we move the right limit to the right we are getting more and more area, and this is positive area since it is above the x-axis.

    What can we say about the derivative of A in the interval [0,2]? By FTC 2, this derivative is the function f, and the function f is just (1/2)x. That means the second derivative of A is 1/2, and this is positive. So the graph of A between 0 and 2 is concave up, and connects (0,0) and (2,1).

    I actually know a formula for A, since A(0)=0 and A´(x)=(1/2)x. It must be (I antidifferentiate and get the correct constant!) A(x)=(1/4)x2. So what I see between 0 and 2 is a piece of a parabola.

    Between 2 and 4, the function keeps increasing. But the slope of the function, A´, is postive: it seems to be a line segment from (2,1) to (4,0). But the derivative of that is –1/2, so the second derivative of A is negative. A is increasing and concave down. I think the graph is must look like what is displayed to the right.

    In fact, between 2 and 4, A´(x)=-(1/2)x+2 (I got this wrong in class, and I hope it is correct here!). I also know that I have accumulated 1 unit of area by the time we get to 2, so that A(2)=1. I can solve this initial value problem: antidifferentiate to get A(x)=-(1/4)x2+2x+C. Use A(2)=1 to get C: -(1/4)22+2(2)+C=1, so C=-2, and the formula for A(x) in this interval is -(1/4)x2+2x-2. We can check this by plugging in x=4, so the result is -(1/4)42+2(4)-2=-4+2(4)-2=2 which is correct.

    Another thing which is not obvious is that the function A is differentiable. Since A is defined by two nice formulas, it isn't very surprising that A´ exists inside the two halves of the domain. In fact, A is differentiable at x=2, also. This can be checked with effort by looking at the difference quotient (there is a similar check in the workshop problem solution inserted in the Rutgers edition of the textbook). But I am willing to believe that A´(2) exists, because:

    • If A(x)=(1/4)x2 then A´(x)=(1/2)x so A´(2) should be 1.
    • If A(x)=-(1/4)x2+2x-2 then A´(x)=-(1/2)x+2 so A´(2) should be 1.
    I bet that the slope of the tangent line at 2 is 1.

    Movement of a particle
    Let's assume that the velocity of a particle traveling on the x-axis is given as a function of time: v(t)=4t2-t4 for t between 0 and 5. (As far as I know, this is a rather non-physical example, but I hope that the questions and ideas contributing to the solutions are correct and useful.)

    1. For which values of t is the particle traveling to the right? To the left?
      "Travel to the right" in terms of velocity means "v(t)>0" and of course "to the left" is "v(t)<0". Since v(t)=4t2-t4=t2(4-t2). Notice that t2 is non-negative, so that 4-t2 has the sign information. We should consider 4-t2 on the time interval [0,5]. Of course, 4-t2=0 when t is +/-2. So (considering everything) v(t)>0 when t is between 0 and 2 and v(t)<0 when t is between 2 and 5. Notice that velocity is 0 when t=0 and t=+/-2.
    2. At what time is the particle farthest to the right?
      The particle travels to the right when t is between 0 and 2, and travels to the left when t is between 2 and 5. So the particle is farthest to the right when t=2.
    3. Suppose that at time 0 the particle's position is 12. What is its rightmost position?
      Well, displacement in an interval [t1,t2] means the difference in position at time t2 and at time t2. So the displacement is the definite integral of the velocity over the interval [t1,t2]. So the displacement over the interval [0,2] (2 is when the particle is farthest to the right) is 02v(t) dt=024t2-t4dt=(4/3)t3-(1/5)t5|02=((4/3)(23)-(1/5)25)-(0[all 0 when t=0)=(64/15). We get the position when t=2 by adding the initial position, which is 12, to this result. So the position when t=2 is 12+(64/15).
    4. What is the displacement of the particle during the time interval [0,5]?
      The displacement is 05v(t) dt: this is the difference in the position at time 5 and the position at time 0. So we compute: it is 054t2-t4dt=(4/3)t3-(1/5)t5|05, and this turns out to be (4/3)25-(1/5)55, which is -625+(4/3)25: a quite negative number, so the particle moves really left in the interval.
    5. What is the distance the particle travels during the time interval [0,5]?
      This is, to me, the most ticklish (in the sense of "delicate") question. The particle travels right from 0 to 2 and then left from 2 to 5. The displacement is not the same as the distance traveled. The distance traveled is how the particle goes from 0 to 2 and then (added on) how the particle goes from 2 to 5. The integral from 0 to 2 will be positive and the integral from 2 to 5 will be negative. But we need the sum of the amounts of each, with the signs stripped off. Hey, here are two math ways of writing what we want:
          05|v(t)|dt (notice the absolute value sign!)
      The reason for the minus sign is that velocity is negative between 2 and 5. So subtracting the integral of velocity adds the distance traveled then.
      I asked my silicon buddy to compute 05|4t2-t4|dt and it did this without much complaint. It used about .17 seconds, which is quite a lot for a simple definite integral. But in that time, it determined the intervals where what's inside the absolute value sign is negative or positive (that's what probably took most of the time), and it chopped the interval into two pieces, switched signs on the formula in one of the intervals, and applied FTC 1 to each of two integrals. The answer was 7003/15.
    Here is a sort of picture of the movement of the particle. The picture is qualitatively correct, but the distances are very wrong. I just want you to "see" what the movement might look like.

    A "complicated" integral
    Let's "compute" 01cos(x17)x16dx. If we want to apply FTC 1, we need to find an antiderivative. Well, the integrand is certainly not random. There is a pairing which I hope you notice after more than 90% of a calculus course. The x17 inside the cosine function, and the x16 multiplying the outside of the cosine function. Since we are trying to identify a function whose derivative is cos(x17)x16, to me a natural guess is something like sin(x17. If we make that guess, then the derivative of sin(x17) is cos(x17)(17x16) using the Chain Rule. But we don't want the multiplicative constant 17, so, just as in a bunch of examples we've already done, we fix up our initial guess to get (1/17)sin(x17). And this works, and then, using FTC 1, the definite integral's value is (1/17)sin(x17)|01=(1/17)sin(1).

    I'm actually not too interested in the specific value of the integral, but more in how to make the guessing process works. Of course, the integrand was set up so we could guess. But the process works often enough that people have make some notation which makes it easier to do. Here is what they would write in this case:

    In x-landcos(x17)x16dx
    Guess a uu=x17
    Compute dudu=17x16dx
    What about the notation?
    Well, if u=x17, then du/dx=17x16. This is one of the traditional notations for derivatives ("Leibniz notation"). If we think that du/dx is a fraction, we could multiply by dx and get the equation above. One reason people like Leibniz notation is that it works well with this method of antidifferentiation which is called substitution. Although I know that the derivative is a limit and not a fraction, I certainly use the substitution method very freely, and I don't worry about separating the du and the dx. It works! Notation should help, and this notation works because the Chain Rule is correctly used.
    Adjust du to match
    what's in the integral
    Go to u-landcos(x17)x16dx= cos(u)(1/17)du=(1/17)cos(u) du
    Antidifferentiate(1/17)cos(u) du=(1/17)sin(u)+C
    Return to x-land(1/17)sin(x17)+C

    Of course, this example is constructed so that things work straightforwardly. A chunk of my job is to show you a range of examples and try to help you learn how to recognize situations where this method of antidifferentiation, called the substitution method, will work. Your job, on the other hand, is to practice a bunch of suitable examples (problems in section 5.6).

    What doesn't work
    Earlier in this lecture we wanted to compute 49(3x2-1/sqrt(x))2dx and one suggestion, which I stifled, was to try substitution. If we take u=3x2-1/sqrt(x), then du=(6x+(1/2)x-3/2)dx. I don't think substitution will be very successful since we don't have all that stuff in the original integral.

    Begin to prepare for the final exam
    Several students have sent me e-mail requesting advice about preparing for the final exam. Here is some preliminary information, and I will try during the coming weekend to write something more complete.

    The exam will be cumulative and cover the entire course. The exam will be principally written by the course coordinator, not by me. But I've been the course coordinator for Math 151 several times, and therefore have some idea of what a final exam for this course will look like. I will write more about this, but here is what you should know immediately:

    Although the exam will cover the entire course, it is likely that it will disproportionately overweight (there will be more coverage!) what we have done since the second exam: antiderivatives, definite integral, uses of the definite integral, methods of computing the definite integral, etc. There are several reasons for this disproportion: first, the material has not yet been tested, and second, for many reasons both theoretical and practical, this is the most important stuff in the course. If you wish to have a chance at doing well on the final exam, start now and do every suggested problem in the sections starting with 4.9. Begin doing this now! Please do not delay.

    You should also accumulate questions to ask during the last recitation meeting, and to ask the recitation instructor and the lecturer. Don't let "details" and "ideas" slide by --- you should always assume that exactly those textbook problems or lecture examples that you can't do or don't understand are exactly what will be asked on an exam.

    Tuesday, November 27 (Lecture #24)

    I tried to repeat a few essential details from the previous class. Many people were not present because they needed to leave early so they could get home for Thanksgiving. As we all know, New Jersey is HUGE. Its north-to-south length is more than 3·1015 angstroms (there are 10 angstroms in a nanometer).

    Last time we saw that if we had a function f defined on an interval [a,b], then we could introduce a rather elaborate collection of objects which are used to analyze f's behavior on all of [a,b]. This is a link to definitions and a discussion of these objects. If [a,b] has a partition P (which break up [a,b] into a number of subintervals) and a collection of sample points (the text's intermediate points) C, then we defined the Riemann sum of f using P and C:
    If the length of the longest subinterval-->0, then (for the functions we will consider in this course!) the Riemann sums approach a limit which is called the definite integral of f from a to b and written abf(x)dx.

    Definite integrals have some very useful properties which people use frequently.

    Property (*) and consequences
    Suppose a<b<c. Then abf(x)dx+bcf(x)dx=acf(x)dx and I will call this statement (*). I am happy if you believe this statement. I hope that the picture to the right is enough verification. The picture has red for area below the horizontal axis and green for area above the axis because I wanted to remind you how area is counted for the definite integral. It is possible to prove this statement from the definition with Riemann sums.

    Unexpected consequences of (*)
    While most people are willing to believe (*), there are some results which may take you some time to learn to use. For example, in the equation
    what if we change b to a? Then we seem to get
    and the only way this could be true is if
    Most people find this easy enough to believe: this quantity is the area of a "region" with width equal to zero, and some height, and such an area should be 0. So any definite integral whose upper and lower limits are the same must be 0.

    O.k., if you believe that one, then I will make another change in the equation:
    Change c to a in this equation, and use the fact that the right-hand side will become aaf(x)dx which we already "know" is 0. So we get:
    and this certainly means that baf(x)dx=abf(x)dx
    so that when the limits of a definite integral are interchanged, then the value's sign is changed!

    Suppose f is the function whose graph is shown to the right. It is piecewise linear, and made of four line segments. Then all of the values of these integrals:
        02f(x)dx    -11f(x)dx    3-1f(x)dx    44f(x)dx
        13f(x)dx    -12f(x)dx    30f(x)dx    0-1f(x)dx
    are contained in this collection of numbers:
        -1    0    1/2    1    3/2.

    Property (**)
    This is extremely useful when you are confused and need to make some sort of estimate. Suppose, somehow, you know that all of the values of f(x) on the interval [a,b] are between m and M. That is, you know m<=f(x)<=M for all of those x's. Then (look at the graph!) the "area" (actually the definite integral) will be trapped inside a box. So the result is:
    This is because the wiggly region contains the smaller box and is contained by the larger box. It looks silly, but the estimates can really be useful in checking computations.

    Yet another example ...
    Here is a final example of how to compute an area (or a definite integral) in the most direct fashion: by getting the area of a collection of approximating rectangles, and then analyzing what happens as the partition gets "finer".

    The example is the area under y=x2 from 0 to 1. This means the area enclosed by the x-axis, y=1, and y=x2. It is an area in the first quadrant. We will approximate the area, which is the definite integral 01x2dx by a collection of Riemann sums. Here f(x)=x2.

    1. Partition [0,1] into N equal subintervals, where you should think that N is some large positive integer. Since the original interval has length 1, each subinterval will have length 1/N. The actual partition points will begin 0/N, 1/N, 2/N, 3/N, 4/N ... and will end with (N-1)/N and N/N.
    2. Get sample or intermediate points. In this computation, I will choose the right-hand endpoints of each subinterval as the sample points. So in the first subinterval [0/N,1/N], the sample point will be 1/N. In the second subinterval [1/N,2/N], the sample point will be 2/N. In the third subinterval [2/N,3/N], the sample point will be 3/N. ... In the Nth subinterval [(N-1)/N.N/N], the sample point will be N/N.
    3. Compute the function values at the selected points. These numbers are the heights of the approximating rectangles, and they are f(1/N)=12/N2 and f(2/N)=22/N2 and f(3/N)=32/N2 and ... f((N-1)/N)=(N-1)2/N2 and f(N/N)=N2/N2. Wow!
    4. Multiply each of the heights by the widths of the rectangles. Each of the widths is identical, however, so things aren't so bad. The results for the areas of the rectangles are:
      12/N2·(1/N) and 22/N2·(1/N) and 32/N2·(1/N) and ... (N-1)2/N2·(1/N) and N2/N2·(1/N). More wow!
    5. Add 'em up. So this is:
      More compactly written, the Riemann sum is j=1Nj2/N3. I combined the powers of N "downstairs". You could also pull out the 1/N3 because it multiplies every term in the sum, and then the result would be (1/N3)j=1Nj2.

    If we had N=5, the approximation would be (1/53)(1+4+9+16+25) which is (1/125)(55). The mysterious part is the 55, of course.

    A magic formula
    Well, everyone knows (it is in the book on page 317, so ... it is possible to know it!) that j=1Nj2 is (1/3)N3+(1/2)N2+(1/6)N.
    We actually checked that if N=5 is plugged into this formula, the result is 55. That's nice. So maybe the formula is correct.

    Where do these formulas come from? Why are they correct?
    There are lots of amazing formulas like this, and they were discovered initially to allow people to compute definite integrals exactly (the definite integrals first occurred as areas or volumes). Next year I hope to teach a first-year seminar and show people how to discover such formulas, and, after they are discovered, check that they are correct. After all, we only checked that the sum of squares formula was correct for N=5. I have no desire to check directly that it is correct for N=238, yet I do believe it will give the correct result. So if you are a first-year student next fall, then ...

    Getting the exact value of the definite integral
    Since we know that j=1Nj2 is (1/3)N3+(1/2)N2+(1/6)N and that the Riemann sum is this divided by N3, then the value of the Riemann sum is (1/N3)((1/3)N3+(1/2)N2+(1/6)N) which is (1/3)+(1/2N)+(1/6N2) and certainly as N-->, this must approach 1/3. So 01x2dx=1/3.

    Change the problem: make it harder!!!
    These techniques seem to be (they are!) very intricate. Let me show you how to solve this problem by first making it more difficult. The people who discovered that the (seemingly) more difficult problem can be solved very neatly really had a lot of intellectual courage.

    So instead of studying 01x2dx, we will study the function
    We really want to know A(1), but we'll consider this more general problem anyway.

    What do we know?
    Well, there is one value of the function that is extremely easy to compute: A(0)=00x2dx=0.

    That seems to be the only simple thing. But this is a CALCULUS course, and maybe we should try to differentiate A(b). This, which certainly should seem almost ludicrous, turns out to be the successful approach.

    In order to find A´(b) we will need to consider A(b+h)-A(b) and then divide the result by h.
    Certainly (*) implies that
    so that A(b+h)-A(b) is bb+hx2dx.

    But we can use (**) now. f(x)=x2 is increasing on [b,b+h], so for all x in the interval, b2<=f(x)<=(b+h)2. So, with m=b2 and M=(b+h)2, we see that
    since h is the width of the interval. Now divide by h:
    Notice that both estimates -->b2 as h-->0. This means that A(b) is differentiable, and that A´(b)=b2.

    An initial value problem
    We have a function, A(b), with the following properties:
        A(0)=0 (initial condition);
        A´(b)=b2 (differential equation).
    This is an initial value problem. All antiderivatives of b2 have the formula (1/3)b3+C. If A(b)=(1/3)b3+C then A(0) must be C, but the initial condition states that A(0)=0, so C=0. Therefore, A(b)=(1/3)b3.

    Now solve the original problem
    Since A(b)=(1/3)b3, then A(1)=(1/3). No problem!

    Hey, things are easier than they look
    In fact it turns out that we didn't need to figure out the C, because the value we need to compute the integral from 0 to 1 is A(1)-A(0), and any antiderivative will give the same result (the "+C" adds and subtracts, and doesn't change the answer).

    The Fundamental Theorem of Calculus, version 1
    If F´=f, then abf(x)dx=F(b)-F(a).
    In fact, the notation usually used "abbreviates" F(b)-F(a) by F(x)|x=ax=b. And "Fundamental Theorem of Calculus" is also usually abbreviated as FTC.

    Here is how the original area desired would be computed.

    1. Let's find the area under y=x2 from 0 to 1.
    2. Recognize that this is the same as computing 01x2dx.
    3. We know an antiderivative of x2: it is (1/3)x3.
    4. FTC then declares that the definite integral (1/3)x3|x=0x=1 and this is (1/3), the value of the area desired.

    The area under a bump of sine
    What is the area undo a bump of sine? Well, first think a bit: the idea of bump maybe means the area between two places where sine is 0 and where the curve is above the x-axis. So the area would be computed by the definite integral 0Pisin(x) dx. Usually having some estimate of the size of a quantity, even before computing, is a good idea. Otherwise you may make some simple error, and get some result which is a huge mistake.

    One overestimate of this definite integral is obtained by looking at a rectangular box which contains the area. The width of the box is Pi and its height is 1, so the integral should be less than Pi·1=Pi. An underestimate of the area could be gotten by looking at two triangles under the graph. Each has base Pi/2 and height 1, so the total area of the two triangles is 2·(1/2)·(Pi/2)·1=Pi/2 (the 1/2 comes from the formula for the area of a triangle). One student suggested 0 as an underestimate. While that is certainly valid, I like under- and overestimates which are closer to the exact answer (as long as I don't have to work very hard!). Now we know
    Pi/2<=0Pisin(x) dx<=Pi.

    To compute this area exactly, we use FTC. We need an antiderivative of sine, and that happens to be -cos(x). So:
    0Pisin(x) dx=-cos(x)|0Pi=-cos(Pi)-(-cos(0))=-(-1)-(-1)=2. The (potential!) confusion of minus signs occurs often when applying FTC, so some care is needed. The exact area is 2, and this is certainly between Pi/2 and Pi.

    A fake computation: the "area" under a bump of cosine
    I then tried to rush ahead and compute the "area" under a bump of cosine. Or, rather, I compute this definite integral: 0Picos(x) dx. I asserted that the result should be ... well, should be ... something.

    Here is the computation using FTC, since an antiderivative of cosine is sine.
    0Picos(x) dx=sin(x)|0Pi=sin(Pi)-(sin(0))=0-0=0.
    What happened to the "area" under the bump? We computed a definite integral, and that's the limit of Riemann sums, and they count geometric area under the x-axis negatively. The graph of cosine on the interval [0,Pi] exactly balances the positive over-the-axis region with a negative under-the-axis region. So it is 0 both computationally and geometrically!

    I sketched the curve y=x2(x-1) from 0 to 1 as shown to the right. I asked what the area was inside the bump. I observed that the function was negative, so the definite integral would be negative, but the area would be positive.

    A solution might go like this:
    01x2(x-1)dx=01x3-x2dx=(using FTC)=(1/4)x4=(1/3)x3|01=(1/4)-(1/3)=-1/12.
    Therefore the area is 1/12.

    Friday (=Tuesday), November 21 (Lecture #23)
    A question of blood...
    I asked how we could estimate how much blood the heart pumps out to the body, drawing an especially silly picture on the board (look left). Knowing the quantity of (hopefully!) oxygenated blood which is pushed around the body can be a useful part of health assessment. How can we measure the blood flow through the aorta, the major artery emerging from the heart? A better diagram of a heart is to the left. The heart is a very complicated and wonderful pump.

    The Aztec solution
    One method might be to take a knife, preferably an obsidian knife, and rapidly cut through the chest wall, sever the aorta, and cause the blood that would flow through it to fill up a bucket. In about five minutes we'd probably have a good idea of how much blood flows through the aorta. There are several criticisms possible of this "protocol". One criticism might be that, well, maybe the person involved might not be good for much afterwards. Another criticism could be that this would not show the actual pumping capacity of the aorta, because maybe within a minute or two some ... difficulties would occur. In some sense, this Aztec protocol is the ultimate in what might be called destructive testing. By the way, obsidian is "a dark glassy volcanic rock formed from hardened lava" and was actually used for certain similar purposes in that culture.

    Another solution: cardiac catheterization
    The American Heart Association and the NIH (National Institutes of Health) both have lots of information about this. Here are some quotes from these sources:

    Very much simplified ...
    Imagine a thin plastic coated wire inserted into your circulatory system and threaded up into the aorta. At the end of the wire is a sort of microscopic windmill: a bunch of little fans which spin as the blood flows. As they spin, they move a bunch of wire around a magnet. Since magnetism+movement+wires=current ( a wonderful fact!), electric current is created and passed back through the wire into a measuring device. This is certainly very much simplified. Now I will further simplify and imagine some numbers. This is not reality, but a math course! The following numbers are invented, and the whole purpose of this "exercise" is to get you to think about what's going on.

    Elapsed timeFluid flow rate
    1 min30 cm3/min
    2 min20 cm3/min
    4 min40 cm3/min
    5 min25 cm3/min
    The blood flow will be given in cm3/min (cubic centimeters per minute) and the time will be measured in minutes since the beginning of the procedure. The information learned is displayed to the right.

    This isn't much information. But how can we use what we've learned? A standard way to use these numbers is the following. It is certainly very approximate in this case.

    At the end of the first minute, we're told that the flow is 30 cm3/min. Hey, we could estimate how much blood is pumped by making the assumption that the flow rate is constant during the time interval involved. Then we see that 30 cm3/min·1 min=30 cm3 (flow rate multiplied by duration) would be pumped. I am not asserting that this is necessarily correct. It is an estimation. We are using the information, and the method is very simple. Similarly, during the second minute, we can estimate that 20 cm3/min·1 min=20 cm3 is pumped. The flow rate changed during the second minute. We're faced with a slightly different "challenge" in using the third line of the table. It has elapsed time (since the start of the procedure) as 4 minutes. So the duration has now changed from 1 minute to 2 minutes. (Why? Well, y'know, in the real world things maybe don't work as you'd like. Maybe the machine broke, or the technician didn't pay attention, or ... in any case, we need to work with the data we have!) So during those two minutes, we have (flow rate·duration) an estimate of 40 cm3/min·2 min=80 cm3 blood pumped. Finally, in the last minute of this procedure, we estimate that 25 cm3/min·1 min=25 cm3.

    Let me abbreviate just a bit. The units are repetitive and I will omit them (this is a math course!). So here are the essential details of the computation:
    That's what we're computing.

    If we wanted a picture of the computation, maybe what is shown to the right gives you some idea. The dots represent the actual known data. The area of the rectangles shown is given by the product of the appropriate flow rate and duration. Yes, there are other possible approximations using the data given. We will study some of them in Math 152, but here only a rather simple-minded approach will be used.

    More information
    Elapsed timeFluid flow rate
    30 sec32 cm3/min
    1 min30 cm3/min
    1 min 30 sec28 cm3/min
    2 min20 cm3/min
    2 min 30 sec–6 cm3/min
    3 min–8 cm3/min
    3 min 30 sec34 cm3/min
    4 min40 cm3/min
    4 min 30 sec43 cm3/min
    5 min25 cm3/min
    Now let's refine the data. Let me suppose that we have no "technical" problems, and that we are recording the flow rate every 30 seconds. We could imagine the numbers given in the table to the right. Please notice that the data we've already obtained is part of this table -- I am imagining that we're measuring the same process as before, only we are sampling things more often. Here we have readings every thirty seconds with no data points are omitted. So we actually have 10 pairs of numbers.

    There is another complication which has occurred, and I hope you notice it. Some of the flow rates (2 min 30 sec and 3 min)are negative. Could this happen? Well, actually maybe yes. The heart is a terrific pump but maybe sometimes things don't always synchronize well or work perfectly: maybe the heart valves don't close all the way or don't close at precisely the correct time in the pumping cycle, etc. This complicates our task. We could ask what the heck we want to measure in this situation, and several answers are possible. We might want to measure the total pumping effort of the heart. Or we might want to know the net amount of oxygenated blood which is pushed out into the body. Let me look at the second quantity. In that case, the negative flow rates affect the total blood pumped because, indeed, that amount in that time should be subtracted, just like the sign of the flow rate indicates, from the total.

    What should be computed here? If we do flow rate multiplied by duration, we should be consistent (the units!) and realize that, for example, 3 min 30 sec is 3.5 minutes. So the numerical computation here is
    And this estimate, using this data, is 119 cm3. This is much less than the earlier figure, I guess because the data at 4 minutes over a duration of 2 minutes led to a big overestimate.

    A picture of what we computed is to the right, which I sort of laid over the old bar chart whose outlines where visible are indicated with dashed lines. I put in the new data points, and then the computation above is indicated by the areas of these rectangles. But the total area of the rectangles which are below the horizontal axis is subtracted from the total. Can you "see" the change from 155 to 119?

    Ideally ...
    You might imagine that ideally we make lots and lots of measurements (in fact, this is more like what is done, because I think there is a computer recording the flow rates very frequently and then making the computations that I'm discussing in such a silly way). What should we think about? We could have lots and lots and lots of data points. And then the rectangles get very thin. And then the data points look very much like a curve. The number that's wanted is the total blood pumped out to the body, but the picture which might result could look a lot like what is shown here. I have again overlaid it with the rectangles from the previous discussion. So it seems there might be some kind of flow rate curve, and that the blood pumped out to the body, the net blood out, is the area of the colored regions above the curve minus the area of the colored region below the curve.

    I hope that you can see some resemblance between this and the discussion of the area of the hand which we did previously. Here we are approximating quantities of liquid, and there we were trying to approximate areas. It turns out that this computational idea can be applied in many situations, and it is used in essentially all fields of engineering and applied science. I now need to tell you the official vocabulary.

    Start with the following: a function f defined on an interval [a,b].

    What this means geometrically ...
    To the right is my attempt to show you what a Riemann sum might look like. But I've made it quite simple. There are only seven subintervals, and the curve which is the graph of f(x) is also not too complicated. The dots on the x-axis denote the sample or intermediate points, and the vertical lines above them are the lengths f(cj). The boxes have areas which are equal to the pieces of the Riemann sum, f(cj)xj. The f(cj) is the height and the xj is the width of a rectangle.
    An example
    To the left is a graph of a function, y=f(x), where f is defined on the interval [0,5]. f's graph is two line segments, the first horizontal from (0,1) to (2,1) and the second tilted from (2,1) to (3,0), followed by the lower half of a circle of radius 1 centered at (4,0), a circular arc going from (3,0) to (4,-1) to (5,0).
    Here are some partitions, choices of sample points, and the resulting Riemann sums.
    Partition 0, 2, 3, 5.
    Sample points 1, 3, 4.
    The function values at the sample points are:
    f(1)=1, f(3)=0, f(4)=-1.
    Riemann sum 1(2-0)+0(3-2)+(-1)(5-3)=0.
    Choosing the sample point at 3 gives a zero contribution to the Riemann sum.
    Partition 0, 2, 3.5, 5.
    Sample points 1, 3, 4.
    The function values at the sample points are:
    f(1)=1, f(3)=.5, f(4)=-1.
    Riemann sum 1(2-0)+.5(3.5-2)+(-1)(5-3.5)=1.25.
    Partition 0, 2, 3.5, 4.5, 5.
    Sample points 2, 2, 4, 5.
    The function values at the sample points are:
    f(2)=1, f(2)=1, f(4)=-1, f(5)=0.
    Riemann sum 1(2-0)+1(3.5-2)+(-1)(4.5-3.5)+0(5-4.5)=2.5.
    The definition allows the sample point to be on the boundary of the subinterval, so two adjoining subintervals could possibly share their sample points. This is a bit wasteful in terms of information, though.
    Also notice that the rectangle on the rightmost interval has height 0.
    Partition 0, 1, 2, 3, 4, 5.
    Sample points .7, 1.44, 2.5, 3.5, 4.75.
    The function values at the sample points are:
    f(.7)=1, f(1.44)=1, f(2.5)=.5, f(3.5)=-sqrt(3/4)-.866, f(4.75)=-sqrt(7/16)-.661.
    (The numbers for f's values at 3.5 and 4.75 come from the equation of the lower unit semicircle, -sqrt(1-x2), suitably translated to the right.)
    Riemann sum 1(1-0)+1(2-1)+.5(3-2)+(-.866)(4-3)+(-.661)(5-4)=.973.
    Notice that putting another sample inside the interval from 0 to 2 didn't improve the approximation very much. If taking samples is difficult or expensive (and it might be in real life!) then probably increasing the sampling where functions wiggle a lot rather than where the values are (relatively) constant is a good idea. This is, in fact, what the approximation strategies inside most calculators try to do.

    Part of the impression I would like to leave with you, even with just a few examples, is that Riemann sums can be quite complicated objects. They can be irritating to compute, and there are so many of them that understanding them may be difficult.

    If the partition gets finer and finer ...
    Suppose that the largest length of the subintervals defining the Riemann sums gets shorter and shorter. Then if you look at the picture, you might hope that all the Riemann sums, no matter how you select the intermediate points (just so they stay inside their correct subintervals, though) will approach one specific number. In the case of the function whose graph is shown above, my candidate for specific number would be the area of a 2-by-1 one rectangle (2) added on to the area of a 1-by-1 right triangle (1/2) and then minus the area of half a circle of radius 1 (-Pi/2). So for that specific function, the Riemann sums will get close to 2.5-Pi/2.

    The definite integral
    Suppose some more general function f is defined on an interval [a,b}, and suppose either that f is continuous or piecewise continuous (that is, continuous except for a few jumps). Then as the maximum length of the subintervals used in the Riemann sums for this function get smaller, the values of the Riemann sums will approach a unique number which is called the definite integral of f on the interval [a,b], and the notation used is abf(x)dx.
    Here a is called the lower limit and b is called the upper limit. The integral sign, , is a sort of stylized abbreviation of the word Sum. The function f(x) is sometimes called the integrand. The "dx" is there to remind you of the length of little subintervals.

    The exams were returned. Here is one version, with some answers. The grading is discussed here.

    Friday, November 16 (Lecture #22)
    An exam was given, with answers now available, along with a discussion of the grading.

    Tuesday, November 13 (Lecture #21)
    We are about to jump into something that will seem very different. But there will be a remarkable connection between this topic and what we have already studied which will be revealed in a few lectures. I will begin with a seemingly silly question.

    Hand area
    What is the area of my hand? There was, of course, some laughter. But the question and other similar inquiries, while superficially silly, is actually quite important. Appropriate drug doses are sometimes calculated in terms of BSA, body surface area. This is especially important when prescribing drugs for infants or drugs with strong side-effects (for example, some drugs used in cancer treatment). And knowledge of body surface area is important in trying to understand heat balance when people exercise. There are many formulas for BSA.

    So my question is much more modest: how can we compute the area of my hand? We discussed this. There are many methods, but if you think about area, there are really not very many shapes whose area we can compute exactly. The most important and simplest shape is a rectangle, where we know the area is the product of the length and width. We should try to take advantage of this. So look at an image of my hand, and ... well, after discussion we came up with this (mostly at the suggestion of Mr. Chang):

    1. Superimpose a square grid over an image of my hand.
    2. Put boxes inside.
    3. Put boxes outside.
    4. Count the boxes totally inside. Count the boxes which contain any (even little!) piece of the hand image (these are the "outside" boxes).
    5. Compute the area of the inside boxes and the area of the "outside" boxes.
    6. The hand area will be between the inside box area and the outside box area.
    Let me "implement" (!) this and get some information.

    I'll suppose first that I put the image of the hand in a square grid which is, say, 1 mm (millimeter) on a side. I don't know if this is physically too reasonable, but, what the heck, I could get something that looks like the picture to the right.

    Then I could count the grid boxes which are entirely within the hand image. There was some objection to this, because, you see, I could always be a little clever (?) and maybe sort of count some boxes which are almost entirely in and then compensate by dropping the count by 1 if I count a bunch of those.

    I commented that I wanted a process (an algorithm) which could be done totally "mechanically", leaving nothing to choice, and which was extremely simple-minded.
    I'll suppose first that I put the image of the hand in a square grid which is, say, 1 mm (millimeter) on a side. I don't know if this is physically too reasonable, but, what the heck, I could get something that looks like the picture to the right.

    I think if I count correctly there are a total of 18 boxes entirely inside the hand image.

    Now for an overestimate. Let me start with the boxes which are entirely inside, and then add on any boxes that have even the slightest "tinge" (?) or hint of the hand image inside. Again, maybe there are objections: we could be cleverer (?) and sort of estimate how much of a box is inside, etc. But then we would have to make decisions, and some of those decisions maybe could be difficult. Here, if we have a simple-minded scheme, I'll bet that any two people will come up with the same answer. At least we get a result that's stable, and probably doesn't depend on the observer too much.

    There are 18 all-inside boxes, and (if I count correctly!) 60 more boxes are needed to cover all parts of the hand image. The total of the outside boxes is therefore 78.

    The result?
    If I believe that area of bigger shapes is bigger, then I see that the area of the hand is between 18 mm2 and 78 mm2. And there is a huge gap between these over- and underestimates of 60 mm2.

    More information?
    How can we improve this? That is, how can we get over- and underestimates which are closer to the actual area of the hand, so the gap will be smaller? We discussed this. The idea of improving the estimates is not totally obvious, and the whole procedure that's being discussed in this lecture evolved over many centuries in at least 4 different cultures (in China, Greece, India, and perhaps also in Egypt and associated cultures). Lots of smart people thought about it.

    We get try a grid with more subdivisions: this is called refining the grid. Again, I believe that Mr. Chang thought that we should just halve the size of the square sides. Maybe the result is something like what is shown to the right.

    Now the idea is maybe not so easy to implement, at least for human beings whose eyesight and counting ability is not so perfect. But we can try anyway.

    Well, if I have counted correctly, there are 63 new squares which are completely inside the hand image. This is in addition to the 18 old squares. But, wait: the new squares have sides which are half the length of the old squares. So each of the new squares has one quarter the area of the old squares. So actually we have now succeeding in underestimating the area by 18+(63/4)=25 and 7/8.

    Now let's get an overestimate. We need to add on all of the little squares which contain any part of the hand image. If I have counted correctly (!!) there are 118 of them. That means we need to add one quarter multiplied to 118 to the previous underestimate in order to get the corresponding overestimate. So 118/4 is 29 and a half. Add this on to the previous 25 and 7/8, and the overestimate is 45 and 3/8.

    Underestimate, overestimate, and gap
    Let me summarize what we have done:

    1-by-1 squares187860
    1/4-by-1/4 squares25.87545.37529.5
    And so on ...

    The secret is that this sort of approach is just about all we can do with real physical quantities. You can really only make better (or worse!) estimates: over- or underestimates. The ideas outlined here, taking finer and finer grids, will lead to overestimates which shrink and underestimates which increase, so that (we hope!) the gap between them -->0: refine until you get as close as you want. And this is about all we can do. I don't think the situation is hopeless, but I do think it is compicated.

    The neat idea of asking for the area of "my hand" is something I borrowed from Professor J. Rosenstein of our math department. I think it is neat. It is a question which brings up important ideas immediately and clearly.

    An exact computation
    Mr. Spero asked if we could do an exact computation, and this I was happy to try, since many people seemed rather distressed about the statements made previously. So I will present a simple example of a general idea. The general idea is very successful, although the simple example may seem to be overelaborate. I will try to compute the exact area "under" y=x for the interval [0,1]. (Yes, I am aware that we know this area -- the ideas are what is important -- as soon as we get the ideas we'll be able to do much more complicated computations.)

    How to begin
    We will try to get over- and underestimates of the area We could divide the interval [0,1] into 4 equal parts using 0 and 1/4 and 2/4 and 3/4 and 1, which I'll choose to call 4/4.

    Then put rectangles underneath y=x in each subinterval. We will be slightly economical and put the largest rectangles we can under the y=x, but we will be slightly simple-minded and only look at rectangles which have sides parallel to the coordinate axes. Then we compute the area of these rectangles:
      0     1      1     1      2     1      3     1  
     --- · --- +  --- · --- +  --- · --- +  --- · --- 
      4     4      4     4      4     4      4     4
    The 1/4's come from the width of the rectangles, and the varying heights (even including the somewhat absurd 0/4) from the heights of the rectangles.
    Now we need an overestimate, and the true area will be "captured" between these two estimates. Again, we will use the interval divided into 4 equal pieces, and rectangles with sides parallel to the coordinate axes which just contain the region of interest to us. So the area of these rectanges is:
      1     1      2     1      3     1      4     1  
     --- · --- +  --- · --- +  --- · --- +  --- · --- 
      4     4      4     4      4     4      4     4
    Finally, if we want to see how much gap or error there is between these over- and underestimates, we can look at the area that is inside the big rectangles and outside the small rectangles. The result (because y=x is so simple!) is 4 squares, each with side length 1/4. So the gap is 4(1/4)(1/4)=1/4.

    The whole setup
    We could do something similar with other numbers of subdivisions of [0,1]. Here are some numerical results:

    # of
    4.37500 .62500.25000
    10.45000 .55000.10000
    25.48000 .52000.04000
    100.49500 .50500.01000

    What really happens, in general
    Here it helps to get a little bit abstract. Suppose we divide the interval [0,1] into n equal pieces, where you should think that n is some large integer. Then the underestimate will have total area (using heights over the left-hand endpoint of each subinterval shown in the diagram to the right)

      0     1      1     1      2     1      3     1          n-1    1
     --- · --- +  --- · --- +  --- · --- +  --- · --- + ··· + --- · --- 
      n     n      n     n      n     n      n     n           n     n
    In fact, we can "massage" this underestimate a bit and get for the underestimate:

    The overestimate uses the heights over the right-hand endpoint of each subinterval, and then the overestimate is:

    The gap between the over- and underestimates is n·(1/n2)=1/n.

    So what happens as n-->? Since the gap is 1/n, this will -->0. So let me look at the overestimate. This is (1+2+3+...+n)·(1/n2). Let's get a formula for 1+2+3+...+n.

    A formula
    You may remember the trick (maybe done in high school when you studied arithmetic progressions). The trick is writing two versions of the sum in oppositite order. Look:

     1 +   2   +   3   +   4   + ... + (n-1) + n 
     n + (n-1) + (n-2) + (n-3) + ... +   2   + 1
    Now add the sums columnwise. So we get 1+n which is n+1 and 2+(n-1) which is n+1 and ... all of the column sums are the same (that's right because the top is going up by 1 each step while the bottom is going down by 1 each step). There are n of the n+1's, so the total of all of the numbers is n(n+1). But this is two copies of the sum we want, so we need to divide by 2.

    I think we have verified that 1+2+3+...+(n-1)+n is [n(n+1]/2. And the overestimate of the area, is this divided by n2. So a formula for the overestimate is [n(n+1]/[2n2]. What happens to this as n-->? Well, we have studied such limits, and we could either divide top and bottom by n2 or we could use L'H twice. In any case, the limit will be 1/2.

    A spectacular result?
    I have now proved (done an exact computation!) that the area of the triangle is 1/2. And I guess I knew that. But this whole approach is a template for more complicated reasoning.

    We will need some notation. Lots of sums occur when we do area computations, and also many other computations. So a special way of writing these sums which takes less space has been developed. The Greek capital letter sigma, , is used. Here are some examples:

    Examples of summation notation
    Written outAbbreviated

    The notation, once you get used to it, is wonderfully economical. I should remark that the summation index, which is j in all of the examples above, doesn't really matter. That is j=13j4 and k=13k4 both mean the same thing, which is 14+24+34.

    How to approximate area
    Suppose that we have a function, f, defined on an interval [a,b] and we want to compute the area under y=f(x). That's the area bounded by the x-axis (y=0) and the lines x=a and x=b on each side, and by the curve y=f(x) on top.

    We can get approximations to the area of by dividing the interval into n equal parts (people don't always use equal parts, but we're just starting out!). Then we can look at "sample points" in each subinterval, compute f's value at those sample points, and multiply those heights by the widths of the subinterval to get areas of rectangles. Then (sigh!) finally take the sum of these rectangles, and that will be an approximation to the area we want to compute.

    The textbook explains in detail how to write the approximating sums. In particular, it identifies three sums corresponding to three choices of sample points: left and right endpoints of subintervals, and the midpoints of subintervals. Below are the formulas, which you should not memorize, and what are supposed to be pictures of a piece of the graph of y=f(x) over (relatively) tall and (relatively) narrow subintervals with the areas of the relevant boxes displayed. Since the interval has length b-a, the width of each subinterval is {b-a}/n. Do not memorize but do try to understand.

    endpoints of
    endpoints of
    Midpoints of

    I don't remember! I think it was something like "Compute j=13(j2+{1/j})". This is (95/6).

    Maintained by and last modified 11/13/2007.