No calculators will be allowed;

**Diary entries for classes
from Wednesday, September 3 to Monday, October 6**

**Diary entries for classes
from Wednesday,
October 8 to Wednesday, November 12**

Date | What happened (outline) | ||||||||||
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Wednesday, December 10 |
I discussed Monday's QotD. I mentioned that seeing x dx many
people would almost immediately try the substitution
u=x^{2}. That doesn't exactly "work" but adjusting the
substitution to u=x^{2}+a^{2} along with the
realization that the derivative of a^{2} is 0 since a is a
constant. I remarked that I knew perhaps three or four
ways to find an antiderivative for the integrand, which students
would learn in Math 152.
Then I computed the definite integral
I repeated that the definite integral; representated alimit of Riemann
sums, and that using that representation, many quantities ofinterest
in applications can be recognized as definite integrals. Then the FTC
asserts that definite integrals can be computed by antiderivatives, if
we happen to know the relevant antiderivative. This distinction (the
I remarked that an antiderivative of x
Motivated by experience with FTC, we
might decide to study the function F(w)=
Look at G(w)=F(3w). By the Chain Rule, G'(w)=F'(3w)·3, and this
means G'(w)=[1/(3w)]·3=1/w. So F(w) and G(w) have What's the point of all this? First, a pedagogical point: I wanted to review some of the most important issues of the course. Second, it is really true that hardly anyone cares about logs for the purpose of computing these days (our little silicon friends do that just fine!) but the use of logs in finding antiderivatives won't go away. (In fact, many weird anduseful functions are defined by antiderivatives!)
The I hope that people will send me answers to the various sets of review problems and that, if they like, they will attend the review session, office hours, and send questions via e-mail. Recognition of attendance was distributed to appropriate individuals, along with a prize containing a prize.. | ||||||||||

Monday, December 8 |
## Area between two curvesConsider the curve defined by y=x*(x+1)*(x+2)=x^{3}-x^{2}-2x. What is the "finite"
or bounded area between this curve and the x-axis? This question is
not terribly well written, but I am trying to ask what the total area
is of the region caught inside the two "bumps" shown. Is this total
geometric area exactly equal to the definite integral
_{-1}^{2}x^{3}-x^{2}-2x dx?
Well, between -1 and 0 the curve is on top of the x-axis. In fact,
_{-1}^{0}x^{3}-x^{2}-2x dx=(by
FTC)
(1/4)x^{4}-(1/3)x^{3}-x^{2}|_{-1}^{0}=0
(the value when
x=0)-[(1/4)(-1)^{4}-(1/3)(-1)^{3}-(-1)^{2}](the
value when x=-1)=-[(1/4)+(1/3)-1]=5/(12), a positive number, as it
should be.
I did a bunch of textbook problems.
What about The antiderivative becomes (1/3)(x ^{2}+a^{2})^{3/2}+C. The C's cancel out
in a definite integral computation. So we substitute in x=a:
(1/3)(2a^{2})^{3/2} and subtract off what we get from
x=0:(1/3)(a^{2})^{3/2}. The result is therefore
(1/3)(2a^{2})^{3/2}-(1/3)(a^{2})^{3/2}.
This simplifies (if you must!) to
(1/3)(2^{3/2}-1)a^{3}.
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Wednesday, December 3 |
I began by computing the definite integral of x^{7} from 0 to
4, and, of course, making my first error of the lecture. Oh
well. I will try a new method of writing integrals on the web. I will
use a little picture of an integral sign. Heh, heh. Here goes:
_{0}^{4}x^{7}dx=(1/8)x^{8}|_{x=0}^{x=4}=(1/8)4^{8}.
This computation is routine. The definite integral describes a certain
area, a roughly triangular-shaped one, bounded by the x-axis, and
y=x^{7}, and x=4. The line x=0 hardly matters, since the
x-axis and the function graph intersect there.
I then tried another "simple" problem:
I discussed problem #2 of section 5.3 in detail. This problem gives
the graph of a curve f, and defines g by g(x)=
## Numbers are numbers are numbersI then discussed numbers. I asked what people could tell me about these numbers:43
43
43
43
43 After some jeering (I felt bleak) we agreed that these
numbers were all the same number.
jeer n : showing your contempt by derision [syn: {jeering}, {mockery}, {scoff}, {scoffing}] v : laugh at with contempt and derision; "The crowd jeered at the speaker" bleak adj 1: offering little or no hope; "the future looked bleak";Then I asked what people could tell me about these numbers:
_{3}^{5}[x^{3}]/[3+cos(x^{2})]dx
_{3}^{5}[u^{3}]/[3+cos(u^{2})]du
_{3}^{5}[t^{3}]/[3+cos(t^{2})]dt
_{3}^{5}[v^{3}]/[3+cos(v^{2})]dv
_{3}^{5}[s^{3}]/[3+cos(s^{2})]ds
These are all definite integrals, and they are all just
numbers. Of course they are all equal. Again, the letters inside the
integrand are dummy variables or bound variables. They are similar to
the local variables in subroutines of a computer program, and they
have no meaning outside of the integral itself. I guess that the
assembly of symbols
_{3}^{x}xdx would make sense:
antidifferentiate, get (1/2)x^{2}, evaluate at x and subtract
off the value at 3 to get (1/2)x^{2}-9/2. But this is
considered bad grammar: people try not to use the same variables
inside the integral as in the limits -- it is difficult to understand.
I then tried to do some simple exercises with part of the FTC. It was
something like this: if f(x)=
How about if f(x)=
We can even do one further variation. Suppose F(x)=
I tried to convince people that on the horizontal axis, if B is to the
right of A (A<B) then the distance from A to B is B-A. I then went
on to assert that on the vertical axis, if B is above A (A<B) then
the distance from A to B is B-A. It doesn't matter much what signs A
and B have: if A<B, the distance from A to B is B-A.
This was a sort of introduction to the following problem: sketch
y=5-x If (x,y) is on both curves, and if y=5-x ^{2} and
y=2x^{2}-7, we then know that 2x^{2}-7=5-x^{2}
so 12=3^{x}, and x is +/-2. Then y is
5-(+/-2)^{2}=1. Look to the right for a rather distorted
picture (the vertical and horizontal axes have rather different
scales!).Now think about the area between these two curves. We can imagine lots and lots of vertical lines dividing this area up into pieces which are almost rectangles (there's a bit of error at the top and bottom). Look at each almost rectangle: it is dx wide, and it is top(x)-bottom(x) high. So the area is (top(x)-bottom(x))dx, and we add up these areas from x=-2 to x=2. Therefore the total area which we would like is _{-2}^{2}(top(x)-bottom(x))dx=_{-2}^{2}((5-x^{2})-(2x^{2}-7)dx=_{-2}^{2}12-3x^{2}dx=12x-x^{3}|_{x=-2}^{x=2}=(24-8)-(-24-(-8))=32.
And that's the area.
We then had an interlude in which we discussed the final. /interlude/ [Theatr] a. a pause between the acts of a play.Please look at the material I have prepared to help the class study for the final.
The | ||||||||||

Monday, December 1 |
Today we disclose the climax of the course. I will begin with a
simple example. Our goal right now is to compute the area of the
region in the plane bounded by the x-axis, x=2, x=3, and the curve
y=x^{5}. This is a definite integral:
S_{2}^{3}x^{5}dx/ Oh, I give up: this
is int(x^5,x=2..3). The last collection of symbols is easier to type,
and the traditional integral symbol isn't well indicated by S,
anyway. On the right is a picture of the region whose area I will try
to compute. Please note that although the picture I have drawn is
mostly qualitatively correct, I have certainly used very different
units on the vertical and horizontal axes (2^{5}=32 and
3^{5}=243). I will arbitrarily take the year 1600
as the dividing point in the two approaches to computing definite
integrals that I'll discuss
(the 1660's is probably more accurate historically). I'll begin with
the "old" style.
## Getting the answer directly from the definition (before ~1600)Here we will go through a four-step process: subdivide, approximate, sum, and limit. This process is, in fact, quite important in many applications. It is the reason that definite integrals represent many different ideas.
We divide the interval from 2 to 3 into n different equal parts. Here
you should think of n as some very large positive integer. (The
picture shows n=6, not terribly large -- imagine n=10,000!) Then each
piece of the interval will have width (3-2)/n=1/n. We will approximate
the area over the first subdivision, from 2 to 2+1/n, by a rectangle
whose height is the value of the function at the right-hand
endpoint. Similar things happen if we choose the left-hand endpoint or
the middle or ... any choice, in fact. So the right-hand height is
(2+1/n)
^{5}(1/n)+(2+(2/n))^{5}+(2+(3/n))^{5}+...+(2+(n/n))^{5}(There are n terms in this sum.)
By the way, the process we are discussing here would be familiar to appropriate representatives of various classical civilizations, including Greece, China, India, and north Africa. The are wonderful medieval Chinese manuscripts which have diagrams that look exactly like our definitions of the definite integral, with much commentary written in Chinese surrounding the diagrams. One
way to analyze the elaborate sum above begins by changing it
algebraically. For example, the first term,
(2+1/n)
I assure you that this result is
But now what's the next step? Divide by n
I don't think this specific computation has much importance, but I
went through it to show how pre-1600 technology would work. I
don't think they had ## The new stuff! (after ~1600)
Let's summarize what we have learned about A(R): ^{6} is
729 and 2^{6} is 64, this answer is the same as (729-64)/6
which is (665)/6, agreeing with the answer we got using the old technology!
Here I will use the notation in the textbook. Also I will refer to what follows as FTC.
I then did a collection of problems from the text. They were: - 5.3 #21: What is the definite integral of 4x+3 from 2 to 8?
(Asked in the text using usual math symbols: the long S for integral
sign, etc.).
**Solution**According to FTC we just need to*guess*an antiderivative. I*guess*F(x)=2x^{2}+3x+83. You can verify my guess by checking that F'(x)=4x+x, which it does. Now in FTC, a=2 and b=8, so that the definite integral must be F(8)-F(2)= [2·8^{2}+3·8+83]-[2·2^{2}+3·2+83]. Notice, please, that the silly*83*is added and subtracted: it cancels out. In computing definite integrals we may omit the silly constants.
**You can't necessarily omit these silly arbitrary constants in other types of computations, such as initial value problems.** - 5.3 #28: What is the definite integral of cos(theta) as theta goes
from theta=Pi to theta=2Pi?
**Solution**According to the FTC, we just need to find an antiderivative of cos(theta), and then ... well, sin(theta) is one antiderivative. And then sin(2Pi)-sin(Pi) is 0. Does this mean there is*no area*between cosine and the horizontal axis? That sounds weird. If we look at the graph, we can "see" the area. But, golly, in fact**(IMPORTANT!)** If we look at the sums defining the definite integral, we can see that when f(x)<0, the sums will be negative: the definite integral weights area*negatively*when the curve is below the axis. - 5.3 #37: what is the definite integral of 6/sqrt(1-t
^{2}) as t goes from 1/2 to sqrt(3)/2?
**Solution** We used what is standard notation. First, though, guess an antiderivative: the suggestion was made that I look at 6arcsin(x), which does work. Then I wrote 6arcsin(x)**|**_{x=1/2}^{x=sqrt(3)/2}. The vertical bar is notation devised especially for the definite integral computation, and means: substitute in the upper "limit" and evaluate, and then substitute in the lower limit, evaluate, and subtract. After some effort, we saw that this was 6·(Pi/3=Pi/6). - 5.3 #8. Here g(x)=int(ln(t),t=1..x) and the text asks for g'(x).
**Solution** so by FTC g'(x)=ln(x). - 5.3 #10. Here g(u)=int(1/(x+x
^{2}),x=3..u) and the text asks for g'(u).
**Solution** so by FTC g'(u)=1/(u+u^{2}).
^{2},x=3..10)=int(q^{2},q=3..10)=int(w^{2},w=3..10)=int(z^{2},z=3..10)
The | ||||||||||

Monday, November 24 |
What will be done now is the definition of the definite integral. This
is a complicated object, and is the mathematical object which
corresponds to quantities which have properties similar to those we
discussed last time. I will concentrate on area in what follows
(because then I can draw simple pictures), but everything I'm doing is
applicable to many many many other things as well.
So again: the problem of area. In what follows I'll be looking at a region bounded by y=f(x), x=a, x=b, and the x-axis. How can we efficiently and accurately approximate the area of this region? We will use rectangles as we did last time, only now I'll give you the official names of everything. Here we go: - First, we will make a
**partition**of the interval [a,b]. This means taking a finite collection of points in the interval: a=x_{0}<x_{1}<x_{2}<...<x_{n}=b. We take n+1 points which chop up [a,b] into n subintervals. x_{0}is always supposed to be a and x_{n}is always supposed to be b. - Select
**sample points**in each of the subintervals. Your textbook refers to these points as x^{*}j, and this must be in the j^{th}subinterval, which is the subinterval [x_{j-1},x_{j}]. - Finally, form the
**Riemann sum**: this is the sum of f(x^{*}_{j}) multiplied by x_{j}-x_{j-1}as j runs from 1 to n. So it is a sum of n terms, and each term consists of a value of f multiplied by the length of the subinterval. The textbook has a discussion of*sigma notation*and you should look at it. I would like to concentrate on the ideas right now.
So for the non-continuous function f(x) we just analyzed, we can find a partition all of whose Riemann sums are very close to 9. Let us try something harder.
any choice of sample
points would lead to a Riemann sum between the upper and the lower
sum, so we indeed have a way of approximating areas within any
positive error we can specify.
Of course, there is notation: an elongated It is a true fact (much better than a false fact!), that every continuous function has a definite integral. But there are functions which are not continuous, such as the function of example 1 above, which also have definite integrals.
I did two very examples that are "simple" if you recognize the
geometry supporting them:
The
The area shown is either a trapezoid or a triangle and a rectangle. In
either case, the area, the definite integral from 0 to 3, is 2.5. Half
of that area is 1.25, and the area of the triangle is 1/2, so that we
need .75 from the area of the rectangle. Since the rectangle is 1 unit
high, we easily deduce that A must be .75 more than 1, so A=1.75.
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Wednesday, November 19 |
I remarked that we would be doing something significantly surprising
intellectually in the balance of the course. So far we have studied
various nice problems involving differentiation. Things like graphing
and optimization (min/max) and slopes of tangent lines and
... stuff. It turns out that the kind of things we've been studying
have great application to what looks like a collection of completely
different problems. It is and it will be amazing!
Please start reading chapter 5 of the text. I will cover today and Monday the material in 5.1 and 5.2, in somewhat different ways than the text does. I may even go a bit further. I began with ## The problem of areaI drew a blob on the board, and asked how much area is inside the blob. I told people that they could use any method they wanted to answer the question. Some students whose minds were already preconditioned by calculus gave rather strange answers, but no one gave me a direct method untilMr.Colquit remarked essentially that we could
cut up the board and then weigh the blob and compare its weight to the
weight of a board of known area, such as a rectangular board. So here
is a direct method: cut it out!.
We will try to be a bit more indirect, more sneaky, and learn
properties of area which will let us - Area should assign a region in the plane a non-negative
number.
**Comment**This certainly seems reasonable. But there are things in the plane which have zero area: one point, or a line both should have*area*equal to 0. - If one region, R
_{1}, is inside another region, R_{2}, then the area of R_{1}is less than or equal to the area of R_{2}.
**Comment**The areas could actually be equal. For example, R_{1}could be R_{2}with just 1 point taken out! - Two regions which are congruent must have the same area.
**Comment**Again, this seems reasonable, but then the word "congruent" needs to be dealt with. The on-line dictionary I use says "congruent" means "coinciding exactly when superimposed" and if clarity is wanted, then the words there need to be explained further. In the plane, two shapes or blobs are congruent if we can move them one onto the other. Sometimes "flips" are allowed. - If a region R is made up of two regions R
_{1}and R_{2}which don't overlap, then the area of R is equal to the sum of the area of R_{1}and the area of R_{2}.
**Comment**In fact, as was pointed out, we can weaken the hypothesis a bit here. The two regions R_{1}and R_{2}can overlap as long as the overlapping takes place only on the boundary. - The area of a rectangle is its length multiplied by its width.
**Comment**Without this assertion, assigning area to be 0 to every region would obey all the previous rules! Of course that's a silly example, but it does show that some reference must be made to "normalize" area, to give it units or something.
With
Now I can approximate inside this piece of the blob and outside the blob by
rectangles, and since I know the area of a rectangle, I can get some
Of course this estimate is not so good, but we could improve it with some sort of scheme like this:
So we can get a collection of approximations which get closer and closer and closer to the true value of the area, and maybe that's good enough. (It is, in many cases.) As I remarked, another instructor told me of a wonderful way of bringing home the idea of area: look at your hand and ask, what is its area? Giving a "good" answer to this may be difficult and involve quite a lot of approximation. Now let's look briefly at ## The problem of blood flowI looked at blood flowing through the heart, and asked how much blood would flow through the heart in a given amount of time, say a minute. Knowledge of this could be rather important in study of certain conditions or diseases or whatever. One rather destructive method would be tocut it up: that is, just slice into the
heart and let the blood drip to fill a bucket for a minute. Well, this
method might not be too satisfactory for several reasons which you can
you think about yourself. But how do people answer such questions? You
insert a catheter into the appropriate blood vessel (Google lists
about 135,000 links in response to the search words "blood flow
catheter" so there are ample references). At the end of the catheter
there's a mechanism, which you can think of as a sort of paddlewheel,
and the blood flow sets it spinning, and the spinning makes an
electric current which is passed up the catheter and measured. How is
the actual blood flow computed? It turns out that the blood flow obeys
rules which are totally analogous to the area rules.
RULES A THROUGH E now state, sort of:
- The blood flow is a non-negative number.
**Comment**Assume for simplicity that the blood flow in a vessel goes only in one direction. In fact, in certain vessels of the body there are "mechanisms" preventing back flow! - Blood flow over longer periods of time is at least equal to blood flow over shorter periods of time which are "inside" the longer periods.
- If the flow rate is identical over two periods of time (for example, the same portion of two heartbeats) then the flow is also the same.
- The blood flow during a sum of two intervals which may overlap only at "the edges" is the sum of the flow during the intervals considered separately.
- If the flow rate is constant, then the blood flow is the product of the duration and the flow rate.
without filling a bucket!).
By the way, I found a number of authentic pictures of the heart on the web and decided I didn't need to illustrate this discussion with what could be a grim and upsetting picture. And now ## Aliens land in Piscataway and leave a barThis is somewhat speculative. The type of bar I am referring isnot one selling alcoholic beverages (!) but a long and very
very heavy bar, a bar which is 100 meters long (metric aliens!) and
made of a number of distinct substances. Additional conditions are
that the bar is too heavy to move, that we can scrape and sample it,
and that we would like to determine its total weight, or, rather, its
total mass. Here maybe is a picture, with the different shapes/colors
representing substances of perhaps vastly differing densities.
Since the bar is geometrically very simple, we can just take a sample scraping somewhere on it, find the density, and approximate the total mass by multiplying the density recorded by the geometric shape's volume. But this is a very rough approximation, since there are different substances and the bar could be complicated. So what we could do is periodically sample the bar, find the densities, and multiply by chunks of the volume. It seems likely that we will get a better estimate of the total mass if we take more samples.
The alert student will observe that in these imaginary measurements
and computations we are being guided by metallurgical (I checked the
spelling!) analogs of Finally ## A last look at FrancineIn this case, Francine is traveling north on the Garden State Parkway. She leaves mile 0 in Cape May at 9:00 AM. You have only the following "equipment" (it is another "thought experiment"):You can look at the speedometer on her car and you can look at your watch. You are asked to tell where she is (what mile on the GSP) at 10:00 AM. What procedure can you follow? Here is one: look at the speedometer at, say, 9:30, record the miles per hour, and multiply by 1 (hour, that is). This gives an estimate of her position. A more "proactive" (means: "creating or controlling a situation by taking the initiative") method would be to take several measurements (say, record the speed every 5 minutes) and then multiply each speed reading by 5/60, and add these up. This would likely give a better estimate of the position she will be in.
Again, if you are alert, you will see in all this we are again using
the analogs of
## Big newsAny quantity which obeys the analogs ofRULES A
THROUGH E can be computed (either approximated efficiently,
or in a large number of cases, computed exactly) using the
methods we will learn. There are many many such quantities, ranging
from such physical "objects" as force and work to quantities in
economics. And it is really amazing (to me, at least!) that
what we've done so far can be applied to this problem. It is a bit
difficult to "see" derivatives and slopes in all of these problems,
without being shown where to look.
So the idea will be to take a quantity, break it up into small, more
easily approximated parts, and then to add these approximations up,
and the resulting sum will give a good approximation (or, even,
sometimes, an equality!) to what we want to compute. This is a method
of
## Exam 2 returnedI returned and discussed the results of the second exam. I apologized for the excess length, but not for the level of the problems. The difficulty of the problems was probably appropriate, but the length was not: I think I had about one problem too many -- I should have had 8 rather than 9. But I additionally remarked that the performance of students on some of the exam was clearly below the level expected. Students must learn how to study.
I forgot to ask a |

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Maintained by
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