In reverse order: the most recent material is first.

Monday, December
11 | (Lecture
26) |
---|

I also mentioned that the word *mantissa* is currently used in
computer architecture. As used there, it refers to the fractional part
of a floating point number.

Another aspect of substitution was addressed. Suppose a
*definite* integral is given. Should the limits in x also be
changed to limits in u, or should the "return to x-land" always be
done and then x limits used? I tried to give a number of examples in
order to illustrate the two possibilities. I usually go back to x-land
and use the original limits because I make mistakes, sometimes many
mistakes, and the return to x means I don't need to change the
limits. Other people have other opinions.

I computed some areas between two curves. I used the general
formulation of _{x=Left}^{y=Right}Top(x)-Bottom(x) dx and
illustrated this using examples where Top and Bottom have different
signs.

I did a very few examples where the area is computed dy, so the
formula above becomes _{y=Bottom}^{y=Top}Right(y)-Left(y) dy.
This is discussed in section 6.1.

Wednesday, December
6 | (Lecture
25) |
---|

I discussed in detail how in "old days" students in highschool learned how to use high-accuracy tables of mantissas of common logarithms (logarithms base 10) in order to approximately compute multiplications. This skill may be obsolete since the introduction of simple and cheap calculators.Mantissan. [L., an addition, makeweight; of Tuscan origin.] (Math.) The decimal part of a logarithm, as distinguished from the integral part, or characteristic.

Does it necessarily follow that logarithms themselves are obsolete?
While at first glance this may seem to be true, I "analyzed" the
following function:

L(x)=_{t=1}^{t=x}(1/T)DT.

We showed that L´(x)=1/x using FTC. Then we used the Chain Rule
and FTC to see that if A is some fixed positive number, and
W(x)=L(Ax), then W´(x)=1/x also (almost a miracle in that the
effect of the A's cancel!). Then the Mean Value Theorem states that
W(x)-L(x)=a constant since the derivatives of L and W are
identical. If we put x=1, L(1)=0 and W(1)=L(A). Therefore the constant
is L(A), and we see that since W(x)=L(Ax)BB,

L(Ax)=L(A)+L(x)

The "profile" function used to define L (that is, 1/t) is not a very
complicated function.
So if we are interested in computing definite integrals, we'd better
be able to compute L, and the equation above may be used to simplify
computations of L.

Indeed, if we use the following `Maple`
instruction:

> f:=n->evalf(add((1/(1+(j/n)))*(1/n),j=1..n));Then the value of the function f(n) is an approximation of a Riemann sum for the definite integral from 1 to 2 of 1/t, with n subdivisions, with the right-hand endpoint of each subdivision used. Results look like this:

Number of subdivisions | Approximate value of the sum |
---|---|

10 | 0.6687714032 |

100 | 0.6906534305 |

1,000 | 0.6928972431 |

10,000 | 0.6931221812 |

100,000 | 0.6931446806 |

Of course, as I mentioned in class, these are approximations to ln(2),
whose 10-digit value is 0.6931471806. By the way, even the computer
gets tired computing Riemann sums and the last entry in the table
above takes about three-quarters of a second on a moderately fast
PC. Next semester we'll see numerical methods which give much better
accuracy with much less computation. But the point I was trying to
make here was that the **conceptual idea**
of logs was useful, even though the **computational
need** for logs may be obsolete.

There are many other integrals which must be computed, and which satisfy quite weird equations that come from the Chain Rule and FTC. So this whole development is something you will likely see again.

The remainder of the lecture was devoted to a certain method for obtaining indefinite integrals: substitution.

We began by studying (x^{3}+5)^{400} x^{2} dx.

This is not a random integral, at all. If I omitted the x^{2}
or changed it much, the integral would be quite difficult to do "by
hand". The integrand (the function to be integrated) is "just" a
polynomial, so there is no great theoretical obstacle to computing
it. The practical difficulty of obtaining a formula for the
antiderivative is what I wish to address.

One way of computing the indefinite integral is to "unfold" or
"expand" (x^{3}+5)^{400}. By hand (even by computer!)
this takes a bit of time and storage space! In fact, we recognized
that we could do this by sort of guessing and reversing the Chain
Rule. It is important to remember, though, that the derivative of a
product of two functions is **NOT** equal to
the product of the derivatives!

So we guesses. I wrote the Chain Rule and we identified the pieces of the computation. Then I introduced the Substitution Method, which is essentially a "bookkeeping" strategy to keep track of constants as we go from "x-land" to "u-land".

Here (x^{3}+5)^{400} x^{2} dx
is in x-land. If u=x^{3}+5, then du/dx=3x^{2} and
du=3x^{2}dx and (1/3)du=x^{2}dx. The entire integral,
translated into u-land, is
then
u^{400}(1/3)du. This is easy to compute in u-land, and
the result is (1/401)u^{401}(1/3)+C. Now we go back to x-land,
and the final answer is (1/401)(x^{3}+5)^{401}(1/3)+C.

I computed some other indefinite integrals with this method:

- x sqrt(4-x
^{2}) dx (use u=4-x^{2}) - cos(3x-7) dx (use u=3x-7)
- [(x+1)/(x
^{2}+2x+117)] dx (use u=x^{2}+2x+117).

We then computed the indefinite integral of tan(x). Since tan(x)=sin(x)/cos(x), the substitution u=cos(x) leads us to the answer -ln(cos(x))+C. Like any other indefinite integral, this can be checked using differentiation. And this indefinite integral again shows that logs will be needed in calculus!

Then I "discussed" two problems from section 5.5 of the text:

#58, which was _{0}^{1}x e^{-x2}dx.
The antiderivative can be computed with u=-x^{2}.

#56, which was _{0}^{2}dx/(2x-3)^{2}. I remarked that
this was an error in the text (a misprint!). 2x-3=0 when x=3/2, and
therefore a simple graph should convince you that the definite
integral (the "area") must be infinite. If the upper bound of the
integral is changed from 2 to 1, then the substitution u=2x-3 will
allow us to compute the value of this definite integral.

Monday, December
4 | (Lecture
25) |
---|

Number of subdivisions | Approximate value of the sum |
---|---|

10 | 0.7105093416 |

100 | 0.6714629470 |

1,000 | 0.6671601350 |

10,000 | 0.6667164623 |

100,000 | 0.6666716592 |

Maybe these results which I computed with `Maple` make the assertion that the limit of the
Riemann sums exists and equals 2/3 more believable.

Then I asked what people could tell me about these numbers:

These are all definite integrals, and they are all

I then tried to do some simple exercises with part of the FTC. It was
something like this: if f(x)=_{2}^{x}
cos(t^{7})dt, then f'(x)=cos(x^{7}).
And then if f(x)=_{2}^{x}
cos(s^{7})ds, then f'(x)=cos(x^{7}). The "dummy
variable" strikes again!

How about if f(x)=_{x}^{3}
cos(s^{7})ds? Here we need to think a bit about definite
integrals. Here's a fact:

**Adding areas fact**

If a<b<c, and if f is continuous on the interval
from a to c, then
_{a}^{b}f+
_{b}^{c}f=
_{a}^{c}f.

So adjoining areas can be added up (something many of us learn as
infants, and it is a rather subtle fact!). But we also agree that
_{a}^{a}=0: that
is, a region with 0 width should have 0 area. Suppose we make some
changes in the adding areas fact, and transform c into a. Then the
right-hand side becomes 0, and we see that
_{a}^{b}f+
_{b}^{a}f=0, which is the same as
_{a}^{b}f=-_{b}^{a}f.
Therefore if we want to be able
to add areas and if we want to recognize that lines have area=0, we
must also acknowledge that interchanging limits causes a - sign to
prefix the integral. Therefore we can do this:

if f(x)=_{x}^{3}
cos(s^{7})ds, then f(x)=**-**_{3}^{x}cos(s^{7})ds so that f'(x)
must be -cos(x^{7}). Amazing! (Well, not really, but ...)

We can even do one further variation. Suppose F(x)=_{-2}^{x}sqrt(1+w^{6})dw. Then we know
that F'(x)=sqrt(1+x^{6}) by FTC. What if I ask the following
ludicrous question:

Suppose G(x)=_{-2}^{cos(x)}sqrt(1+w^{6})dw. What is
G'(x)?

How can we think about this? Since we know F'(x), and G(x)=F(cos(x)),
this is *just* (!?) an application of the Chain Rule (!):
G'(x)=F'(cos(x))·(the derivative of
cos)=sqrt(1+(cos(x))^{6})·(-sin(x)). Whew! This takes
some getting used to.

Wednesday, November
29 | (Lecture
24) |
---|

I talked about some simple situations where specific definite integrals could be evaluated due to the simplicity of the regions involved (a part of a circle or a triangle or a trapezoid). The definition of the definite integral as a limit of Riemann sums means that the

**A paradigm shift**

Several cultures invented approaches similar to what we've done here
to compute quantities such as area and volume and mass and
moments. These cultures (which were located in areas corresponding to
present day China, India, Egypt, and Greece) all used the same general
approach: chop up the quantity into smaller pieces, approximate the
pieces, take the sum, and then "compute" the limit. Of course the
notation was different, but the ideas were fundamentally the
same. Very clever algebraic and numerical techniques were used. See
section 5.2 for information about this. It turns out that studying
what seems to be a *harder* problem (!) earns both conceptual
and computational advantages. So I attempted to do that for the
specific case of f(x)=(3/2)x^{2}+(1/2)x. I wanted to compute
the area "under" this curve from 1 to 2. This is the area of the
region bounded by x=1, x=2, the x-axis, and y=f(x). A direct approach
is possible, but instead I studied the following problem:

Suppose A(r) is the area bounded by x=1, x=R, y=f(x), and the x-axis. What can we learn about A(r)?We saw that A(1)=0 and then that A´(R)=(3/2)R

**The Fundamental Theorem of Calculus**

I stated and commented on the Fundamental Theorem of Calculus
(FTC). Then I applied it to about half a dozen problems in section 5.3
of the textbook.

`Maple` data and its construction

I also gave some data which I had computed using
`Maple`. `Maple` is a program which is available on the
`Eden` system and on many other computer systems around campus,
since Rutgers has a site license. It is quite useful.

Here I defined the function for> f:=x->(3/2)*x^2+(1/2)*x; 2 f := x -> 3/2 x + 1/2 x > f(1);A 2 > f(2); 7

The> add(f(1+(j/50))*(1/50),j=1..50); 43001 ----- 10000 > evalf(%); 4.300100000

Number of subdivisions | Approximate answer | Predicted error |
---|---|---|

50 | 4.300100000 | <.1 |

500 | 4.255001000 | <.01 |

5,000 | 4.250500100 | <.001 |

50,000 | 4.250050000 | <.0001 |

500,000 | 4.250005000 | <.00001 |

We used FTC to see that the "true" value of the definite integral was 4.25 (maybe not too much of a surprise). I mentioned that the last number (which involved a half-million subdivisions!) took about 16 seconds to compute. I consider that a lot of time.

Monday, November
27 | (Lecture
23) |
---|

Given a function f(x) defined on an interval [a,b], I wanted to compute the area "enclosed" by y=f(x), x=a, x=b, and the x-axis. I discussed these technical words and phrases:

- Partition of [a,b].
- Sample points.
- Riemann sum.

I analyzed Riemann sums for a piecewise defined function, something like f(x)=2 if x<4 and f(x)=5 for x>=4, on the interval from 2 to 6. We were able to understand "any" Riemann sum, and we showed that if the length of the largest subinterval in the partition was small, then the Riemann sums would approach what we naively might call the area "under" the curve.

I then moved on to (3/2)x^{3}+(1/2)x on the interval
[1,2]. First we looked at a Riemann sum which divided the interval
into 50 equal parts, and we showed by a rather simple (but clever!)
geometric argument that we could estimate the "error" of that specific
sum from the "true value" of the area. Then I turned things around,
and asked if we could find a Riemann sum which was within 1/100,000 of
the true value. We did this.

I then defined the definite integral. This is all in section 5.2 of the text.

**A homily**

Before I returned the second exam, I made some remarks which I attempt
to record here. I apologize if these comments are too direct and too
personal, but I believe they are truthful and may be necessary.

**The exam**

With the possible exception of the last problem, all of the problems on this exam were routine. Students should have looked at the problems and realized that they had done ten problems just like them (or should have done ten such problems!). Most of the problems were copied directly from the textbook or review material. I graded 6 of the 8 problems on the exam. Here are some specific remarks.

The first problem was the most routine. When I looked at a solution to that problem, I could tell instantly if a student knew what to do. The problem asks for the max/min values of a function on an interval. The function, a low-degree polynomial, is simple. So: differentiate the polynomial, find the critical points, compare the value of the function at the endpoints and the critical points in the interval. This is perhaps 4 or 5 lines of computation. Anything else is overwork and misunderstanding.

The seventh problem, a graphing problem, was worth the most points and was the most intricate. It is analogous to an essay in a composition course. The solutions written by many students contained contradictory ideas, which I found very difficult to understand. The problem asks for coherent, deductive reasoning and achieving this should be possible.**Personal**

I repeat what I said in class. I don't think any law*forces*students to learn calculus. So perhaps there's little reason to attend lectures or recitations. The recitations are likely to be the most helpful part of class attendance. The instructor is very approachable. But many students don't come to these classes and this is likely to be an important reason for their lack of success.

At the beginning of the semester, I declared that this course is different from a high school class chiefly in that it requires much more effort and responsibility from students. For example, I would expect that in a high school class, perhaps 30 to 40 optimization problems could be discussed in class, either by the instructor or with students working in groups. There was time to discuss only 4 or 5 of these problems during the lectures. This difference must be made up by*student effort*: you can't learn this material by sitting and watching the lecturer and/or the recitation instructor only, no matter how sympathetic or learned they are. An exaggerated metaphor: would you learn to play the trumpet by watching and listening to Wynton Marsalis?

Students who want to study and successfully understand almost any scientific, engineering or technical subject must learn basic calculus. If you don't learn this stuff, you won't be able to learn that stuff: very simple! Effort and**practice**are most important. Yes, intelligence and talent are good, but I really believe that almost all students have sufficient intelligence and talent. Persistence is the most important factor in learning mathematics.**Competitiveness**

One reason to learn this material, aside from what I think is its intrinsic beauty and interest, is that such knowledge qualifies people for good jobs. The initial salary and benefits package for many engineering jobs requiring a bachelor's degree is quite good. On the other hand, look at the deal from the company or employer side: why should a person be hired if they can't do the job? Please realize that in a few countries I can name, there are more than a quarter*million*people graduating each year who can write solutions to the problems on the second exam quite well. You should be worried if you're going to compete with them.

Does Math 151 give sufficient support for learning? My part is chiefly to give prepared lectures and to write appropriate exams. I try to support student learning in other ways, also. But much -- most! -- of the learning will follow from student effort. As you go through life you'll need to take more and more responsibility for your own learning.

Suppose a person knew how to program in, say,`Visual Basic`20 or 30 years ago. That was good. But if that person insisted, for a whole career, on only doing what always was done, that career wouldn't have been very rewarding. Much of the time you'll need to learn new things, new ideas, new methods. Sometimes this will involve lots of work. I know, even in my own teaching "racket", that I've had to learn new ways of thought and new methods, even for subjects which I would hope I understood fairly well by now. You will need to periodically reinvent yourself.**Exam grades**

The exam grades were quite low. You can improve your grade. In recitations on Thursday, November 30, students will have 30 minutes to earn 30 points, which will be added to their second exam grades. The topics will be the same as those tested by the second exam,*along with*antiderivatives. Good luck, really, seriously. I will succeed as a teacher if*you*succeed as a student.

Monday, November
20 | (Review for
exam #2) |
---|

- Technical computations:
- The chain rule
- Implicit differentiation

- Consequences of the tangent line
- Linear approximation, tangent line approximation, linearization, "differentials"
- Newton's method
- L'Hopital's rule

- The Mean Value Theorem. The theory:
- If f´=0 in an interval, then f=a constant in that interval (so two functions with the same derivative must differ by a constant)
- If f´>0 in an interval, then f is increasing in that interval.
- If f´<0 in an interval, then f is decreasing in that interval.

- Graphing a function
- The first derivative: its sign indicates {in|de}creasing behavior.
- The second derivative: its sign indicates concave {up|down} on the graph.
- Critical points; local and global max and min; inflection points

- "Stories"
- Related rates
- Optimizing a function (max/min in an interval)

Monday, November
20 | (Lecture
21) |
---|

**Again ...**

I discussed how to "solve" the initial value problem
f´(x)=sqrt(1+x^{3}) with f(0)=2. From the qualitative
aspect (using direction fields) I knew that f(1)>f(0)=2. What if we
wanted more precise information? I showed how we can do "simple
arithmetic" motivated by the Mean Value Theorem, and approximate f(1)
as closely as we wished.

I began with

We will try to be a bit more indirect, more sneaky, and learn
properties of area which will let us *approximate* what we
want. What properties does "area" have?

- 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 **RULES A THROUGH E** we can begin
an organized campaign to compute or approximate areas. For example,
here is how we could compute the area of a typical blob. I will use
the **RULES** in the analysis that
follows.
I would first slice the blob by line segments to get something which
had at most one curvy side.
I will be done if I describe to you how to compute the area of each of
the pieces. I will specialize to one of the pieces, and assert that
similar things can be done to all of the other pieces. So:

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
*estimate* of the area of the piece of the blob.

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.

** Maintained by
greenfie@math.rutgers.edu and last modified 10/16/2006.
**