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The following summarizes a class by MIT math professor David Jerison. (Bibliography information given at the end.) Unless otherwise noted, all images are from either screen shots of the video or from the class pdf handout. My own commentary is found between brackets of some kind.

David Jerison

Single Variable Calculus (Course at MIT)

Class 18: Definite Integrals

Brief summary:

The definite integral allows us to calculate the area under curves and the cumulative sums of functions.

Summary

This is an introduction to integral calculus, and this lesson’s topic is definite integrals. The idea of definite integrals can be presented in a number of ways, but Jerison will begin with the geometrical interpretation, specifically the area under a curve. Another perspective on the integral is ‘cumulative sum’ .

We will consider a curve under which are boundaries a and b,

and we will be concerned with the area under the curve and within those boundaries.

We also need other information, namely the bottom line, the x-axis, and the curve, defined in terms of the function f(x).

For out notation for this, we have an integral.

But it has limits on it. It starts at a and ends at b.

Then we write in the function f(x)dx.

This is the definite integral. Geometrically it is interpreted as the area under a curve. To compute that area we will follow three steps:

1. Divide region into “rectangles”

2. Add up area of rectangles

3. Take limit as rectangles become thin: “infinitesimally thin, very thin” [00.05.30]

Pictorially, we have a and b, and we have our curve.

We then chop the x-axis into little increments.

Then we chop it up into rectangles with some staircase pattern. “In some cases the rectangles overshoot; in some cases they're underneath. So the new area that I'm adding up is off. It's not quite the same as the area under the curve.”

So our region is here.

But it is missing this part here.

And it includes these extraneous pieces here.

But as the rectangles get thinner and thinner, the small off-parts shrink until they are negligible.

Our first example:

We will make a = 0, but to see the pattern, we will make b be arbitrary.

So we will draw it. Here is the parabola.

The piece under the curve stops at b.

We now divide the area into n pieces. We get to choose whether we want them directed to the right or left, and thus whether we overshoot or undershoot the curve. In this case we choose to direct them to the right.

Now we need to write formulas for these areas. Rectangles are base times height And our intervals are equal. The base length would then be b/n.

So now we make a table to look for patterns

The first place after zero is b/n. It is an x value.

The height is f(x), which is x^{2}. So in this case it would be (b/n)^{2}.

The next rectangle is 2b/n, and thus its height is (2b/n)^{2}.

And thus for the third: 3b/n, and (3b/n)^{2}.

Now we need to add up the areas. The area of each is the base times the height of each one, or each row of our column. And we add each together. The final term is nb/n, because there are a total of n rectangles.

The pattern is that we have the same base plus the first height, then plus the second height, with each new height having 2b, 3b, 4b, [+1b] up to nb in the numerator. So the rectangles get taller and taller with the last being the biggest.

Our formula begins complicated. But the limiting value is simple. What we notice is that each term in this series of additions has (b/n)(b/n)^{2} or (b/n)^{3}. So we will factor that out from each of the terms to get:

Now we will take the limit as n goes to infinity. The quantity that is hard to understanding is this massive quantity here.

We will now make a modest, miniscule change. We will write 1 as 1^{2}.

Now we will use a trick. It is not entirely recommended, but we will talk about that later. To understand how big this quantity is, we will use a geometric trick. We will draw a picture of the quantity by building a pyramid. The base will be n by n blocks. [This will at first be a bit confusing, because we starting by wanting to know the area under a curve, now we are trying to find the volume of a solid figure. We should first not confuse the slabs in the pyramid that we will build as being equivalent to the rectangles under the curve. Let’s think separately from the curve for a moment. Lets say we have a pyramid whose base is a square with length 5. We are also assuming the base length is the same as the height, for reasons we will see in a second. We can then find the area of this pyramid with the equation 1/3(base)(height) = 1/3(5^{2})(5) = 1/3(25)(5) = 1/3(125) = 125/3 ≈ 41.67.]

Then the next layer up will be n-1 by n-1.

“So the total number of blocks on the bottom is n squared. That's this rightmost term here.”

We did not write it in before, but let’s now write the next-most last term, n-1^{2}. It is the second layer of the pyramid.

Our first drawing was from the top view, but we can also picture it from the side view, with the bottom slab being n units long.

We keep piling up the slabs until we get to the top layer,

which is one giant block of stone, here as 1^{2}.

In our staircase pattern, we continue up to this top block.

[[Let’s for a moment note an ancient puzzle, Democritus’ staircase cone. We begin with a cone, and we cut it into slabs that get smaller and smaller until we reach the smallest. If we combine these slabs again, we would expect to obtain a cone. But the problem is determining their sizes. If they were all the same size, we would rebuild it as a cylinder. The sides would be smooth and even, but the shape would be wrong. If they were different size, we would get a staircase shape. The sides would be rough, but the shape would be right. If we take an infinitesimal view, the smallest parts are actually very tiny changes. Combining them will produce a cone with the right shape and with smooth sides.

(Moving diagram by Corry Shores, made with Open Office Draw and Unfreez)

]]

We will now use a trick to estimate the size of this figure; “it's sufficient in the limit as n goes to infinity”. [17.02] We can imagine the solid form underneath the staircase. It is an ordinary pyramid inside, and we know the formula for its volume since we know the volume for cones.

The formula for the volume within that space of the normal pyramid is 1/3(base)(height)

The base is a square of length n, meaning that the base is n^{2}. [In our original sequence there are n terms, each of which gets a slab. Apparently we are standardizing n as a unit of length and not just a tally of terms, (later he explains that the height of each slab is 1) and so] the height is n, giving us:

[Now notice that the pyramid lines go under the staircase edges, meaning that the volume of the pyramid will be less than the sum of the staircase.

] We found that whole sum is bigger than 1/3(n^{3}).

The pyramid’s sides have slope 2, meaning you go 1/2 half over each time you go up 1.

We can also trap the staircase on the other side too, by drawing a parallel line going out 1/2 more on each side, making the base (n+1)(n+1). And it will go up 1 higher.

We then use the same formulation for finding the area of the larger pyramid, and we can place all three into relation:

So we trap that volume between these two quantities:

Now we are ready to take the limit. First recall that we were summing the series of rectangles.

We then factored out (b/n)^{3}.

[We will now distribute (b/n)^{3 }to each of the terms, giving them a common denominator n and common factor b3, which we factor out in the numerator.]

{So the idea seems to be here that previously we excluded (b/n)^{3 }from our pyramid formulation. Now we are adding it back, which will change our inequality. Previously it was:

So now we will divide this inequality by n^{3} (later we will also add the b^{3}). When we do so, the largest term on the right will be reduced. It begins (n + 1)^{3 }/ n^{3} . Perhaps then converts to [(n + 1)/n][(n + 1)/n] [(n + 1)/n], with each term also being [(n/n)+(1/n), reducing altogether to [1+(1/n)]^{3} .} So we now divide the inequality by n^{3}, obtaining:

Now, look at the left-most part of the inequality. As n goes to infinity, we have 1/(infinity) or 0, meaning the left-most part of the inequality is now 1/3.

[Again recall our notation for the integral, here meaning its limits are a and b.

And we said that it is finding the sum for the function f(x)dx between a and b:

It seems now we might read this as the sum of all differentials from a and b, that is, the sum of the areas of all the very tiny rectangles whose x values have diminished arbitrarily or infinitesimally small. Our function was f(x)=x^{2}. We divided the curve from a to b into n rectangles with base b/n. Then we used the (base)(height) formula, and placed the sequence into a series of additions. The height, in accordance with the function, is the x value squared, and since the x value is determined by how many bases we have moved from a, the second one for example would be 2(b/n) or 2b/n, and thus its height is (2b/n)^{2}, and so on for rest of the series to the nth rectangle.

We then factored out (b/n)^{3}.

We then wanted to find a value which was approximate to the right side of the formulation. We did that by reconceving the sequence of values as being successive slabs in a square pyramid. Under this interpretation, we found values for the volumes of slightly smaller and slightly larger pyramids.

It is not entirely clear how everything is rectified between our original area under the curve and our current formulations and approximations of a pyramid based on the curve’s rectangles’ values. However, we still now bring back the other parts of the original formulation which were factored out, namely, the (b/n)^{3 }. But we began first with just the 1/n^{3}. This gave us now the inequality:

We then made n go to infinity, meaning that in this pyramid conception we are increasing the slabs to infinity, but in the curve-rectangle conception it means we are increasing the number of rectangles to infinity. This means there are an infinity of very tiny rectangles with miniscule areas. But since it is very close to infinity, the right-most part of our inequality tends toward 1/3. So now we add the b3 factor that we had been excluding, and we get 1/3(b^{3}). Hence} after adding b^{3}, which was so far excluded during our other numerical operations, we get:

We will now look at summation notation, to help compress all the complicated elements of our computations.

{Before we continue with Jerison’s account, let’s first briefly review summation notation from Edwards & Penney.

_{i}. The i (called the summation variable, summation index, or running index) is the variable part of the terms, and it is substituted firstly with

*a*(which is what the i = 1 means), and it is subsequently substituted with the successive integers. The sequence ends when the i value reaches the n value.

_{1}*a*, and each substitution is squared. The substitutions continue until reaching 10. Then all the terms are added, to produce 385. The variable-letters are arbitrary, so we may note it different ways:

_{1}}

Jerison will first give an example:

“So, the general notation is the sum of a_{i}, i = 1 to n, is = a_{1} + a_{2} + ... plus a_{n}. So this is the abbreviation.”

“And this is a capital Sigma.”

[Now recall the value that was difficult to determine and required the complicated procedure.

We will now give the summation notation for it.]

Our previous formulation is equivalent to:

[Shown more specifically:

]

“And so, this quantity here, for instance, is (1 / n^{3}) times the sum i^{2}, i = 1 to n. So that's what this thing is equal to.” [We have 1/n^{3} at the beginning, because the summation is for the terms of the numerator, and that sum in total is placed over the denominator n^{3}. In other words, the series begins with 1. All the terms are squared, and the series ends with n, the nth or final term. All of this is over n^{3}]. What we showed in our calculations was that this tends toward 1/3 as n goes to infinity.

Now recall again that previously we had a very long sum:

We can render this into summation notation thus:

“one way of writing it is, it's the sum from i = 1 to n […]”

“[…] of, now I have to write down the formula for the general term. Which is (b/n)(ib/n)^{2}.”

[Again, we originally had:

Here b/n was the same factor in each case. But the coefficient in the numerator of the right-side part of each term in the series increased from 1 to n by increases of 1 each time. The summation notation is saying that this variable begins with 1 in the first term of the series and increases incrementally to n in the final term of the series.]

And we can still factor out the (b/n)^{3} like before.

[Note first this expression of the distributive property in Edwards and Penney:

]

On account of the distributive law, we factor it out like this:

We will do another example. The function is f(x) = x. If we draw it, it is a line with slope 1.

[The x value is b. Since the slope increases evenly, the height at that point will be b too, and the area will be of a triangle with base b and height b.]

[As the area of a triangle 1/2(b)(h),] the area of this triangle is 1/2(b)(b).

And thus 1/2(b)^{2}

We do not need to do any elaborate summing here, because we already know how to find this area. [Since there are no irregularities of a curve falling outside the regular polygonal area, we do not need to make infinitesimal approximations].

Example 3: f(x) = 1

If we stop it at b, then we are just interested in this area.

0 to b,

We know it is height 1. The area for a rectangle is base times height, so the area here is:

And thus it simply equals b.

We will now look at the pattern of the function and the area under the curve.

In our prior examples, we found that:

To help see the pattern, let’s convert the left side terms to powers of x.

And let’s render the third row, right side, to match the format of those above it:

So let’s now guess for the next higher term [which according to the pattern will of course be b^{4}/4]

Jerison then notes how Archimedes was attempting something similar for finding the area of a parabola, but his method was so complicated that it confused mathematicians for a long time and blinded them to this simple pattern. [See the Boyer discussion of Archimedes method and see Archimedes’ ‘Quadrature of the Parabola’]

“That's a reasonable guess, I would say. Now, the strange thing is that in history, Archimedes figured out the area under a parabola. So that was a long time ago. It was after the pyramids. And he used, actually, a much more complicated method than I just described here. And his method, which is just fantastically amazing, was so brilliant that it may have set back mathematics by 2000, years. Because people were so, it was so difficult that people couldn't see this pattern. And couldn't see that, actually, these kinds of calculations are easy. So they couldn't get to the cubic. And even when they got to the cubic, they were struggling with everything else. And it wasn't until calculus fit everything together that people were able to make serious progress on calculating these areas. Even though he was the expert on calculating areas and volumes, for his time. So this is really a great thing that we now can have easy methods of doing it.” [33.10 – 34.15]

Now, we will not need to labor to make pyramids as we did before. But before we get to those simpler methods next class, we will need a little more practice with the notation for definite integrals. We will look at Riemann sums, which provide the general procedure for definite integrals. So we consider this function with limits a and b.

We then break it into increments called Δx.

How many pieces are there? If there are n pieces, then the general formula will be Δx = 1/n times the total length. So it would be b – a / n.

We will only allow ourselves one bit of flexibility: we will pick any height of f in each interval. We pick any value in between the intervals, call them c_{i}, and their level is f(c*i*). That is the rectangle that we choose.

So we pick f(c_{i}),

and we are going to add them all up.

It is the sum of the rectangles, because f(c_{i}) is the height and Δx is the base.

This is reminiscent of Leibniz notation for the integral. As Δx is replaced by dx at the limit:

The rectangles get thin as Δx goes to 0.

This is called a Riemann sum. In fact, our first example was a Riemann sum.

Example 4, a non-area example.

This example will show that integrals can be interpreted as **cumulative sums**.

We will consider a variable t which is time, in years. And we will consider a function f(t), which is dollars per year. And the unit dollars per year is a borrowing rate.

This example will show that there is a good reason for this dx that we append onto the definite integrals.

“It allows us to change variables, it allows this to be consistent with units. And allows us to develop meaningful formulas, which are consistent across the board.” [41.30]

So, we are borrowing money every day. Thus Δt is 1/365 [because change in time is one day.] “ That's almost 1 / infinity, from the point of view of various purposes.” [42.13] These are the time increments during which we borrow. But the rate [of how much you borrow each day] varies, because sometimes you need more money and other times less. So how much did we borrow? So consider day 45, which corresponds to t = 45/365. Our borrowing rate is f(45/365) [the amount of dollars for that fraction of the year. If it were the last day, it would be full amount.] Then we multiply that by change in time.

And Δt here is the change of one day, so

This final amount [right side of equation] is the final amount that we borrow in dollars. [I am not sure why we have day 45 if we are finding the final amount for all days. Perhaps either in this case he means total up to that day, or perhaps we only need to arbitrarily select one in order to figure out the whole.] We can render this into summation notation thus:

This formulation above gives us total amount borrowed. This will be very similar to the integral from 0 to 1 of f(t)dt.

It is equally important to model the amount that we owe the bank at the end of the year. The interest compounds continuously. We start with P as the principle. After time T, we owe Pe^{rT}, where r is interest rate, at 0.05/yr. [I do not know what the e is.]

To know how much we owe at the year:

We borrowed these amounts here [each day we borrowed a little, adding to the principle]

We owe more. We owe e to the power r [the interest rate] times the amount of time left in the year, which is 1 – i/365. (Or 365 minus i days left).

And since our formulation is e to the rT, we need to substitute it in.

[We need to add this up for each successive day] so we sum it:

This sum is essential the integral from 0 to 1.

[It seems in the following that we substitue t for i/365, and Δt becomes dt.] “The delta t comes out. And you have here e^{r(1 - t)} , so the t is replacing this i/365, f(t)dt”

“That is just an example. See you next time.”

David Jerison. ‘The Definite Integral’ Class 18 of 18.01 Single Variable Calculus. Fall 2006.

http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/

Videos: http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/video-lectures/

Lecture Notes:

http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/lecture-notes/

MIT OpenCourseWare.

Some summation notation from:

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