CHAPTER 07.01: PRIMER ON INTEGRAL CALCULUS: Accumulation Function: Example
In this segment, we're going to talk about accumulation function. To talk about accumulation function is important for two reasons. So one is that it reviews what you have learned in your integral calculus. It also forms . . . gives you a good idea of the first fundamental theorem of calculus but more importantly it shows you that how the numerical methods have been introduced to you in the integral calculus class itself, and in fact, that becomes your first introduction to integration, as opposed to just what we talk about in analytical calculus. So let's take an example here. What we're going to do is we're going to show that if you are going to integrate t squared from 0 to x, you're going to get x cubed, divided by 3, and you have see these equations given all the time, that you are given tables of integrals, and in one of those tables of integrals, this might be one of the integrals which is given to you. But how do we really get . . . how do we go from here to here? That's what we're going to show by the accumulation function example in this case. So this is how it's going to work. If we have this function, f of t equal to t squared, as a function of t, and what I want to do is I want to go all the way up to x, I'm going to start from 0 and go to x, I want to find out what the area under the curve is, that's what I want to be able to do. And the way I'm going to do it is I'm going to break this . . . I'm going to take this 0, comma, x, I'm going to partition into n equal segments, that's what I'm going to do first. So I'm going to partition it into n equal segments, which basically tells me that the delta t which I have is x divided by n. So if I'm going to break it up into n equal segments, I'm going to get delta t equal to x divided by n, because that's the length of the segment . . . total interval, and I'm dividing it by n. So what that means is that if this is 0, this will be . . . this will be t0, and this will be t1, and all the way up to, this will be tn, that's what I'm going to get. And what I'm going to do is I'm going to use the right-hand Riemann sum. I'm going to use the right-hand Riemann sum to calculate my areas under the curve, so the last one will look like this. And again, t0, for example, is 0, t1 is nothing but x divided by n, then t2 is 2 x, divided by n, all the way up to tn, which is nothing but x itself, so because they are equally space, so the space between t0 and t1 is x divided by n, between t1 and t2 is x divided by n, and what I am doing is that in order to find out what the sum . . . what the area under the curve is, I'm going to use the right-hand Riemann sum. So I'm basically going to calculate this area, I'm going to calculate this area, and I'll have t3 here, for example, I'm going to calculate that area, and then the last area which I will have, and that I'm going to . . . that I can use as an approximate value of the integral going from 0 to x, for this particular function of t squared, but if I make infinite such rectangles, so if I choose n going to, approaching infinity, then the value of the area under the curve will be exactly the same as the actual area under the curve. So that's the whole point what I wanted to show, but what we have to appreciate here is that if n is a finite number, I am basically doing a numerical method here of finding the area under the curve. So it's by choosing n going to infinity that I'll be able to get the exact value of the area under the curve, and that's what forms the analytical form of that particular part of the calculus. So let's go ahead and write down what the total area is. So if I'm going to look at the area, it'll be equal to the value of the function at t1 times delta t, so that will be the first area which I am calculating, because all I am doing is calculating the value of the function at t1, because I am using the right-hand Riemann sum, and multiplying by the width, so that becomes the area of the rectangle. The area of the next rectangle will be f of t2 times delta t, and the area of the last rectangle will be f of tn times delta t. So again, keep in mind that when we are using this right-hand Riemann sum, we are choosing the right point as the sample point, and also that we are partitioning it in equal segments, so those are two things which you've got to think about, because in Riemann sum, you can have sample point anywhere in the partition, and all the partitions do not need to be of equal width. So, but in the right-hand Riemann sum, the way we are doing it is that we are choosing the right point of the segment, of the partition, and then also we are choosing equal widths there with delta t, and also we know that delta t is x divided by n. So this I can write as a summation, i is equal to 1 to n, f of ti times delta t. But what is ti? I know that ti is like t0 is 0, t1 is x divided by n, t2 is 2 x, divided by n, it is nothing but i x, divided by n, times delta t, and what is delta t? delta t is nothing but x divided by n. So because I know that t0 is 0 times x, divided by n, t1 is 1 times x, divided by n, and so on and so forth, so that's what the argument of the function f is, and then x divided by n is simply the value of delta t. So here, we're going to get summation, i is equal to 1 to n, and what is the function? The function is simply the square of the argument, which will be i times x, divided by n squared, times x divided by n. Now, keep in mind that I could have straightaway gone to this particular step right here. All I'm trying to show you is that you can do the same process for any other function f, like for example, if you had integral of t cubed, or sine x, or whatever it is, you could do that. The only is that you have to be able then be able to sum the series exactly, but this kind of procedure will work for integrating any kind of polynomial function which somebody throws at you in order to be able to get what the exact . . . exact expression for the integral is. So from here, what I am getting is summation of, i is equal to 1 to n, I'm getting i squared x squared, divided by n squared, x divided by n, and since x and n are constants in this case, i squared is the only thing which will stay inside the summation. So in this case, I will get equal to x cubed, divided by n cubed, summation, i is equal to 1 to n, i squared. And you can find out this summation from any handbook, and you will find out that this summation here will turn out to be n times n plus 1 times 2 n plus 1, divided by n cubed . . . divided by 6 . . . and divided by 6, that's what you get for the summation of i squared. What I can also do is I can now see that, hey, this is what I'm going to get for the value of the area under the curve if n is a finite number. So if somebody told me that I'm going to break 0 to x into, let's suppose, five segments, all I have to choose is n equal to 5, and I will get some approximation of the area under the curve. But as we said that if I'm going to choose n to be infinity, that means that I'm going to have these infinite rectangles, then in that case, the exact value of the integral is simply the limit of an approaching infinity of this summation, i is equal to 1 to n, the value of the function at particular sample points, in this case it is the right sample point at the end of the partition, f of ti times delta t. So I already calculated this for our integral right here, which is this, so this limit of n approaching infinity, x cubed, divided by n cubed, times n times n plus 1, 2 n plus 1, divided by 6. Since the limit is being taken on n, I can take x cubed, divided by 6 outside, so x cubed, divided by 6 I can take outside, and I'm going to say limit of n approaching infinity, now, what is inside? I'm going to expand this factored form, I get 2 n cubed, plus 3 n squared, plus n, that's what I get from the numerator here, the denominator is n cubed. And, so now I'm going to take the limit of this expression as n approaches infinity. So I'm going to get equal to x cubed divided by 6, limit of n approaching infinity, I'm going to divide the numerator by n cubed and the denominator by n cubed, so I'm going to get 2, plus 3 by n, plus 1 by n squared, that's what I'm going to get. So as n approaches infinity, this part is going to become 0, and this part is going to become 0, so I'll be left with what? x cubed, divided by 6, times 2, and that gives me x cubed, divided by 3. So that's how you are showing that the integral of 0 to x, t squared dt is equal to x cubed, divided by 3. And you can do this for any other polynomial function in a similar fashion, or any other function, but you may have problems with getting the exact value of the summations, you may have to use Taylor series for trigonometric and some transcendental functions to be able to follow the same procedure. It might be algebraically a little bit more complicated, but it can be done. And that is the end of this segment. |