>> Because the calculations are somewhat algebra-intensive, I have chosen a somewhat simpler function for us to deal with here. It's just a variation on the one that we had before: it's "f of x is 1/5 x squared." Notice the, the beginning of the interval is at 1, and the end is at 4, as we saw before. Now to make the calculation of area, it's the limit, as n approaches infinity, of the sum of a number of areas. And it's set up like this: it's a sum as i approaches -- excuse me -- as i runs from 1 to n of -- and here's the width and here is the height of the various rectangles. Now notice the counting mechanism, i, appears right here. Now this is the form that we saw a little bit earlier. And we're just going to go through and massage this according to the a and the b that we have here in the information about our function. Let's simplify this business about the width and f of whatever this expression is. Let's come over here. The width is b minus a over n. Now the b, the end of the interval, is at 4. The beginning of the interval, the a value, is 1. So 4 minus 1 is 3. 3 over n, then, is the width of each of the little rectangles. Okay, we'll replace that in just a minute. Now for the height, it's f of a plus -- now this is the width, you see, times the counting mechanism. So it is f of, start at the, the end, and we have the width times 1, and then the width times 2, and the width times 3. You see for all of the different rectangles that we're finding the area of. And by the way, we could set this up, we could, we could figure out a way to, to assure ourselves that we're finding the height at the left side, on the left side of our rectangles rather than the right side. But I've just chosen to do it on the right side, and it turns out that it doesn't make any difference. But at any rate we have the, let's see, the b minus a over n is the 3 over n. The a, the left end of our interval is at one. And now we just clean this up a little bit. And 1 is n over n, and this is 3i over n, and bringing these together we have n plus 3i over n. Now I just chose to write this as a single term here, because it, it, I think it's a little bit easier to deal with when we go through this squaring process. Now we have to do that because it's f of this fraction. So it means in the f function replace x with the fraction. Now the function was 1/5 x squared, so we replace x with the fraction, you see. And now we evaluate, and squaring here, we get -- well, let's see, in the numerator, it's n squared plus 6ni plus 9i squared and the denominator, n squared. So we have this. Now I have chosen to rewrite this as, I'm taking this n squared and I'm just putting it out here, 1 over 5n squared times this trinomial. Alright now, so this is an expression for f of the right side of all of those rectangles, and so we can come back over here and make, we can make replacements. And it's right here in our, in our limit process here, the width, b minus a over n. I'm just looking over there. Oh, we said that it was 3 over n. So it's 3 over n, and then times -- now f of all of this business we simplified, and down over there it's 1 over 5n squared times n squared plus 6ni plus 9i squared. Now here's where we'd like to just look at this part, this summation business involving that counting mechanism. We would like to manipulate this a bit in order to remove the counting mechanism. And it turns out that we can do that. Now, it's a little bit algebra-intensive as I mentioned before, but when situations like this occur, it's just time to amplify the organizational skills. There's nothing terribly difficult about it except the organization of it. The, all of these things that we're going to do, all of these operations that we're going to perform have been performed many times before --