Archimedes: the area under a parabola

About 23 centuries ago, a great mathematician named Archimedes was fascinated by the concept of length, but also area and quite a few others. And one of the problems he studied in connection with conics, which he found fascinating– well, who wouldn’t– is the following. He considered the problem that I’m now going to describe. You look at the parabola y equals x squared, that curve in the plane. You look at just its part between x equals 0 and x equals b. And now you look at the domain that is under that graph y equals x squared, but over the x-axis. That domain, let’s call it d. And the problem is to determine the area of the domain d.

And we’re now going to solve this problem exactly in the way that Archimedes solved it so long ago. We know the area of a rectangle. That’s about all we really know, initially, about areas. A rectangle– well, if it has two sides of length c and d, then its area is the product, c d. We’re going to use just that fact to solve our problem. We’re going to use a method that was introduced by Archimedes that he applied to a wide variety of problems, and it’s called the exhaustion method. It consists of approximating the domain of interest by a family of rectangles in a finer and finer way. This process is based on the partitioning of the underlying interval.

Let me explain what that means. For example, suppose you have the interval 0, b, which is the interval underlying our problem, and you wish to partition it into 8 subintervals. I take capital N equals 8. I chop the interval up into eight equal subintervals. And the points which define these subintervals, I call them x1, x2, up to x7. In fact, to make our notation more complete, I define the first point of the interval, the point 0, as the point x0. And the final point, b, I call it x8. So this would be a partition into eight subintervals. And the points involved are called the nodes of the partition. Now, this is what Archimedes does.

He looks at his domain, and he takes the underlying interval, and he partitions it, for example, into eight subintervals. There are the nodes. On each subinterval, he constructs a rectangle which is based on the x-axis and which go as high as possible, but without crossing the graph y equals x squared. So you push it up to the max. And now I wish to consider a typical node and the rectangle that is defined just to the right of it. In the picture, it’s x5. But a typical node is something of the form x sub i, and it’s a point n over n times i, where i is a number between 0 and n minus 1.

That rectangle– the one between x5 and x6 in the picture– what is its area? Well, the length of the base is clearly b over capital N, because each of my subintervals has that length. How am I going to find the area? I nee to know the height of the triangle. And here I use the fact that the top left corner of the rectangle is a point on the curve y equals x squared. But the x part of that point, the x-coordinate, is bi over n. Therefore, the y-coordinate is the square of that, and that gives me the height of that rectangle. So I know the base. I know the height.

I can calculate the area of that particular rectangle that starts at xi. It’s the product of b over n times the height I’ve just found. Let’s retain that fact now. The area of the rectangle on the subinterval between xi and xi plus 1 is given by b cubed i squared over n cubed. The total area of all these rectangles is obtained by taking the sum of all these terms as i ranges from 1 to capital N minus 1. And the expression you see there is simply an expression for that sum.

Now, long ago, we proved by induction a formula for the sum of the first n squares where lowercase n was a positive integer. We proved that 1 squared plus 2 squared, and so on, up to little n squared is given by a certain formula in terms of n. We’re going to apply that formula to the expression that you see above between the braces. So we’re going to apply it, and we’re going to get this as an answer. Now, you can check that this is what you get. And when you apply it, bear in mind that you’re applying the general formula where little n is given by capital N minus 1. So you make that substitution.

And in getting the final form of this formula, I’ve cancelled out a capital N from the top and the bottom. So here’s a formula that we need to look at and retain– that formula. Now, it turns out that when the partition gets very fine as the capital N becomes very large, the family of rectangles that we defined in the way that we did approximates the domain d in a better and better fashion. And in the limit, it really gives you the domain. And it turns out that the total area of the rectangles as the capital N gets very large approaches a certain limiting value that is the area of the domain d.

Here was our expression for the total area of those rectangles. What does it approach? What does it converge to when capital N gets very large? Well, to see what that is, divide out the n squared at the bottom, 1 in the n minus 1, and 1 in the other term, and rewrite your expression this way. Now you can see that when capital N gets very large, the factor 1 minus 1 over n, for example, really becomes 1, and the next one becomes 2. So at the end, what you wind up getting is b cubed over 3. That’s the answer. That’s the area of the domain d.

Archimedes found this, and it is a totally amazing result, for the times, to be able to make this calculation. Now, are we called upon to remember this answer? Well, not at all. Because it turns out that any first-year calculus student can easily find this answer in a minute using the tools of calculus. But the method– the method of exhaustion by rectangles is a very valuable one which is still used today and used to calculate many areas, and in fact, volumes and other things. It gives rise to a number of formulas, some of which are worth memorising, some of which you may know, and that we’ll look at in the next segment.