2.1 The Area Under a Power Function

Our discussion will not apply to negative values of ,
since we make frequent use of the fact that for all non-negative
numbers and

Also is not defined when is negative.

The figures for the argument given below are for the case , but you should observe that the proof does not depend on the pictures.

Let be a positive integer, and for , let .

Then for , so the points divide the interval into equal subintervals. For , let

If , then for some index , and , so

Hence we have

and thus

If , then and so . Hence, for all , and hence

so that

Now

and

Since the boxes intersect only along their boundaries, we have

and similarly

Thus it follows from equations (2.1) and (2.2) that

The geometrical question of finding the area of has led us
to the numerical problem of finding the sum

We will study this problem in the next section.

Notice that
. Thus equation
(2.4) can be written as

Now we will specialize to the case when .
A direct calculation shows that

There is a simple (?) formula for , but it is not particularly easy to guess this formula on the basis of these calculations. With the help of my computer, I checked that

Also

Thus by taking in equation (2.6) we see that

On the basis of the computer evidence it is very tempting to guess that