So, the displacement of an object is given by the product of the velocity and elapsed time, and is represented by the area of the shaded rectangle in the figure.

We give an alternative interpretation of the definite integral and make a connection between areas and antiderivatives.

### More than one perspective

We’ll start with a question:

Find the displacement of the object over the time interval .

This scenario will give us a very rough approximation of the displacement, but it is a good starting point. This scenario is illustrated in the figure below.

In our next step we will assume that the velocity is constant during two-second time intervals and . For each of these two intervals, we assume that the velocity is equal to the velocity at the beginning of the time interval.

In this case, the object would move by during the time interval , and during the next two seconds, the object would move by .

If we let denote the length of the time interval, we can approximate the displacement and write

Using sigma notation, we write Since we evaluate the velocity at the sample points , , we can also write

This is a left Riemann sum for the function on the interval , when . This scenario is illustrated in the figure below.

Under this assumption, during the first second, the object would move by ; during the second time interval of one second, the object would move by , during the third second, the object would move by , and during the last second, it would move by .

During the entire time interval, the object would move by

Using sigma notation and the fact that , we can write Again, we denote the sample points by , , and write This is a left Riemann sum for the function on the interval , when . This scenario is illustrated in the figure below.Under this assumption, during the first half-second, the object would move by ; during the second time interval of half a second, the object would move by , during the third half- second interval, the object would move by ,etc.

During the entire time interval, the object would move by

Using sigma notation and the fact that , we can writeUsing the standard notation for sample points, we can write

This is a left Riemann sum for the function on the interval ,when . This scenario is illustrated in the figure below.

We continue by approximating the displacement of the object assuming that the velocity is constant during -second time intervals, with velocity equal to the velocity at the beginning of each time interval.

Under this assumption, during the first -second interval, the object would move by ; during the second such time interval, the object would move by , during the third time interval, the object would move by

, etc.

During the entire time interval, the object would move by

Using sigma notation, notation for the length of each time interval, , and notation for sample points, we writeand This is a left Riemann sum for the function on the interval .

What happens when we let ?

Approximations get better and better, because we approximate the velocity with a constant over smaller and smaller time intervals. Therefore, it seems reasonable to write

By the definition of the definite integral, we can write

This gives us another interpretation of the definite integral. Namely,

“definite integral of the velocity over = the displacement over .”

On the other hand, the displacement of an object during the time interval is given by

The change in position can be written in terms of , the position function of the object But, we know that is an antiderivative of . Therefore, If , we get So, the position function is given by Now we can compute the displacement

Much more important than this computation is the fact that we were able to express the displacement as a definite integral and compute the definite integral of using the antiderivative of .

Is this always so?

Is it true that for any continuous function and its antiderivative , we can compute the definite integral of on the interval by this simple formula