Definite integrals arise as the limits of Riemann sums, and compute net areas.

In the previous sections, we’ve been approximating the area under a curve by treating the region as a collection of rectangles.

The benefit of doing this calculation with still variable, is that it’s now easy to find the approximation for every possible value of ! If we wanted to approximate the area under using rectangles, we could substitute into this formula. . If we wanted to use rectangles, we can substitute into it. .

Why would we want to be able to plug in different values of ? Remember how we made the approximation better and better? By taking more and more rectangles! The actual area is given in the limit as .

All of the functions we have been examining have been positive. What changes if is negative in the region?

Let’s look at on with using midpoints. , so the regular partition is , , , and , giving us midpoints , , , and .

In the Riemann sum the first two terms and are the areas of the first two rectangles. Notice that and are negative! They are not the heights of their rectangles. They are the opposite of the heights. That means and are not the areas of those rectangles. They are the opposite of those areas. The Riemann sum is adding the areas of the first two rectangles, then subtracting the areas of the last two rectangles.

This same situation occurs whenever our function is negative. The areas of the corresponding rectangles are being subtracted, not added.

The Riemann sum is not calculating the approximate area, but the net area.

Remember that when we setup our Riemann sum to approximate the areas, we had a choice on what sample points to use. We could use left-endpoints, right-endpoints, midpoints, a combination of the three, or any other possible points we wished. Different choices of sample points leads to different values for the Riemann sums.

To say it another way, if we have an integrable function, the difference between those sample points disappears once we take the limit, and this limit is the actual net area of the region.

Whenever we are working with an integrable function, the definite integral makes sense, but which functions are integrable? The next theorem tells us that most of the functions we deal with are.

Consider the following graph of : Compute:
(a)
(b)
(c)
Use the formula for the area of a triangle.
Remember, we are dealing with “signed” area here: