Over- and under-estimation of Riemann sums | AP Calculus AB | Khan Academy

Consider the left and right Riemann sums that would approximate the area under y is equal to g of x between x equals 2 and x equals 8. So we want to approximate this light blue area right over here. Are the approximations overestimations or underestimations? So let's just think about each of them.

Let's consider the left and the right Riemann sums. First, the left, and I'm just going to write left for short, but I'm talking about the left Riemann sum. They don't tell us how many subdivisions to make for our approximation, so that's up to us to decide. Let's say we went with three subdivisions. Let's say we wanted to make them equal; they don't have to be, but let's say we do.

So the first one would go from two to four. The next one would go from four to six, and the next one from six to eight. If we do a left Riemann sum, you use the left side of each of these subdivisions in order to find the height. You evaluate the function at the left end of each of those subdivisions for the height of our approximating rectangles.

So we would use g of 2 to approximate for or to set the height of our first approximating rectangle, just like that. And then we would use g of 4 for the next rectangle, so we would be right over there. And then you'd use g of 6 to represent the height of our third and our final rectangle right over there.

Now when it's drawn out like this, it's pretty clear that our left Riemann sum is going to be an overestimation. How do we know that? Because these rectangles, the area that they're trying to approximate, are always contained in the rectangles, and these rectangles have this surplus area. So they're always going to be larger than the areas that they're trying to approximate.

In general, if you have a function that's decreasing over the interval that we care about right over here and strictly decreasing the entire time, if you use the left edge of each subdivision to approximate, you're going to have an overestimate. Because the left edge, the value of the function there, is going to be higher than the value of the function at any other point in the subdivision.

And so that's why for decreasing functions, the left Riemann sum is going to be an overestimation. Now let's think about the right Riemann sum, and you might already guess that's going to be the opposite, but let's visualize that. So let's just go with the same three subdivisions, but now let's use the right side of each of these subdivisions to define the height.

So for this first rectangle, the height is going to be defined by g of 4, so that's right over there. And then for the second one, it's going to be g of 6, so that is right over there. And for the third one, it's going to be g of 8. And so let me shade these in to make it clear which rectangles we're talking about. This would be the right Riemann sum to approximate the area.

It's very clear here that this is going to be an underestimate. We see in each of these intervals that the Riemann, the right Riemann sum, or the rectangle that we're using for the right Riemann sum, is a subset of the area that is trying to estimate. We're not able to; it doesn't capture this extra area right over here.

And once again, that is because this is a strictly decreasing function. So if you use the right endpoint of any one of these or the right side of any of these subdivisions in order to define the height, that right value of g is going to be the lowest value of g in that subdivision.

And so it's going to be a lower height than what you could even say is the average height of the value of the function over that interval. So you're going to have an underestimate in this situation. Now, if your function was strictly increasing, then these two things would be swapped around.

And of course, there are many functions that are neither strictly increasing nor decreasing, and then it would depend on the function and real, and sometimes even it would depend on the type of subdivisions you choose to decide whether you have an overestimate or an underestimate.