Midpoint sums | Accumulation and Riemann sums | AP Calculus AB | Khan Academy

What we want to do in this video is get an understanding of how we can approximate the area under a curve. For the sake of an example, we'll use the curve ( y = x^2 + 1 ).

Let's think about the area under this curve above the x-axis from ( x = -1 ) to ( x = 2 ). So that would be this area right over here. There are many ways that I could tackle this, but what I'm going to do is break up this interval into three equal sections that are really the bases of rectangles. We're going to think about the different ways to define the heights of those rectangles.

So once again, I'm going to approximate using three rectangles of equal width, and then we'll think about the different ways that we can define the heights of the rectangles. Let's first define the height of each rectangle by the value of the function at the midpoint. We see that right over here.

Let's just make sure that it actually makes sense to us. If we look at our first rectangle, right over here, let's just first appreciate that we have split up the interval from ( x = -1 ) to ( x = 2 ) into three equal sections. Each of them has a width of 1. If we wanted a better approximation, we could do more sections or more rectangles.

Now, let's see how we would compute this. The width of each of these is 1. The height is based on the value of the function at the midpoint. The midpoint here is ( -\frac{1}{2} ). The midpoint here is ( \frac{1}{2} ), and the midpoint here is ( \frac{3}{2} ).

So this height is going to be ( -\frac{1}{2}^2 + 1 ). ( -\frac{1}{2}^2 ) is ( \frac{1}{4} ) plus 1, so that's ( \frac{5}{4} ). Therefore, the height here is ( \frac{5}{4} ).

So, you take ( \frac{5}{4} ) times 1; this area is ( \frac{5}{4} ). Let me write that down. So if we're doing the midpoint to define the height of each rectangle, this first one has an area of ( \frac{5}{4} ).

Let me do that in a color you can see: ( \frac{5}{4} ). The second one, same idea, ( \frac{1}{2}^2 + 1 ) is ( \frac{5}{4} ) times the width of 1, so ( \frac{5}{4} ) there.

So let me add that: ( + \frac{5}{4} ). The third rectangle—what's its height? Well, we're going to take the height at the midpoint. ( \frac{3}{2}^2 ) is ( \frac{9}{4} ) plus 1, which is the same thing as ( \frac{13}{4} ).

So it has a height of ( \frac{13}{4} ) and then a width of 1, so times 1, which would just give us ( \frac{13}{4} ). So ( + \frac{13}{4} ) would give us ( \frac{23}{4} ), which is the same thing as ( 5\frac{3}{4} ).

This is often known as a midpoint approximation, where we're using the midpoint of each interval to define the height of our rectangle. But this isn't the only way to do it; we could look at the left endpoint or the right endpoint. We'll do that in other videos.

If we want to do it just for kicks, let's just do that really fast. So if we want to look at the left endpoint of our interval, well here, our left endpoint is ( -1 ). ( -1^2 + 1 ) is 2. ( 2 ) times ( 1 ) gives us ( 2 ). Then here, the left part of this interval is ( x = 0 ).

( 0^2 + 1 ) is 1. ( 1 ) times ( 1 ) is 1. Now here, our left endpoint is ( 1 ). ( 1^2 + 1 ) is equal to ( 2 ) times ( 1 ). Our base is equal to ( 2 ).

So here we have a situation where we take our left endpoints, where it is equal to ( 2 + 1 + 2 ) or ( 5 ). But we could also look at the right endpoints of our intervals.

So this first rectangle here is clearly under-approximating the area over this first interval. Its right endpoint is ( 0 ). ( 0^2 + 1 ) is ( 1 ), so height of ( 1 ), width of ( 1 ) gives an area of ( 1 ).

The second rectangle here has a height that we get from looking at our right endpoint. ( 1^2 + 1 ) is ( 2 ) times our width of ( 1 ); well, that's going to give us ( 2 ).

Then here, our right endpoint is ( 2^2 + 1 ) is ( 5 ) times our width of ( 1 ) gives us ( 5 ). So in this case, when we look at our right endpoints of our intervals, we get ( 1 + 2 + 5 = 8 ).

Eyeballing this, it looks like we're definitely overcounting more than undercounting, so this looks like an over-approximation. The whole idea here is just to appreciate how we can compute these approximations using rectangles.

As you can imagine, if we added more rectangles that had skinnier and skinnier bases, but still covered the interval from ( x = -1 ) to ( x = 2 ), we would get better and better approximations of the true area.