Worked example: Rewriting limit of Riemann sum as definite integral | AP Calculus AB | Khan Academy

So we've got a Riemann sum. We're going to take the limit as n approaches infinity, and the goal of this video is to see if we can rewrite this as a definite integral. I encourage you to pause the video and see if you can work through it on your own.

So let's remind ourselves how a definite integral can relate to a Riemann sum. If I have the definite integral from A to B of f of x, f of x DX, we have seen in other videos this is going to be the limit as n approaches infinity of the sum, capital Sigma, going from i equal 1 to n.

We're essentially going to sum the areas of a bunch of rectangles where the width of each of those rectangles we can write as Δx. So, your width is going to be Δx of each of those rectangles, and then your height is going to be the value of the function evaluated someplace in that Δx. If we're doing a right Riemann sum, we would do the right end of that rectangle or of that sub-interval.

We would start at our lower bound a and we would add as many Δx's as our index specifies. So if i is equal to 1, we add one Δx, so we'd be at the right of the first rectangle. If i is equal to 2, we add two Δx's, so this is going to be Δx times our index. This is the general form that we have seen before.

One possibility you could even do a little bit of pattern matching right over here. Our function looks like the natural log function, so that looks like our function f of x. It's the natural log function, so I could write that.

So, f of x looks like the natural log of x. What else do we see? Well, a that looks like two; a is equal to 2. What would our Δx be? Well, you can see this right over here, this thing that we're multiplying by, is divided by n, and it's not multiplying by an i.

This looks like our Δx, and this right over here looks like Δx * i, so it looks like our Δx is equal to 5/n. So what can we tell so far?

Well, we could say that, okay, this thing up here, the original thing, is going to be equal to the definite integral. We know our lower bound is going from 2; we haven't figured out our upper bound yet. We haven't figured out our b yet, but our function is the natural log of x, and then I will just write a DX here.

So, in order to complete writing this definite integral, I need to be able to write the upper bound. The way to figure out the upper bound is by looking at our Δx because the way that we would figure out a Δx for this Riemann sum here, we would say that Δx is equal to the difference between our bounds divided by how many sections we want to divide it into, divided by n.

So it's equal to (b - a) / n. You can pattern match here. If this is Δx is equal to (b - a) / n, let me write this down. So this is going to be equal to b - our a, which is 2, all of that over n. So b - 2 is equal to 5, which would make b equal to 7.

B is equal to 7. So there you have it. We have our original limit, our Riemann limit, or limit of our Riemann sum being rewritten as a definite integral.

And once again, I want to emphasize why this makes sense. If we wanted to draw this, it would look something like this. I'm going to try to hand-draw the natural log function; it looks something like this. This right over here would be one, and so let's say this is two.

We're going from 2 to 7. This isn't exactly right, and so our definite integral is concerned with the area under the curve from 2 until 7. This Riemann sum you could view as an approximation when n isn't approaching infinity, but what you're saying is, look, when i is equal to one, your first one is going to be of width 5/n.

So this is essentially saying our difference between 2 and 7; we're taking that distance 5, dividing it into n rectangles, and so this first one is going to have a width of 5/n. And then, what's the height going to be? Well, it's a right Riemann sum, so we're using the value of the function right over there, right at 2 + 5/n.

So this value right over here, this is the natural log, the natural log of 2 + 5/n, and since this is the first rectangle times 1, we could keep going. This one right over here, the width is the same, 5/n, but what's the height? Well, the height here, this height right over here is going to be the natural log of 2 + 5/n * 2.

This is for i equal to 2, and this is i equal to 1. So hopefully, you are seeing that this makes sense. The area of this first rectangle is going to be natural log of 2 + 5/n * 1 * 5/n.

The second one over here, natural log of 2 + 5/n * 2 * 5/n. This is calculating the sum of the areas of these rectangles, but then it's taking the limit as n approaches infinity, so we get better and better approximations, going all the way to the exact area.