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

Let's get some practice rewriting definite integrals as the limit of a Riemann sum.

So let's say I wanted to take the definite integral from π to 2π of cosine of x dx. What I want to do is write it as the limit as n approaches infinity of a Riemann sum. It's going to take the form of the limit as n approaches infinity, and we can have our sigma notation right over here. I would say from, let's say, i is equal to 1 all the way to n.

Let me scroll down a little bit so it doesn't get all scrunched up at the top. Let me draw what's actually going on so that we can get a better sense of what to write here within the sigma notation.

Let me do it large. If this is π right over here, this would be 3π/2, and this would be 2π right over here. Now, what does the graph of cosine of x do? Well, at π, cosine of π is -1; we'll assume that's -1 there. And cosine of 2π is 1.

So the graph is going to do something like this. This is obviously just a hand-drawn version of it, but you have seen cosine functions before; this is just part of it. This definite integral represents the area from π to 2π between the curve and the x-axis.

You might already know that this area is going to be—or this part of the definite integral will be negative—and this would be positive. They'll cancel out, and this would all actually end up being 0 in this case. But this exercise for this video is to rewrite this in Riemann sum as the limit as n approaches infinity.

So as a Riemann sum, what we want to do is think about breaking this up into a bunch of rectangles. Let's say—or I should say—n rectangles. So that's our first one right over there. Then this might be our second one.

Let’s do a right Riemann sum where the right boundary of our rectangle, what the value of the function is at that point, defines the height. So that’s our second one, all the way until this one right over here is going to be our nth one.

So this is one; let me write it this way: this is i is equal to 1, this is i is equal to 2, all the way until we get to i is equal to n. Then, if we take the limit as n approaches infinity, the sum of the areas of these rectangles are going to get better and better.

Now let's first think about it. What is the width of each of these rectangles going to be? Well, I am taking this interval from π to 2π, and I'm going to divide it into n equal intervals. So the width of each of these is going to be 2π minus π.

I'm just taking the difference between my bounds of integration, and I am dividing by n, which is equal to π/n. So that's the width of each of these: that’s π/n; this is π/n; this is π/n.

And what's the height of each of these rectangles? Remember, this is a right Riemann sum, so the right end of our rectangle is going to define the height. So this right over here, what would this height be?

This height, this value, I should say, is going to be equal to f of what? Well, this was π, and this is going to be π plus this—the length of our interval right over here or the base of the rectangle.

So we started at π, so it's going to be π plus; this one's going to be π/n times 1. That's this height right over here. What's this one going to be right over here? Well, this one is going to be f of π—our first start—plus π/n times what? We're going to add π/n; 2 times π/n times 2.

So the general form of the right boundary is going to be—in this case for example, this height right over here is going to be f of—we started at π—plus we’re doing the right Riemann sum, so we’re going to add π/n times i.

If we wanted to say it generally—if we're talking about the ith rectangle, remember we're going to sum them all up; what’s their height? Well, the height is going to be, in this case, going to be cosine of π plus—if we're with the ith rectangle—we are going to add π/n times i.

Then that’s the height of each of our rectangles.

And then what's the width? Well, we already figured that out: times π/n. If you want to be careful and make sure that this sigma notation applies to the whole thing, there you have it.

We have just re-expressed this definite integral as the limit of a right Riemann sum.