Worked example: finding a Riemann sum using a table | AP Calculus AB | Khan Academy

Imagine we're asked to approximate the area between the x-axis and the graph of f from x equals 1 to x equals 10 using a right Riemann sum with three equal subdivisions. To do that, we are given a table of values for f. I encourage you to pause the video and see if you can come up with an approximation for the area between the x-axis and the graph from x equals 1 to x equals 10 using a right Riemann sum with three equal subdivisions.

So, I'm assuming you've had a go at it. Now, let's try to do that together. This is interesting because we don't have a graph of the entire function, but we just have the value of the function at certain points. As we'll see, this is all we need in order to get an approximation for the error. We don't know how close it is to the actual area with just these points, but it'll give us at least a right Riemann sum for the approximation.

Let me just draw some axes here because whenever I do Riemann sums, you can do them without graphs, but it helps to think about what's going on if you can visualize it graphically. So, let's see, we are going from x equals 1 to x equals 10. This is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. They give us the value of f of x when x equals 1, when x equals 2, 3, 4, when x equals 7, 5, 6, 7, 8, 9, 10, and they tell us that when x is 1, we're at 6, and we go to 8, to 3, to 5.

So, let me mark these off. We're going to go up to 8, so 1, 2, 3, 4, 5, 6, 7, 8. So, what we know when x is equal to 1, f of 1 is 6. So, this point over here is f of 1, which is the point (1, 6). Then we have the point (4, 8), which will put us right about there. And then we have (7, 3) on our graph y equals x f of x, so (7, 3) would put us right over there, and then we have (10, 5). So, (10, 5) puts us right over there.

That's all we know about the function. We don't know exactly what it looks like. Our function might look like this; it might do something like this. Whoops, I drew a part that didn't look like a function. It might do something like this and oscillate really quickly. It might be nice and smooth and just kind of go and do something just like that, kind of connect the dots. We don't know, but we can still do the approximation using a right Riemann sum with three equal subdivisions.

How do we do that? Well, we're thinking about the area from x equals 1 to x equals 10. Let me make those boundaries clear. This is from x equals 1 to x equals 10. What we want to do is have three equal subdivisions. There are three very natural subdivisions here if we make each of our subdivisions three wide. So, this could be a subdivision, and then this is another subdivision.

In Riemann sums, you don't have to have equal subdivisions, although that's what you'll often see. So, we've just divided going from one to ten into three equal sections that are three wide. So that's 3, this is 3, and this is 3. The question is, how do we define the height of these subdivisions, which are going to end up being rectangles? That's where the right Riemann sum applies.

If we were doing a left Riemann sum, we would use the left boundary of each of the subdivisions and the value of the function there to define the height of the rectangle. So, this would be doing a left Riemann sum, but we're doing a right Riemann sum. We use the right boundary of each of these subdivisions to define the height.

So, our right boundary is when x equals 4 for this first section. What is f of 4? It's 8, so we're going to use that as the height of our first rectangle that's approximating the area for this part of the curve. Similarly, for this second one, since we're using a right Riemann sum, we use the value of the function at the right boundary, which is 7, so the value of the function is 3. This would be our second rectangle, our second division, I guess, used to approximate the area. Last but not least, we would use the right boundary of this third subdivision when x equals 10, where f of 10 is 5.

So, just like that, our right Riemann approximation using our right Riemann sum with three equal subdivisions to approximate the area would be to add the area of these rectangles. The first rectangle is three wide, and how high is it? Well, the height here is f(4), which is 8. So, this is going to be 24 square units, whatever the units happen to be.

This is going to be 3 times the height here is 3; f of seven is 3, so that is 9 square units. Then here, the width is three times the height; f of ten is five, so three times five gets us 15. Our approximation of the area would be summing these three values up. This would be 24 plus 9 plus 15, which gives us another 24. So it's 24 plus 24, getting us to 48.

There you go! Just using that table of values, we've been able to find an approximation. Now, once again, we don't know how good of an approximation it is. It depends on what the function is doing. There's a world where it could be a very good approximation. Maybe the function does something like this. Or maybe the function does something like this, where in this situation, this might be a very bad approximation. But we can at least do the approximation using a right Riemann sum just using this table.