How to Use Riemann Sums to Calculate Integrals

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  • 0:10 When to Use a Riemann Sum
  • 1:02 Riemann Sums
  • 4:15 Using Two Slices
  • 5:46 Calculating Multiple Slices
  • 7:54 Infinite Slices
  • 13:15 Lesson Summary
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Lesson Transcript
Instructor: Erin Monagan

Erin has been writing and editing for several years and has a master's degree in fiction writing.

As a new property owner, you might relish mowing your lawn. Up and down your property you mow and measure out small sections to find the area of your property. In this lesson, you will discover what a Riemann sum approach is and how to calculate an estimated area using multiple slices.

When to Use a Riemann Sum

You just found yourself the best property ever. It sits out on Lake Heaviside, it's just off a road (but not a huge road) and it's right next to a park. Your mom thinks that this area is a little bit pricey for you, but you think of how much land you've got. Honestly, you have no idea how much land it is. So let's try to calculate it. Let's say that the road is the x-axis, and your property extends all the way to the water's edge at the line f(x). The boundary with the park is at x=a, and at x=b the water hits the road (which is the other end of your property).

Using a Riemann sum to calculate land area
calculate integrals

Riemann Sums

So how might you find the area? Turning back to what you learned in calculus, you know that the area of this land is going to be the integral from x=a to x=b of f(x)dx. In this case, a is your lower limit (this is where your land meets the park next door), b is the upper limit (where your land, the water and the road all meet) and f(x) is the integrand (the shoreline of your property). How do we apply this to your property? Your shoreline matches up with f(x) = 50 - x^2 + 5x. You know you need to integrate this from where your property meets the park (x=0) and where your property meets the road and the shoreline (x=10). This is the integral from 0 to 10 of (50 - x^2 + 5x)dx. How do we calculate this? One way is to use a Riemann sum approach. Remember that the integral from x=a to x=b of f(x)dx = the limit as delta x goes to 0 of the sum from k=1 to k=n of f(x sub k)delta x sub k. This is just summing up the area of multiple rectangles. Let's use one slice, one rectangle, to estimate your area. If I do that, then n=1, and I can use the Riemann sum as the sum from k= 1 to 1 of (50 - (x sub k)^2 + 5x sub k)delta x sub k.

In this case, my x sub k is going to be the value of x somewhere in the cake slice. If I only have one slice, I can write this sum as (50 - (x sub 1)^2 + 5x sub 1)delta x sub 1. Where's x sub 1? Let's use a left-sided Riemann sum and say that x sub 1 is on the left side of my slice. That value is x=0. Let's plug in 0 for x=1. If I do that, I end up with (50 - (0)^2 + 5(0)) delta x. So, delta x is the width of this slice, and the width is 10 - 0. That's the right side of the slice minus the left side of the slice. If I simplify the whole thing, I get 50 * 10 to make the area of this slice 500. If I use a Riemann sum with one slice, my area is estimated at 500. Let's write this out in a table and keep track of it for later:

n area
1 500

Using Two Slices

What if I use two slices? Now my Riemann sum goes from k=1 to k=2. My function is still the same, and I still have the delta x term. If I write out the sum, I've got (50 - (x sub 1)^2 + 5x sub 1)delta x sub 1. This is the area of my first slice. I'm going to add that area to the area of my second slice, which is (50 - (x sub 2)^2 + 5x2)delta x sub 2. Now, x sub 2 is a value somewhere in this second slice. Because I'm using a left-sided Riemann sum, I'm going to choose x sub 1 to be on the left side of my first slice and x sub 2 to be on the left side of my second slice. I'm also going to make these slices equal in thickness, so each one will have a width of 5. Plugging in 0 for x sub 1 and 5 for x sub 2, I can solve this for 50(5) + 50(5), which gives us a total area of 500.

Using a left-sided Riemann Sum
left side

Let's put that on our table:

n area
1 500
2 500

I'm still not convinced that two slices is giving us our actual area. Let's take a few more.

Calculating Multiple Slices

Let's say n=5. I'm going to cut my area up into five slices, and I'm going to estimate the area of each one of those slices. Here, my Riemann sum goes from k=1 to k=5 (for 5 slices), my function is still the same and I'm going to make delta x the same width for each of these slices. That means that each delta x is going to equal 2. Again, I'm going to use a left-sided Riemann sum, so the x value for each slice is going to be on the left side of each slice. This means that I'm going to use x sub 1 as 0 for the left side for this first slice, x sub 2 is going to be 2, x sub 3 is going to be 4, x sub 4 is going to be 6 and x sub 5 will be 8. If this doesn't make sense, think about it this way: There are five slices that have to total ten points. Each slice has to go over 2 to do this. If I write in those numbers, 2, 4, 6, 8 and 10, I can see that x sub 4 is at x=6. I plug in my x sub 1, x sub 2, x sub 3, x sub 4 and x sub 5 into my sum. I can solve out the math and I get that the area is (50 * 2) + (56 * 2) + (54 * 2) + (44 * 2) + (26 * 2). I get that my total area is roughly 460.

I'll put that on my table:

n area
1 500
2 500
5 460

So before, the area was estimated to be 500, but now it's 460. What happens if I take even more slices?

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