What is the Trapezoid Rule?

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  • 0:06 Riemann Sum Review
  • 2:30 Trapezoid Area
  • 5:25 Trapezoid Rule
  • 9:26 Lesson Summary
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Lesson Transcript
Instructor: Jordan White

Jordan has been a writer, editor and Web researcher for educational publications since 2010. He holds a bachelor's in English from Western Michigan University.

In this lesson, you'll move beyond simple rectangles to estimate the area under a curve. Get more sophisticated with your approximations and use trapezoids instead of those pesky rectangles.

Riemann Sum Review

You've got a great idea. You're going to put a skateboarding ramp in your backyard! But in order to do this, you need a lot of dirt to go underneath the ramp. How much dirt do you actually need?

Let's say you want the ramp to follow the curve y = x^2 + 1. You want it to go from x = 0 to 2. Let's say that's 0 meters to 2 meters. In order to find out how much dirt you're going to want to put underneath this ramp, you sit back and think about all you've learned in calculus. You say, well, I could use a Riemann sum to estimate the area between my ramp and the ground. Because you don't want to spend all day measuring how high this ramp is, you want to use a Riemann sum with only one slice along this curve. What a Riemann sum would do would be, you'd measure the height somewhere along this curve, and you'd multiply that height times 2 meters. That's the distance in x. The height times the width here would give you the cross-sectional area and tell you about how much dirt you would need.

But let's think about this. If you measure at the far left side, so if you have a left-side Riemann sum, you'd get an area of 2 because the height on the left side is 1 multiplied by the width, which is 2, which gives you an area of 2. If you use a midpoint Riemann sum, your area would be estimated at 4. If you use a right-side Riemann sum, the cross-sectional area would be estimated to be 10. This sounds absolutely fantastic, but none of these projections look right.

Area of a trapezoid flipped on its side
Trapezoid Flipped On Its Side

I mean, really? Does this ramp look like anything flat? Like this? Do you have any reason to believe that the area estimated by the left-side or right-side Riemann sum would be anywhere close to the actual cross-sectional area? Well, what about this midpoint? Does that look right? You just don't know. So, rather than taking multiple slices and doing a Riemann sum with two different areas, you decide to not use rectangles and instead estimate this area with a trapezoid.

Using Trapezoids to Estimate Area

Let's review. The area of a trapezoid is equal to the height of the trapezoid times the average of the two parallel sides. So if you have a trapezoid that looks like this, we're going to multiply the height times w sub 1, plus w sub 2, all divided by 2 (Area = (Height) * (w sub 1 + w sub 2) / 2). w sub 1 is this length here, on the short side, and w sub 2 is the long side. If we flip this trapezoid on its side the area is still the same, but now we've got height going along horizontally and w sub 1 on the left side of the trapezoid and w sub 2 on the right side of the trapezoid. Well, this looks a lot like our function.

Here's our function, and I can see a trapezoid outline going between the left side, 0, and the right side, 2.

In this case, w sub 1 is the height on the left side, w sub 2 is the height on the right side, and my height is actually the distance between 0 and 2 on the x-axis. If I plug these points into my area formula, I get that the area equals the value of the function on the left side plus the value of function on the right side all divided by 2 times my delta x. That's the difference between the left-side value of x and the right-side value of x. So in this case, it's 2 - 0.

Using a trapezoid to estimate area on the graph of the function
Using Trapezoids to Estimate Area

Let's plug in the points for y = x^2 + 1 from 0 to 2. f(x) on the left side is equal to f(0). f(x) on the right side is f(2) - all of that divided by 2 times my delta x, which is 2 - 0. If I plug 0 into my function y = x^2 + 1, I get 1. If I plug 2 into that function, I get 5. So my area becomes ((1 + 5) / 2) * 2, and that's just 6.

I could get an even better estimate by dividing this into two slices, and take the trapezoid area of two different slices and add them up to get the total sum. If I do that, then my first area goes from f(0) to f(1), so my delta x is 1 - 0, and my second area goes from f(1) to f(2), so my delta x is 2 - 1. If I plug in the values for f(0), f(1) and f(2), I find that the area under the curve is estimated to be 5.

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