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Course: Calculus, all content (2017 edition)>Unit 4

Lesson 10: Trapezoidal rule

Midpoint and trapezoidal sums in summation notation

Estimating the area under a curve with trapazoids instead of rectangles can give a closer approximation. Created by Sal Khan.

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• Isn't the formula for estimating area using the midpoint, and for using trapezoids the same?
• the difference is that the midpoint method takes the average of the two x values, while the trapezoid method takes the average of the two y values
• Which of those is the most accurate?
• Some times it depends on the function being modeled, but in general, the trapezoid rule is most accurate when Δx is larger. As Δx becomes smaller and smaller, to the infinitesimally small dx, all the different forms converge to the same value.
• Which method gives you the most accurate approximation?
• For a fixed value of `n`, none of the approximations is best for all functions. For example, if you give me `n`, I'll say approximate

1+cos(&pi;x)

between `0` and `2n`. That puts the rectangle boundaries at the even numbers between `0` and `2n`, and for any even integer `x`, `1+cos(&pi;x)` is `1+1`, or `2`. So the left, right, and trapezoidal approximations all look like `n` rectangles of base 2 and height 2, for a total of `4n`. The midpoints of all the boundaries are all the odd numbers between `0` and `2n`, and for any odd integer `x`, `1+cos(&pi;x)` is `1-1`, or `0`. So the midpoint approximation is `n` rectangles of base 2 and height 0, for a total of `0`. All of these approximations are pretty terrible. (The correct value is `2n`.) We have rigged the function so that all the boundary and midpoint values are at extreme values. For fixed `n`, you can always concoct a function that will make any of the approximations look very good or very bad.

By the way, this example shows why Jazon's claim that "the average of the left and right approximations is exact" cannot be correct. The average here is `4n`, which is far from exact.
• For left riemann sums, could you not also have the subscript i=1 be i=0? Seems it would be slightly clearer than having the sum begin at xi-1.
• That's purely a matter of choice in how you use the notation. Some people find it more confusing to begin counting from 0, which means subscripts don't match normal counting (for example, the fourth item is x sub3, not x sub4.
• I don't understand why we call it "x to the n - 1"

Why is it -1? Can someone please explain it to me? I never understood this.
• Because we define i=1 under the sigma sign, but we want to evaluate the rectangle starting at x sub 0 not x sub 1, so we say x sub n-1.
• Under what circumstances does the sum of the trapezoids approach the definite integral of the function? How many partitions are needed?
• The sum of the trapezoids approaches the value of the definite integral of the function as the number of partitions approaches infinity. This is true for any of the Riemann sums and is the basis for the definition of the definite integral.
• Isn't taking the definite integral still more accurate? How are these techniques useful?
(1 vote)
• Those two principles exist for separate reasons. The trapezoidal riemann sum (as well as LRAM, MRAM, and RRAM) are just approximations of area, and as you said, they are all less accurate than a definite integral. However, they do prove the existence of a definite integral because as the number of intervals of a rieman sum increases to infinity (provided that the bounds remain the same), the riemann sums become the integral. By fundamental theorem of calculus, the integral is the same as the antiderivative, and calculating the antiderivative is , again, far more accurate than any riemann sum.

In short, yes the definite integral is more accurate, but any riemann sum help proves the existence of an integral when the intervals increase to infinity.
• What does the x sub i equall or mean