If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

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.

Want to join the conversation?

  • purple pi purple style avatar for user Nick Proulx
    Isn't the formula for estimating area using the midpoint, and for using trapezoids the same?
    (44 votes)
    Default Khan Academy avatar avatar for user
  • stelly blue style avatar for user Julian Delgadillo Marin
    Which of those is the most accurate?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • leaf blue style avatar for user Stefen
      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.
      (16 votes)
  • leaf yellow style avatar for user Narek Kazarian
    Which method gives you the most accurate approximation?
    (10 votes)
    Default Khan Academy avatar avatar for user
    • old spice man green style avatar for user jpl
      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(πx)

      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(πx) 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(πx) 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.
      (6 votes)
  • leaf green style avatar for user Darcy
    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.
    (4 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Creeksider
      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.
      (4 votes)
  • leaf yellow style avatar for user Narek Kazarian
    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.
    (2 votes)
    Default Khan Academy avatar avatar for user
  • orange juice squid orange style avatar for user jonah.yoshida
    Under what circumstances does the sum of the trapezoids approach the definite integral of the function? How many partitions are needed?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • leaf blue style avatar for user Stefen
      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.
      (2 votes)
  • blobby green style avatar for user Eric Cunningham
    Isn't taking the definite integral still more accurate? How are these techniques useful?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • hopper cool style avatar for user Thomas Nguyen
      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.
      (5 votes)
  • female robot grace style avatar for user Strings to Eternity
    Hey what is we use the left boundary uptil half of the graph ie uptill (Xn /2) and the right boundary after this to (Xn) wouldn't we get a way better approximation ?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user nithinjainson
    What is the mathematical difference between the Rectangular rule and Trapezoidal Rule?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • leaf blue style avatar for user Stefen
      The video is clear about the physical difference. The mathematical difference is that the trapezoidal approximation usually results in an estimate of area that is closer to the actual value of the area.
      (2 votes)
  • male robot hal style avatar for user war machine 13
    why is the green n block slanted and the other ones are not
    (1 vote)
    Default Khan Academy avatar avatar for user
    • orange juice squid orange style avatar for user True Mark
      I assume you are talking about the fourth graph/diagram in bottom right corner. In that particular graph Sal is estimating the area by using trapezoids rather than rectangles. They are 4-sided shapes with a pair of parallel opposite sides.

      The "slant" is dictated by the average slope of the function between the two points that your dealing with (eg- x0 and x1). The n-"block" is more slanted because the function (if you look) has a steeper slope at that interval. Essesntially you're following the shape of the graph, (except you're using straight lines instead of curves)...hope that made sense
      (2 votes)

Video transcript

In the last few videos, we've been approximating the area under the curve using rectangles, where the height of each rectangle was defined by the function evaluated at the left boundary. So this would have been the first rectangle. Then the second rectangle would look something like this. And then we'd go all the way to the nth rectangle would look something like that. And we saw-- so this is the first rectangle, this is the second rectangle, and we'd go all the way to the nth rectangle-- and so we saw that the way that you would take the sum of all of these rectangles in order to approximate the area is that you would get the sum from i equals 1 to n. And so i is essentially a count of which rectangle we're dealing with. And what we're going to do is multiply the height times the base. So the height of each rectangle, the height of rectangle one, in this case, was the function evaluated at x0. The height of rectangle two was the function evaluated at x1. The height of rectangle n was the function evaluated at x sub n minus 1. So the height of a rectangle i is going to be the function evaluated x sub i minus 1. If i is 2, then we're evaluating it at x1. If i was 2, then this would be the function evaluated at x1. So it's the left boundary. And then we have to multiply it times the width. And in the last few videos, and in this video, we will assume that all of the rectangles have equal width. Now, we'll call that equal width delta x. And to find it, we just have to take the total distance that we're going in the x direction. So it's going to be b minus a divided by the number of rectangles we want. So it's going to be times delta x. Now, you might imagine that this is not the only way to take the sum using rectangles, or this is not the only way to take the sum or approximate the area using some type of geometric shape. For example, we could have created rectangles where the height is defined by the rightmost boundary. So let's define that. So here's our first rectangle. And we're defining the height by the right boundary of the rectangle. So this right over here is rectangle one, and its height is f of x1. And then for this one right over here, we take the right boundary. The right boundary defines that height. If we go all the way to the-- this is rectangle two-- if we go all way to the nth one, we use the right boundary to define the height of the rectangle. So in this case-- this is the nth rectangle-- how would we write this sum? Well, it would be the sum-- which is, remember, we're just trying to approximate the area under the curve-- from i is equal to 1 to n. So i is a count of each of the rectangles. And so the height of the first rectangle is f of x1. The height of the nth rectangle is f of x sub n. So this height right over here is f of x sub n. So the height of the i-th rectangle is going to be f of x sub i. Whatever the rectangle number is, we take the x sub that same number and evaluate the function there. That gives us the height. And we multiply that times delta x. So the difference between this and this here, for the i-th rectangle, we use x sub i minus 1, so the left boundary. Here we use the right boundary, f of x sub i. Well, we don't have to stop there. Instead, we could use maybe the midpoint between the two boundaries instead. So, for example, over here we could we could use the midpoint between x0 and x1 to find the height of the rectangle. So this is right over here. This is f of x0 plus x1 over 2, just the midpoint between these two points, to define the height of the rectangle. So it would look something like that. And the next one, we would look at the midpoint to define the height. And we go all the way to the nth one, and we define the midpoint between its two sides of the rectangle. So the function evaluated there tells us how high our rectangle should be. And it would look something like that. And so what would this sum look like? Well, once again, we would count each of our rectangles, so i equals 1 to i equals n. i is which rectangle we're working on. So this is the first one, this is the second one, and this is the nth one. And the height isn't just going to be f evaluated x sub i minus 1 or f evaluated x sub i. It'll be the function evaluated at the midpoint between the two-- x sub i minus 1 plus x sub i, all of that over two, and then times delta x. The delta x's are the same in every one of these scenarios. Now finally, let's try to break out of approximating only with rectangles and get a little bit more creative. Why don't we try to approximate with trapezoids? So let's try to do that. So what we could have here is the left part of the trapezoid. The height is f of x sub 0. So this is f of x sub 0. And then the right side of the trapezoid is f of x sub 1. And then what would be the-- here, and let me do that for all of them. So that would be the first trapezoid. Then the second trapezoid would look like this. This one looks almost like a rectangle, but we assume that the top isn't completely flat. And then we go all the way to the nth one. And this should be clear that we're dealing with a trapezoid. All the way to the nth one will look something like that. So how would we calculate this area, the area of the trapezoid? Well, you just have to remember that the area of a trapezoid is just the average of the heights of the two sides times the base. So in this case-- and let me write it out a little bit. So the area right over there is going to be the average of the heights. So it's going to be f of x sub 0 plus f of x sub 1, all of that over 2. And then we're going to multiply that times delta x. So that would be the area just of this one right over here. We took the average of the two heights and multiplied that times the base. Now, if we wanted the sum of the areas of all these trapezoids, and we wanted to write in general terms, we could just write it's the sum-- once again, we're going to count the trapezoids. So this is the first trapezoid, this is the second, all the way to the nth trapezoid. So it's i equals 1 to i equals n. And the height of each trapezoid, we're going to use the function evaluated at the left boundary, x sub i minus 1, the average of the function evaluated at the left boundary and the function evaluated at the right boundary, and we're going to take the average of that and then multiply that times the base. So the whole reason why I wanted to do this is to show you there's multiple, multiple ways of doing this. And in fact, if you wanted to get really general, you could even have different widths. But then that gets a little bit more confusing. But really, just to show you that you might see some of this fancy notation in your calculus book or in your precalculus book. But all it's doing is summing up the areas of trapezoids and rectangles depending on whether they're using the right boundaries of the rectangle to define the height, the left boundaries, the midpoint of the left and the right boundaries, or they could even construct trapezoids.