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Definite integral as the limit of a Riemann sum

Definite integrals represent the exact area under a given curve, and Riemann sums are used to approximate those areas. However, if we take Riemann sums with infinite rectangles of infinitely small width (using limits), we get the exact area, i.e. the definite integral! Created by Sal Khan.

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  • male robot johnny style avatar for user Lasha Jibghashvili
    Has definite integrals something to do with derivatives? Or its completely different thing?
    (31 votes)
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    • leaf green style avatar for user janismac
      integrals and derivatives are complementary, they undo each other.
      if you take the indefinite integral of any function, and then take the derivative of the result, you'll get back to your original function.

      In a definite integral you just take the indefinite integral and plug some intervall (left and right boundary), and get a number out, that represents the area under the function curve.

      Important distinction:
      an indefinite integral gives you a function
      a definite integral is a calculation that gives you a numeric result
      (129 votes)
  • piceratops ultimate style avatar for user sahsan
    Why wouldn't people use circles to approximate areas under curves?
    (15 votes)
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    • piceratops ultimate style avatar for user kitwaal
      Technically you could, assuming that you'd be using circles small enough to meet your requirements for accuracy. Think about this in three dimensions- you can fill a volume with spheres or balls and then summing up the volume of the spheres. Circles don't fit together very neatly, and this is actually pretty mathematically hard to figure out if you wanted to be rigorous about it, but give it a try both ways. In any case, if you can use circles to approximate the area under a curve you'll pretty much know this subject well enough to do anything.
      (40 votes)
  • mr pink red style avatar for user Mr. Johnson
    Let's say you have a function y=x^2 and you want to find the area under the curve along the interval -3 to 3 and you're delta x was 6 (or any finite number). Could you get not just an approximation, but the exact area under the curve if you took the left side approximation and added it to the right side approximation and then divided that sum by 2? In other words, would the amount that you overestimated the area under the curve be exactly the same amount that you underestimated the area under the curve? Would the value of the overestimation (in either the left-side approximation, or right-side approximation) be equal to the value of the underestimation? My guess is that if the function has a vertical line of symmetry AND the interval along the x-axis extends an equal distance away from the line of symmetry in both directions (in our example -3 to 3 with the line of symmetry being the y-axis) that the amount of overestimation and underestimation would be the same and when you took the average of the two amounts you would be left with the exact value of the area under the curve. Thoughts?
    (6 votes)
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    • male robot hal style avatar for user Ben Willetts
      The quick answer is no, that wouldn't work. :-)

      You are correct that it would work for a very limited number of cases on a very limited number of functions, but it wouldn't actually work for the example you give, because the function is curved. That means the average of the right approximation and the left approximation is NOT the same as the actual value of the area -- we can see this with the function x^2 by averaging the value at 0 (which is 0) and the value at 1 (which is 1). The average of 1/2 is not the same as the area under the function, which is 1/3.

      Because the number of cases this would work for is so limited, it's easier just to use integration for everything -- and this approach has the advantage that it always works.
      (10 votes)
  • duskpin ultimate style avatar for user Prakash Kalluri
    Apart from the rectangular bars, there is some area left under the curve. What about that area?
    (6 votes)
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    • male robot hal style avatar for user Jesse
      If you use finitely many Riemann rectangles, you have an approximation to the area under the curve. It is only through taking a limit as the number of rectangles increases ad infinitum (and their width shrinks to zero) that you will obtain the exact area under the curve.
      (9 votes)
  • duskpin ultimate style avatar for user #1 βooκs ρroρonεnτ
    Wouldn't an infinitely small Δx be notated with δx or just δ?
    (3 votes)
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  • leaf green style avatar for user Ishan
    Can we replace n appraoch infinity by delta x approch zero? If not then why so?
    (5 votes)
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    • leaf green style avatar for user rathisushant5
      Remember that delta x and n are related to each other.(delta x=(b-a)/n).So if we approach n to infinity, delta x will approach 0(a very small value-not 0) or if we do the other way round, that is approach delta x to 0, n will approach infinity (because n=(b-a)/delta x).In both cases we will get the same result.
      (3 votes)
  • male robot hal style avatar for user Yash
    What is the correct way of saying that delta x becomes extremely small (approaches 0) in Calculus context - delta x becomes infinitely small or delta x becomes infinitesimally small? I know it's a trivial thing, but I wish to use the correct terminology.
    (3 votes)
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    • leaf green style avatar for user kubleeka
      It's neither. We allow Δx to become arbitrarily small. That is, we can take Δx as close to 0 as you wish. The term 'infinitesimal' can help guide intuition, but it's a defining property of the real numbers that nothing is infinitesimal.
      (8 votes)
  • stelly yellow style avatar for user 24tinat
    This sounds like a weird question but why isn't dx just delta x? Delta x is the width, and dx is also the width, just infinitely smaller. So why can't be just write dx as delta x?
    (1 vote)
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  • blobby green style avatar for user crispinsea
    How did Bernard Riemann get all the properties of the integral from the Riemann sum? For example how did he know the definite integral from a to b of f(x) is F(b)-F(a). from the sum because it is infinite. Or if you take the derivative of the indefinite integral you get the integrand (how do you take the derivative of a limit and sigma notation)?
    (4 votes)
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    • leaf grey style avatar for user Qeeko
      Those are some very good questions. Any good book on calculus or one on elementary real analysis treating the Riemann integral should answer your questions (the details are too lengthy for me to type up here).

      Let me point out two subtle facts. Firstly, a function may possess an anti-derivatve, yet fail to be (Riemann) integrable. This fact is often overlooked, especially at the elementary level. What is more, even if ƒ is an integrable function on [a, b], and we define the function F on [a, b] by

      F(x) = ∫ [a, x] ƒ(t) dt,

      the integral going from a to x, F need not be differentiable. In other words, it need not be the case that F'(x) = ƒ(x) for all x - one can show that F is differentiable at x if and only if ƒ is continuous at x. Moreover, one can show that the set of points in [a, b] at which ƒ is discontinuous has measure zero, so F is differentiable almost everywhere (in a technical sense).
      (2 votes)
  • piceratops tree style avatar for user lions 2013!!
    is it possible to use Simpson's rule even when n is not an even number?
    (4 votes)
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Video transcript

We've done several videos already where we're approximating the area under a curve by breaking up that area into rectangles and then finding the sum of the areas of those rectangles as an approximation. And this was actually the first example that we looked at where each of the rectangles had an equal width. So we equally partitioned the interval between our two boundaries between a and b. And the height of the rectangle was the function evaluated at the left endpoint of each rectangle. And we wanted to generalize it and write it in sigma notation. It looked something like this. And this was one case. Later on, we looked at a situation where you define the height by the function value at the right endpoint or at the midpoint. And then we even constructed trapezoids. And these are all particular instances of Riemann sums. So this right over here is a Riemann sum. And when people talk about Riemann sums, they're talking about the more general notion. You don't have to just do it this way. You could use trapezoids. You don't even have to have equally-spaced partitions. I used equally-spaced partitions because it made things a little bit conceptually simpler. And this right here is a picture of the person that Riemann sums was named after. This is Bernhard Riemann. And he made many contributions to mathematics. But what he is most known for, at least if you're taking a first-year calculus course, is the Riemann sum. And how this is used to define the Riemann integral. Both Newton and Leibniz had come up with the idea of the integral when they had formulated calculus, but the Riemann integral is kind of the most mainstream formal, or I would say rigorous, definition of what an integral is. So as you could imagine, this is one instance of a Riemann sum. We have n right over here. The larger n is, the better an approximation it's going to be. So his definition of an integral, which is the actual area under the curve, or his definition of a definite integral, which is the actual area under a curve between a and b is to take this Riemann sum, it doesn't have to be this one, take any Riemann sum, and take the limit as n approaches infinity. So just to be clear, what's happening when n approaches infinity? Let me draw another diagram here. So let's say that's my y-axis. This is my x-axis. This is my function. As n approaches infinity-- so this is a, this is b-- you're just going to have a ton of rectangles. You're just going to get a ton of rectangles over there. And there are going to become better and better approximations for the actual area. And the actual area under the curve is denoted by the integral from a to b of f of x times dx. And you see where this is coming from or how these notations are close. Or at least in my brain, how they're connected. Delta x was the width for each of these sections. This right here is delta x. So that is a delta x. This is another delta x. This is another delta x. A reasonable way to conceptualize what dx is, or what a differential is, is what delta x approaches, if it becomes infinitely small. So you can conceptualize this, and it's not a very rigorous way of thinking about it, is an infinitely small-- but not 0-- infinitely small delta x, is one way that you can conceptualize this. So once again, as you have your function times a little small change in delta x. And you are summing, although you're summing an infinite number of these things, from a to b. So I'm going to leave you there just so that you see the connection. You know the name for these things. And once again, this one over here, this isn't the only Riemann sum. In fact, this is often called the left Riemann sum if you're using it with rectangles. You can do a right Riemann sum. You could use the midpoint. You could use a trapezoid. But if you take the limit of any of those Riemann sums, as n approaches infinity, then that you get as a Riemann definition of the integral. Now so far, we haven't talked about how to actually evaluate this thing. This is just a definition right now. And for that we will do in future videos.