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.

### Course: Integral Calculus>Unit 1

Lesson 5: Defining integrals with Riemann sums

# 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.

## Want to join the conversation?

• Has definite integrals something to do with derivatives? Or its completely different thing?
• 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
• Why wouldn't people use circles to approximate areas under curves?
• 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.
• 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?
• 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.
• Apart from the rectangular bars, there is some area left under the curve. What about that area?
• 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.
• Wouldn't an infinitely small `Δx` be notated with `δx` or just `δ`?
• The Leibniz notation is dx.
Why? - years on convention.
The x must remain since we need to know with respect to what variable the function is being integrated. This is more obvious once you get into multi variable calculus.
• Can we replace n appraoch infinity by delta x approch zero? If not then why so?
• 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.
• 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.
• 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.
• 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)
• Well, if both ideas were called ∆x, how would you differentiate between the "infinitesimally small width" ∆x and "relatively large width" ∆x? To remove this ambiguity, we have ∆x and dx.
• 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)?
• 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).