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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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• 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? • Wouldn't an infinitely small `Δx` be notated with `δx` or just `δ`? • Can we replace n appraoch infinity by delta x approch zero? If not then why so? • 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. • 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).
• is it possible to use Simpson's rule even when n is not an even number? • I still just don't get how Riemann went from having the sigma notation to translating it to a definite integral? How did he know that the process of definitely integrating a function (integrating using the formulas and then replacing the bounds in the integral and subtracting them) would lead to the same answer? This question is really getting on my nerves because nowhere online have I found an explanation. • Hi Christina
You have two questions here, one of notation, how did we go from Σ to ∫, and the other of how did definite integral come about.

Notation
If you have made it to integral calculus, you must have come through algebra and differential calculus, and if so, you have already seen a change in notation, so I’ll start there.

When you were in algebra you calculated the slope of a line as Δy/Δx. When you got to differential calculus the problem was not the slope of a line, but the slope of a curve at some point. The concept was developed by using a secant line, or average slope. If we wanted to estimate the slope at a point c, we drew a line between two points on either side of c that were – Δy/Δx and Δy/Δx from c. The we asked, what happens as Δy/Δx gets smaller and smaller, that is, the endpoints of the secant line get closer and closer to c. We decided that once the Δ of the Δy/Δx was infinitesimally small, we would change the notation from Δy/Δx to dy/dx, to remind us that we are dealing with infinitesimals (also called differentials). So for the slope of a straight line or secant line we used Δy/Δx but for the slope of a tangent line we use dy/dx.

It is pretty much the same deal on how we went from Σ to ∫. The Riemann sum is a sum of sections whose width is Δx, so we have, in general, Σf(x)Δx. As we make Δx smaller and smaller, until it is infinitesimal, we again change the notation from Δx to dx AND we change the notation of Σ to ∫, that is Σf(x)Δx to ∫f(x)dx. It really is just sort of a visual reminder that we are dealing with infinitesimal changes in x, that is, dx. In case you didn’t know, the integral symbol ∫ is just an elongated S, which stand for sum, so yes, the Riemann sum is the same as the Riemann integral, the only difference is the Δx is infinitesimally small.

The Definite Integral
As far as how the definite integral came about, that happened way before Riemann. The 2 theorems are called the Fundamental Theorems of Calculus. They are talked about here on Khan. This Wikipedia page has proofs of them that do not require math skills above what you should have by now - it will clearly show how the definite integral was "discovered". Here are a few links to get you going:
https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
http://mathforum.org/library/drmath/view/53374.html

I sure hope that helped. If not send me a comment on what you need clarified.
Stefen