# Can Riemann Sums Calculate Area Enclosed by Function?

• sutupidmath
In summary: Therefore, it is not Riemann integrable.In summary, the Reimann sum can be used to calculate the area enclosed between a part of a function and the Ox axes in the interval [a,b) even if the function is not defined at b. However, for the Reimann integral to exist, the function must be bounded on [a,b) and have a measure 0 set of discontinuities. Otherwise, it is not Riemann integrable.

#### sutupidmath

can we use rienmann sum to calculate the area that is enclosed between a part of a function and the Ox axes in the interval [a,b) if the function is not defined at b ?

Yes. An isolated point cannot change the value of the area under a function.

For that matter neither can a large number or even a countably infinite number of isolated points change the value of an integral. (assuming of course that's there's no nasties like the Dirac Delta "function" involved).

Riemann sum. Not Reinmann. Adn to expand on Uart's post. The Riemann integral is the limit as epsilon tends to zero of the integrals [a,b-epsilon], when it exists.

can we use rienmann sum to calculate the area that is enclosed between a part of a function and the Ox axes in the interval [a,b) if the function is not defined at b ?

Yes you can, if the function is bounded on [a, b). And it is equal to the integral of the function [a,b].

For that matter neither can a large number or even a countably infinite number of isolated points change the value of an integral.

This is not true for the Reimann integral, which is why the Reimann integral is utterly worthless, in favor of Lebesgue.

For example, the function:

f(x) = {0 if x is irrational, 1 if x is rational}

has only countably many discontinuities, but it not reimann integrable

Crosson said:
For example, the function:

f(x) = {0 if x is irrational, 1 if x is rational}

has only countably many discontinuities, but it not reimann integrable

No, I'm pretty sure that function is discontinuous everywhere.

Even if the singularities were only at the rationals (and they aren't, as moo points out) they fail uart's restriciton to isolated singularities.

Any bounded function, as long as the set of discontinuities has measure 0, is Riemann integrable. Any countable set has measure 0. As both Moo of Doom and matt grime said, the function you give is discontinuous everywhere. Its set of discontinuities has measure 1.