Can Riemann Sums Calculate Area Enclosed by Function?

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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 ?
 
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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).
 
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:frown:
 
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:frown:

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