Can Riemann Sums Calculate Area Enclosed by Function?

  • Thread starter Thread starter sutupidmath
  • Start date Start date
  • Tags Tags
    Sum
Click For Summary
Riemann sums can be used to calculate the area enclosed by a function and the x-axis over the interval [a,b) even if the function is not defined at b, provided the function is bounded on [a, b). Isolated points do not affect the value of the integral, as neither a finite nor countably infinite number of such points can change the area under the curve. However, the Riemann integral is criticized for its limitations, particularly with functions that are discontinuous everywhere, such as those defined by rational and irrational values. In contrast, a function with a countable set of discontinuities can still be Riemann integrable if its discontinuities have measure zero. Ultimately, discussions highlight the preference for Lebesgue integration over Riemann integration due to these limitations.
sutupidmath
Messages
1,629
Reaction score
4
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 ?
 
Mathematics news on Phys.org
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: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.
 
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.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 0 ·
Replies
0
Views
3K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K