Can Riemann Sums Calculate Area Enclosed by Function?

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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.
  • #1
sutupidmath
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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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  • #2
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).
 
  • #3
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.
 
  • #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 ?

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:
 
  • #5
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.
 
  • #6
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.
 
  • #7
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.
 

FAQ: Can Riemann Sums Calculate Area Enclosed by Function?

What is a Riemann sum?

A Riemann sum is a method for approximating the area under a curve by dividing it into smaller, simpler shapes and summing their areas.

How do Riemann sums work?

Riemann sums use rectangles to approximate the area under a curve. The width of the rectangles is determined by the chosen partition, which divides the interval into smaller subintervals. The height of each rectangle is determined by evaluating the function at a specific point within each subinterval (usually either the left or right endpoint). The areas of all the rectangles are then added together to give an approximate value for the total area under the curve.

Can Riemann sums calculate the area enclosed by any function?

Yes, Riemann sums can be used to approximate the area under any continuous function on a closed interval. However, the accuracy of the approximation will depend on the number and size of the rectangles used.

What is the difference between left, right, and midpoint Riemann sums?

Left Riemann sums use the left endpoint of each subinterval to determine the height of the rectangles, while right Riemann sums use the right endpoint. Midpoint Riemann sums use the midpoint between the left and right endpoints. These different methods can give slightly different approximations for the area under a curve, with midpoint Riemann sums typically being the most accurate.

Are there limitations to using Riemann sums for calculating area?

While Riemann sums can be used to approximate the area under any continuous function, they may not be the most efficient or accurate method for more complex functions. Additionally, the accuracy of the approximation is highly dependent on the partition and number of rectangles used, so careful consideration must be given when choosing these parameters.

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