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- Thread starter andyrk
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Because if [itex]f[/itex] isn't continuous, then [itex]F[/itex] might not be differentiable.

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HallsofIvy

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Right. That helped alot. Thanks. :)premisesare false! There is NO requirement, in the Fundamental Theorem of Calculus (the part that say "if [itex]F(x)= \int_a^x f(t)dt[/itex] then F'(x)= f(x)") that f be continuous. Itmightthat your text book is proving it with theaddedassumption that f is continuous because then the proof is easier. But it can then be easily extended to functions that are not continuous.

So coming back to it one more time, we say that f is continuous just to make the proof and easy and simple enough to be understood, but it (the proof) can also have f which is not continuous and still be able to explain the proof well enough (with just a little bit of complexity perhaps). But we don't look at that case since it makes our life difficult. Is that correct?

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lavinia

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If a function is continuous then its anti-derivative can be found by Riemann integration.

If it is discontinuous then it may not have an anti-derivative or if it does it may not be computable from Riemann integration. However there are functions that are discontinuous whose anti-derivative exists and can be computed from Riemann integration.

The assumption of continuity removes many complexities. That is why continuity is often assumed in the statement of the Fundamental Theorem of calculus.

On the other hand, it is perfectly true that the fundamental theorem works for a more general set of functions though I have no idea what that set is exactly.

Here are some illustrative examples.

- The function on the unit interval that is zero from 0 to 1/2 and 1 from 1/2 to 1. It is discontinuous at 1/2 and its Riemann integral is not differentiable at 1/2.

- The function equal 1 on the unit interval except at 1/2 where it is zero. Its Riemann integral from 0 to x is F(x) = x. But F'(x) does not equal the original function since it is equal to 1 everywhere.

- The function that is 0 on the Cantor set and 1 on its complement. This function has uncountably many discontinuities but is still Riemann integrable. Again the Riemann integral from zero to x is F(x) = x. However, the function that is 1 on the complement of a Cantor set of positive measure is not Riemann integrable.

- The classic function that is ##0## at zero and ##2xsin(1/x) - cos(1/x)## away from zero. It is discontinuous at zero. It has an anti-derivative that can be computed from Riemann integration.

- Wikipedia points out that the function that is 0 at zero and ##2xsin(1/x^2) - 1/xcos(1/x^2)## has anti-derivative ## x^2sin(1/x^2) ## away from 0 and zero at 0 but it can not be found by Riemann integration.

- The function that is zero on the rationals and 1 on the irrationals. It can not be Riemann integrated and it has no anti-derivative

If it is discontinuous then it may not have an anti-derivative or if it does it may not be computable from Riemann integration. However there are functions that are discontinuous whose anti-derivative exists and can be computed from Riemann integration.

The assumption of continuity removes many complexities. That is why continuity is often assumed in the statement of the Fundamental Theorem of calculus.

On the other hand, it is perfectly true that the fundamental theorem works for a more general set of functions though I have no idea what that set is exactly.

Here are some illustrative examples.

- The function on the unit interval that is zero from 0 to 1/2 and 1 from 1/2 to 1. It is discontinuous at 1/2 and its Riemann integral is not differentiable at 1/2.

- The function equal 1 on the unit interval except at 1/2 where it is zero. Its Riemann integral from 0 to x is F(x) = x. But F'(x) does not equal the original function since it is equal to 1 everywhere.

- The function that is 0 on the Cantor set and 1 on its complement. This function has uncountably many discontinuities but is still Riemann integrable. Again the Riemann integral from zero to x is F(x) = x. However, the function that is 1 on the complement of a Cantor set of positive measure is not Riemann integrable.

- The classic function that is ##0## at zero and ##2xsin(1/x) - cos(1/x)## away from zero. It is discontinuous at zero. It has an anti-derivative that can be computed from Riemann integration.

- Wikipedia points out that the function that is 0 at zero and ##2xsin(1/x^2) - 1/xcos(1/x^2)## has anti-derivative ## x^2sin(1/x^2) ## away from 0 and zero at 0 but it can not be found by Riemann integration.

- The function that is zero on the rationals and 1 on the irrationals. It can not be Riemann integrated and it has no anti-derivative

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