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Fundamental Theorem of Calculus

  1. Mar 3, 2015 #1
    In the fundamental theorem of calculus, why does f(x) have to be continuous in [a,b] for [itex] F(x) = \int_a^x f(x) dx [/itex]?
  2. jcsd
  3. Mar 3, 2015 #2
    Because if [itex]f[/itex] isn't continuous, then [itex]F[/itex] might not be differentiable.
  4. Mar 3, 2015 #3
    But why? By definition, F(x) is the antiderivative of f(x), that means F'(x) = f(x). So if f(x) is not continuous in [a,b] then that means that F'(x) is not continuous in [a,b]. But if F'(x) is not continuous that does not mean that F(x) is not differentiable in [a,b] because of continuity of F'(x) is not at all related to differentiability of F(x) and infact its the other way round. Continuity of F(x) is a good measure of differentiability of F(x) since for F'(x) to exist in [a,b] F(x) needs to be continuous in [a,b]. So why does the question of f(x) being continuous in [a,b] still arise?
  5. Mar 3, 2015 #4
    in the above you shown equation it indicates the area of the graph in between a and b for the function f(x). If the grarh is not continuous then there is no meaning of area.i.e. the reason that the function f(x) must be continuous
  6. Mar 3, 2015 #5
    Yes, but we can still find the area under the curve of a function even though the graph maybe discontinuous. Can't we?
  7. Mar 4, 2015 #6


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    It's hard to answer a question in which the premises are 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. It might that your text book is proving it with the added assumption that f is continuous because then the proof is easier. But it can then be easily extended to functions that are not continuous.
  8. Mar 4, 2015 #7
    Right. That helped alot. Thanks. :)
    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?
  9. Mar 4, 2015 #8
    No. I just researched a lot about it. It has to be continuous otherwise the squeeze theorem (which is the basis of the Fundamental Theorem of Calculus) won't be valid.
  10. Mar 4, 2015 #9


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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
    Last edited: Mar 4, 2015
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