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True or false that all real function is an antiderivative

  1. Dec 20, 2003 #1
    is it true or false that all real function is an antiderivative of some real function but neither real function have an antiderivative?

    I still have the doubt!

    Definition(Louis Leithold,The Calculus with Analytic Geometry)
    Antiderivative: F is antiderivative of f in I if F'(x)=f(x) for all x in I.

    The question is all f have some F in some I?
    The question is all f is a G of some g in some I?
    (Real Analysis)
     
    Last edited: Dec 22, 2003
  2. jcsd
  3. Dec 20, 2003 #2

    Hurkyl

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    By "primitive" do you mean that if [itex]f(x) = \int g(x) \, dx[/itex] then [itex]f(x)[/itex] is a primitive of [itex]g(x)[/itex]? (The usual English word for this is that [itex]f(x)[/itex] is an antiderivative or an integral of [itex]g(x)[/itex])


    I'm not entirely sure what you're trying to ask... though it is false that any function is an antiderivative of another function.
     
  4. Dec 20, 2003 #3

    HallsofIvy

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    It's not clear what you mean by "either" function. If you mean the original function in the question and its anti-derivative, then obviously IF "every function had a anti-derivative", then it wouldn't make sense to say that THAT function did NOT have an anti-derivative.
    However, as Hurkyl pointed out, it is not true that every function has a primitive (anti-derivative). For example, the function, f(x)= 1 if x is rational, 0 if x is irrational, does not have an antiderivative.
    It IS true that every bounded function whose points of discontinuity form a set of measure 0 is integrable (has an anti-derivative). In particular every continuous function has an anti-derivative as well as every bounded function with only a finite number of points of discontinuity.
     
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