# true or false that all real function is an antiderivative

by kallazans
Tags: antiderivative, doubt, function, real
 P: 5 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)
 Emeritus Sci Advisor PF Gold P: 16,101 By "primitive" do you mean that if $f(x) = \int g(x) \, dx$ then $f(x)$ is a primitive of $g(x)$? (The usual English word for this is that $f(x)$ is an antiderivative or an integral of $g(x)$) I'm not entirely sure what you're trying to ask... though it is false that any function is an antiderivative of another function.
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,896 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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