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If this is possible, could somebody give me an example of a function $f:[a,b]\to \Bbb{R}$ continuous in $[a,b]$ but that THERE IS NOT a function, namely G, such that its derivative is the funciont f? i.e., such that $$G'=f.$$

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closed as off-topic by Najib Idrissi, Mark Fantini, Claude Leibovici, Joonas Ilmavirta, Sujaan Kunalan Dec 3 '14 at 9:32

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    $\begingroup$ I don't think that this is possible... en.wikipedia.org/wiki/Fundamental_theorem_of_calculus $\endgroup$ – Ferra Dec 2 '14 at 12:17
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    $\begingroup$ See Volterra's function. $($This comment is addressed at the answers below, rather than at the actual OP$)$. $\endgroup$ – Lucian Dec 2 '14 at 20:11
  • $\begingroup$ @Lucian If I'm reading that article correctly, $V'$ (the derivative of Volterra's function) is discontinuous, meaning that the OP's $f$ cannot be $V'$. $\endgroup$ – jpmc26 Dec 3 '14 at 0:06
  • $\begingroup$ @jpmc26: Might be more useful to consider $f = V$. (Some shifting and scaling may be required if $[a,b]$ is "boring".) $\endgroup$ – Eric Towers Dec 3 '14 at 0:10
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This isn't possible: Every such function is a derivative. For any continuous function $f: [a, b] \to \mathbb{R}$, we can define the function $$G(x) := \int_a^x f(t) \,dt,$$ and by the Fundamental Theorem of Calculus, $$G'(x) = f(x).$$ (This assumes that we define the derivative of a function at the endpoint of an interval to be the value of the appropriate one-sided limit.)

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It's impossible. Since $f$ is a continuous function, we can define $$G(x)=\int_a^xf(t)dt$$ and the fundamental theorem of calculus says that $G$ is differentiable and $G'=f$.

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Mathematically, the antiderivative of every such $f$ exists, as in the integral expressions seen in other answers.

Practically speaking, though, most examples of $f$ have no closed-form expression for the antiderivative in terms of elementary functions. (This statement can be made precise with Galois Theory)

One such case is given on the Wikipedia page. Take $$ f(x) = e^{-x^2} $$ When we need a value for $\int_0^tf(x)dx$, we have to do some approximating (Abramowitz and Stegun).

In practice, almost any "tricky" function you make up is unlikely to have a closed-form expression for its integral. I'll make one up now, and I'm willing to bet there is no closed-form expression: $$ f(x)=\sqrt{\tan(\log(x))} $$

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    $\begingroup$ Good job answering what might be the actual question. I believe the usual term for the functions you are thinking of is elementary functions. Simple functions are something different. $\endgroup$ – Matthew Leingang Dec 2 '14 at 17:16
  • $\begingroup$ Good catch, Matthew, and so edited. $\endgroup$ – Brian B Dec 3 '14 at 1:07
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Stated very simply:

A function $f$ that is continuous is (Riemann) integrable. Therefore, it's integral will be a function whose derivative is $f$.

Sometimes, however, the integral is nonelementary! We may not be able to write it down in closed-form.

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I'm not pretty sure but this could be what you are looking for:

http://en.wikipedia.org/wiki/Weierstrass_function

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    $\begingroup$ This is an example of a function that is continuous everywhere and differentiable nowhere, not a function that doesn't have an antiderivative. $\endgroup$ – user155385 Dec 3 '14 at 7:47

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