A function that is not a derivative 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
$$G'=f.$$
 A: 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 $G'=f$.
A: 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:
$$
f(x)=\sqrt{\tan(\log(x))}
$$
A: 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.)
A: 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.
A: I'm not pretty sure but this could be what you are looking for:
http://en.wikipedia.org/wiki/Weierstrass_function
