Suppose we try to define the meaning of the Riemann integral $$\int_a^b f(x)\,dx$$ as follows; the above integral exists iff there exists a differentiable function $F$ on the interval $[a,b]$ such that $F' = f$ there, in which case we define $$\int^b_a f(x)\,dx = F(b)-F(a).$$
What goes wrong?
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Nothing goes wrong, except for the fact that the class of functions you can give an integral is not good enough for even elementary purpose. What would you do with $\int_3^5 \lfloor x\rfloor dx$, for example ?
And one other problem : how do you know, for a given function $f:[a,b]\mapsto \mathbb R$, that it is the derivative of a function $F$ ?
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Expanding on Omnomnomnom's comment above, this definition of integral strips much of the geometric intuition. For instance, with Riemann integration, we know that we are essentially computing the area underneath a curve. Not only is this pedagogically nice, but it is also nice for numerical methods: I can approximate an integral arbitrarily well by using upper and lower Riemann sums fairly easily, even if it is impossible to get an exact answer.
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If I understood correctly the question, then the answer is the continuity as I wrote out in comments. The above is definitely true when the function $f$ is continuous everywhere over the interval $[a,b]$ you want to integrate. If you try to prove the Fundamental theorem of calculus you'll see that the continuity of $f$ is crucial to prove your statement, since as you might know a function can have a finite point of discontinuity and still being integrable over a bounded connected subset of $\mathbb{R}$. Wiki's article I think contains enough information to sorted it out on your own https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus .
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The Riemann integral $$\int_B f(x)\>{\rm d}x\ ,$$ i.e., the limit of Riemann sums $$\sum_{k=1}^N f(\xi_k)\>{\rm vol}(B_k)\ ,$$ of a function $f:\>B\to{\mathbb R}^m$ over a domain $B\subset{\mathbb R}^n$ is a fundamental geometrical and physical concept in its own right.
It is only in the case $m=n=1$ (o.k.: $n=1$) that the FTC allows to connect the limit of Riemann sums with so-called primitives of $f$.
There are actually several problems here.
The first one is somewhat trivial, and has been discussed in other answers. There are some Riemann integrable functions that are not derivatives. Simply consider any function defined on some interval $[a,b]$ continuous everywhere except at a single point where it has a jump discontinuity. By Darboux's theorem, such functions are not derivatives. Note that we want these functions to be Riemann integrable, because a single jump in the function should not affect its integrability.
There's other problems as well. You haven't specified boundedness of the derivative. The function $f:[-1,1] \to \mathbb{R}$ given by $f(x) = x^2\sin(x^{-2})$ and $f(0) = 0$ is differentiable on $[-1,1]$, but its derivative is unbounded. In particular, for all $\delta \in (0,1)$, $f'[(-\delta, \delta)] = \mathbb{R}$. Such functions should not be Riemann integrable. There's no sense of 'area'.
But, even if the derivative is bounded, there's some quite ugly derivatives. For example, the derivative of the Volterra function is bounded, but it is discontinuous on a set of positive measure. For Riemann integrability, it's very desirable for the discontinuity set to be of measure zero. The function should not 'jump' too much; otherwise it can't be well-approximated by step functions.