Is integral operator surjective? Is the linear operator ${\rm A}$ given by $${\rm A}f(t) = \int_0^t  f(x)\ dx$$  onto? It would be onto if the range space of ${\rm A}$ is equal to whatever ${\rm A}$ is mapping to, but I am not even sure what ${\rm A}$ is mapping onto...
 A: Perhaps the more relevant question in this context would be, in which domain are we working? If it's the space of infinitely differentiable functions $C^\infty[a,b]$, then the fundamental theorem of calculus gives us a positive answer.
More precisely, the fundamental theorem of Lebesgue integration fully characterize those functions that are given as an integral of another function as those which are absolutely continuous.
However, judging from your tags, you might be looking at a more common space, say the space of (piecewise?) continuous functions on some interval, in which case the answer is negative. Not every continuous function is absolutely continuous: the linked article contains some examples, the best suited for our needs might be $f(x):=x\sin\frac{1}{x}$ over $[-1,1]$, say. To see why there exists no $g\in C[-1,1]$ such that $f(x)=\int_0^x g(t)dt$, show that a function defined by such an integral is absolutely continuous, then use the sequence $(\frac{\pi}{2}+\pi n)^{-1}$ to show $f$ isn't.
A: I Assume your function is from $C[0,1]$ to $C[0,1]$. Now look at set of all functions in $C[0,1]$  which is having a primitive. 
http://en.wikipedia.org/wiki/Primitive_function
