The anti definite integral Is there such a possibility as defining an anti (definite integral)?
For instance:
$$
S(a,b)=F(b)-F(a)=\int_a^b f(x)dx
$$
The question is what operation $D$ on $S$ produces $f(x)$:
$$
D[S(a,b)]=f(x)
$$

For the simple case of $f(x)=x$, it is pretty obvious that, say $D[b^2-a^2]=x$ because 
$$
\int_a^bxdx=b^2-a^2 \implies D[b^2-a^2]=x+C
$$
where $C$ is a constant.
So at the minimum, absent a general procedure, one can map out manually a few cases.
But, how far can one take this; is there a special class of functions for which the anti (definite-integral) can be "well-defined"?
 A: Note that $S(a,b)$ is a number as is $b^2 - a^2$.  (All definite integrals that exist are numbers.)  This will be a significant problem.  Here are two definite integrals on the same interval with the same value (hence, identical $S(a,b)$).  \begin{align*}
0 &= \int_{-1}^1 0 \,\mathrm{d}x  \\
0 &= \int_{-1}^1 x \,\mathrm{d}x
\end{align*}
So, is $D[0]$ supposed to be $0$ or $x$?
There is also a definitional problem.  From "$D[S(a,b)]$", we should recognize that we get a different $D$ for every choice of interval, but that choice of interval is not present in our notation for $D$.  This is fixable, write "$D_{a,b}$".  Then we have
$$  D_{-1,1}[0] = 0  $$
and 
$$  D_{-1,1}[0] = x  \text{.}  $$
So there is still an unresolvable ambiguity.  (If that ambiguity were magically resolved, we could then approach the interesting question of what to do with $D_{a,b}[S(c,d)]$ where the only given relations among $a$,$b$,$c$, and $d$ are $a \leq b$ and $c \leq d$.  No idea where that would go since we haven't magically resolved the ambiguity.)
There's a sneaky trick hiding in your example.  The integral on $[a,b]$ of $x$ is the number $b^2 - a^2$.  But rather than work with a definite integral, we were instead to consider the accumulation function 
$$  f(x) = \int_a^x t \,\mathrm{d}t  $$
where $a$ is a constant and $x$ is a variable.  This isn't a number; it's the function $(1/2)(x^2 - a^2)$.  (You forgot the one-half in your version.)  It is feasible to get the integrand back from this accumulation function, just differentiate with respect to $x$.  Notice that the choice of $a$ only vertically shifts the graph, so corresponds to adding a constant vertical offset, which is discarded under differentiation.  So the choice of $a$ does not alter the result of differentiation.
The idea that you can differentiate an accumulation function to retrieve its integrand is (the first half of) the fundamental theorem of calculus.
