Is this famous identity for indefinite integrals axiom or theorem? Let say that $A(x)$ describes area under $f(x)$ in the interval $[0, x]$ ($x$ varies). Then, if we sum up all infinitesimally small rectangles each of area $f(x)dx$ for all $x$-values in the domain of $f(x)$ we will get exact area under $f(x)$ from $0$ to $x$.We write this process as $ \int_{0}^{x}f(x)dx = A(x).$
But, we can also easly prove that instant velocity of $A(x)$  for input $x$ is of course $f(x)$ (that is instant change in the added area, $f(x)dx,$ over $dx$ which is indeed $f(x)$). So, $\frac{d}{dx}A(x) = f(x).$
This was for some interval, but if we want $\int_{-\infty}^{\infty}f(x)dx$ for whole number line, my intuition tells me that integral would be some function, not a number (except if x-axis is asymptote to both directions of our function, for example $e^{-x^2}$).
Now, my question.
How can we prove that $$\int_{-\infty}^{\infty}f(x)dx = A(x) + C?$$
I understand that $\frac{d}{dx} [A(x) + C] = \frac{d}{dx}A(x)$ but I can't see how can previous proven facts in calculus prove this. So, is this the definition, axiom or theorem which can be proven? Thanks
 A: From the comments on the question, it appears you are asking for an explanation of why $\int f(x) dx= A(x)+C$. This isn't the same was what appears in the question itself, but I can't make sense of the equation $\int_{-\infty}^{\infty} dx= A(x)+C$. So my answer addresses the interpretation of your question based on the comments.
Define $A(x)=\int_0^{x} f(t) dt$. Then (part of) the Fundamental Theorem of Calculus states that $\frac{d}{dx} A(x) = f(x)$. Now, the notatation $\int f(x) dx$ denotes the family of antiderivatives of $f(x)$. We've just seen from FTC that $A(x)$ is an antiderivative of $f(x)$. If $F(x)$ is another antiderivative, then
$$
\frac{d}{dx}(A(x)-F(x))= f(x)-f(x)=0
$$
It follows from this that $A(x)-F(x)$ is a constant function (by the Mean Value Theorem for example). So $F(x)=A(x)+C$ for some constant $C$.
Conclusion: $\int f(x) dx$  is precisely $A(x)+C$ where $C$ is an arbitrary constant.
Edit: To be more precise about answering your question, I wouldn't call this statement an axiom. Rather it is a corollary of two theorems: FTC and the fact that if a function's derivative is identically 0 then the function is constant.
Edit 2: It should also be emphasized that the notation $\int_{-\infty}^{\infty} f(x) dx$ is very different from $\int f(x) dx$. While the latter represents the family of all antiderivatives of $f(x)$, the former represents the net area under the graph of $f(x)$ over $(-\infty,\infty)$ which is a real number if it exists at all. You can connect this to $A(x)$ if you want to, e.g.,
$$
\int_{-\infty}^{\infty} f(x) dx = \int_0^{\infty} f(x) dx + \int_{-\infty}^0 f(x) dx = \lim_{x\to\infty} A(x)+\lim_{x\to -\infty} A(x)
$$
But you have to be carefuly about the limits existing, and watch for $\infty-\infty$ situations, etc. But this would work for a nice function like $\frac{1}{1+x^2}$.
