TITLE: Is really $f(x)=\int g(x) dx$ a function?
QUESTION [7 upvotes]: I saw many of this kind of questions on some text/question books. Is there any other explanation of this, or is it really wrong as I thought?
Here is a question of that kind: 
If $\displaystyle f(x)=\int x(x^2-a)^2 dx$ and $f(a)=7$ then $f(-a)=?$
Here what is $f(0)$ or $f(1)$? $\displaystyle f(0)=\int 0(0^2-a)^2 d0$ or $\displaystyle f(1)=\int 1(1^2-a)^2 d1$ does not make sense.
For me, a right function need to be as: $\displaystyle  f_c(x)=\int_c^xt(t^2-a)^2dt$
(where $c$ is some constant, can be $0$ as usual).

REPLY [2 votes]: One usually uses the notation $\int f(x) \, dx$ to denote an anti-derivative, or, more rigorously, the collection of all possible anti-derivatives of the function $f(x)$ over some domain that it usually understood implicitly from the context. Thus, one can interpret the statement 
$$ \int x^2 \, dx = \frac{x^3}{3} + C $$
as the statement that all the possible anti-derivatives of the function $x^2$ on the real line are of the form $\frac{x^3}{3} + C$ for some $C \in \mathbb{R}$. This way, one can interpret the expression $f(x) = \int x(x^2 - a)^2 \, dx$ as the statement "$f(x)$ is one of the possible anti-derivatives of the function $x(x^2 - a)^2$ on $\mathbb{R}$" although admittedly, this can be written in a much less confusing way as "$f'(x) = x(x^2 - a)^2$ for all $x \in \mathbb{R}$".
The fundamental theorem of calculus states that if $f$ is continuous on an interval $(a,b)$, then an anti-derivative of $f$ is given by $\int_c^x f(t) \, dt$ where $c \in (a,b)$ is an arbitrary point but the notation $\int f(x) \, dx$ can be used without even knowing what is a definite integral, only knowing what is a derivative.