Is really $f(x)=\int g(x) dx$ a function?

Question

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).

Answer 1

$f(x) = \int g(x) dx$ is very bad notation. The $x$ in the integral sign is a so-called 'dummy variable'. $\int g(x)dx$ is just a number, it does not depend on $x$. The only thing it could possible mean is the function that has constant value $\int g(u)du$. When people erroneously write this down they usually mean $$f(x) = \int_a^x f(t)dt,$$ for some $a$.

Answer 2

You can consider $$ \int x(x^2-a)^2\,dx $$ as one of the functions $f(x)$ such that $f'(x)=x(x^2-a)^2$, which is nonetheless not determined until some further condition is imposed.

You do have a further condition, that is, $f(a)=7$, so the function is determined.

How do you determine it? You know that, for some constant $c$, $$ f(x)=\frac{x^6}{6}-\frac{ax^4}{2}+\frac{a^2x^2}{2}+c $$ and the condition $f(a)=7$ reads $$ \frac{a^6}{6}-\frac{a^5}{2}+\frac{a^4}{2}+c=7 $$ thereby determining $c$.

However, this is not really necessary: the function $f$ satisfies the property that $f(x)=f(-x)$, so…

Answer 3

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.