Maths question

Just checking; is the following always true as a rule?

∫ f(x) dx = ∫[0, x] f(x) dx + C ⇐ ∫[0, x] f(x) dx ∈ ℝ
where C is an arbitrary constant.

(The implied precondition from "⇐ ∫[0, x] f(x) dx ∈ ℝ" part above is required because I know "∫ f(x) dx = ∫[0, x] f(x) dx + C" doesn't work for such cases as f(x) = 1/x etc. because the ∫[0, x] f(x) dx integral for such cases doesn't converge to a real number)
I think it surely must be but I only ask because I would like to put it in my book I am writing but I have thoroughly search the net for it and strangely cannot find this seemingly obvious rule (or implied equivalent) mentioned ANYWHERE over the net even though it seems obvious to me and this made me wonder why; Is something wrong with it? Is it always technically correct but nevertheless for some reason considered bad practice to think in terms of it?

@humy
This is the Fundamental Theorem of Calculus, so sort of, but your notation's a little confused. Basically the definite integral and the indefinite integral only look similar they're defined quite differently. The indefinite integral is basically the anti-derivative operator, it's a way of writing (d/dx)^(-1). The definite integral is the area enclosed by a function f(x), the x-axis and two vertical lines at the limits. The definite integral can be calculated by knowing the anti-derivative and vice versa, but they are separate objects, the fundamental theorem ties them together.

F(b) - F(a) = ∫[a, b] f(x) dx
with F(x) ∈ {F(x) ∈ C1 | f(x) = dF(x)/dx}
C1 is the set of piecewise differentiable functions on ℝ. So, if F(x) is an anti-derivative of f(x), so is F(x) + C.

We're free to select any F(x) that differentiates to give f(x) so in the absence of boundary conditions the function normally quoted as the anti-derivative is just the one with any constant term omitted. F(x) is the representative of an equivalence class of antiderivatives.

For the indefinite integral, better notation is.

∫ f(x) dx = ∫[a, x] f(y) dy = F(x) - F(a) = F(x) + constant.

The leftmost term here should be read as (d/dx)^1 f(x) = F(x) + const. The point is that the derivative is a many to one mapping and so the inverse operator is one to many and we need more information to specify a particular function.

If the representative function F(x) has singularities then since the lower bound of the integral is arbitrary we can just choose it so that F(a) is finite. There's no particular reason for having the lower bound as zero.

https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus

@deepthought said
@humy
This is the Fundamental Theorem of Calculus, so sort of, but your notation's a little confused. Basically the definite integral and the indefinite integral only look similar they're defined quite differently. The indefinite integral is basically the anti-derivative operator, it's a way of writing (d/dx)^(-1). The definite integral is the area enclosed by a funct ...[text shortened]... for having the lower bound as zero.

https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus

Thanks for that.
I will revise the subject in due course but I think I have just decided its best for me to simply avoid saying anything about this aspect of calculus in my book; Just stick to what I know best.