How comes that the integral ( indefinite) of a function f is always a primitive of f , although the domain of the integral is not the same as f's.

Question

The first part of the Funfamental Theorem of Calculus says that any function defined as the integral of f(t)dt from some point a to x is a primitive of f.

Out of this, I conclude that the domain of the integral function is an interval limited by a from the left.

But the domain of function f need not be limited by a.

Hence my question: isn't this at least surprising that the integral function, although its domain need not be identical to the domain of f is guaranteed to be a primitive of f?

I can't give any solid reason why it seems surprising to me, but *I would naturally tend to think that the primitive of a function f should have a domain at least as big as its derivative ( namely, f) .*

My question may include a lot of false assumptions.

Answer 1

You are missing the fact that, if $b<a$, then $\int_a^bf(t)\,\mathrm dt=-\int_b^af(t)\,\mathrm dt$. Therefore, yes, the map $x\mapsto\int_a^xf(t)\,\mathrm dt$ is defined when $x<a$.

Answer 2

You may use that:

$$\int_a^b f(t)dt = - \int^a_b f(t)dt $$

Hence your function can be definied even for $x<a$, so you don't have a left bounded domain.

Answer 3

Let $I$ be an intervall in $ \mathbb R$ and $f:I \to \mathbb R$ continuous. If $c \in I$ and if $F: I \to \mathbb R$ is defined by

$$F(x):= \int_c^x f(t)dt,$$

then $F$ is an anti-derivative of $f$.