A few questions about definite integrals and even functions

Question

$RTP$ $\int_{-a}^a f(x) dx = 2\int_{0}^a f(x) dx$ if f is even

Let's assume that f(x) is even. This means that f(x) = f(-x). Now we can prove $$\int_{-a}^a f(x) dx=\int_{-a}^0 f(x) dx +\int_{0}^a f(x) dx=2\int_{0}^a f(x) dx$$

The logic here, I believe, lies in the symmetry of f(x). An even function is symmetric with respect to the y-axis, and a definite integral translates to the area under a function's curve over some interval. Therefore the area under the intervals $[-a,0]$ and $[0,a]$ must be equal ( or 2 times either of our definite integrals). This is implicit in the proof. Correct?

Second. Is it always justified to write a definite integral in the aforementioned fashion - as the sum of two definite integrals? One containing the first half our or integration interval, another containing the remaining half.

Third. For this proof we weren't told that $0∈dom:f(x)$. Is that not necessary for a proof like this?

Answer 1

1. That's right.
2. Yes, $\int_a^c f(x) dx + \int_c^b f(x) dx = \int_a^b f(x) dx$. Surprisingly this doesn't even require $c$ to be between $a$ and $b$.
3. You do need that, but it's implicit in the definition of $\int_{-a}^a f(x) dx$. Without that you can still say $\int_a^b f(x) dx = \int_{-b}^{-a} f(x) dx$, though.