if $f:(-1,1) \to \mathbb R$ is odd, then each indefinite integral of $f$ is even.

Question

if $f:(-1,1) \to \mathbb R$ is odd, then each indefinite integral of $f$ is even.

My attempt:

Let $\int f(x) dx = F(x)+C$.

$f$ is odd so $f(-x)=-f(x)$.

so $\int f(-x) dx = -\int f(x) dx$

so $F(x)+C = \int f(x) dx = -\int f(-x) dx = -(-F(-x))+C$

so $F(x)= F(-x)$, hence - even.

Is that correct? I'm wondering if I'm missing something here...

Answer 1

Correct, since $\int f(ax)dx=\frac{1}{a}F(x) +c$, where $a\neq 0$ is a parameter.