R-Integral of Odd Functions

Question

Let $f$ odd and Riemann integrable on $[a,b]$

$$\int_a^b f(x)dx= \int_{-a}^{-b} f(-x)-dx= - \int_{-b}^{-a} f(-x)-dx = \int_{-b}^{-a} f(-x)dx = -\int_{-b}^{-a} f(x)dx $$

Could you please check me? If there is a missing please fix me. Thanks in advance

Answer 1

It looks good like this but I would suggest, in order to further improve your given proof, to add some explanations concerning what you have done within the single steps. For example like this

$$\int_a^b f(x)dx\stackrel{(1)}{=}\int_{-a}^{-b} f(-x)-dx\stackrel{(2)}{=}- \int_{-b}^{-a} f(-x)-dx\stackrel{(3)}{=}\int_{-b}^{-a} f(-x)dx\stackrel{(4)}{=}-\int_{-b}^{-a} f(x)dx $$

$(1)$ First of all apply the substitution $x=-x$. The upper border of integration becomes $-b$, the lower $-a$ and the differential changes to $dx=(-dx)$.

$(2)$ Reverse the order of integration, i.e. exchanging the upper and the lower border by adding a minus sign before the integral. $($to be honest this sentences sounds really bad$)$

$(3)$ The minus sign of the differential and the one infront of the integral cancel out.

$(4)$ Finally using that the function $f(x)$ is odd, i.e. $f(-x)=-f(x)$.