Proving that $\int_{-b}^{b}f(x)dx=0 $ if $f$ is an odd function So I split this integral up into two:
$$ \int_{-b}^{0}f(x)dx+\int_{0}^{b}f(x)dx= \int_{-b}^{b}f(x)dx$$
Then I can do the following:
$$ \int_{-b}^{0}f(x)dx+\int_{0}^{b}f(x)dx= \int_{0}^{b}f(-x)dx+\int_{0}^{b}f(x)dx$$
and since $f$ is odd $(f(-x)=-f(x))$ we get 
$$\int_{0}^{b}-f(x)dx+\int_{0}^{b}f(x)dx=0$$
I'm not really understanding why my second is correct other than if I use substitution, i.e. $x=-y$. 

Another approach. Let's assume $f$ is sufficiently regular on a given set. By the fondamental theorem of calculus and the chain rule, we get that
$$
\left(\int_{-b}^{b}f(x)dx\right)'=f(b)-(-f(-b))=0
$$ giving that
$$
\int_{-b}^{b}f(x)\:dx=\text{constant}
$$ by putting $b:=0$, the constant is seen to be $0$. 
