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. $x=-y$. 

Your substitution works without the need to slip the integral.
You have
\begin{align*}
\int _{-b}^b f(x){\rm d}x &= \int _{b}^{-b}-f(-u) {\rm d} u,\,x=-u\\
&=\int _{b}^{-b}f(u) {\rm d} u \\
&=-\int _{-b}^{b}f(u) {\rm d} u\\
&=-\int _{-b}^b f(x){\rm d}x
\end{align*}
Therefore $\int _{-b}^b f(x){\rm d}x =0$. 
So the algebraic idea was since we know $f(-x)=-f(x)$, make the substitution $u=-x$ and see that it can be derived.
