Proving that $\int_{-b}^{-a}f(-x)dx=\int_{a}^{b}f(x)dx$ $f:[a,b]\rightarrow R$ that is integrable on [a,b]
So we need to prove:
$$\int_{-b}^{-a}f(-x)dx=\int_{a}^{b}f(x)dx$$
1.) So we'll use a property of definite integrals: (homogeny I think it's called?)
$$\int_{-b}^{-a}f(-x)dx=-1\int_{-b}^{-a}f(x)dx$$
2.) Great, now using the fundamental theorem of calculus:
$$-1\int_{-b}^{-a}f(x)dx=(-1)^2\int_{-a}^{-b}f(x)dx=\int_{-a}^{-b}f(x)dx$$
This is where I'm stuck. For some reason I think it might be smarter to skip step 2, to leave it asL
$$-1\int_{-b}^{-a}f(x)dx$$ 
because graphically, we've "flipped" the graph about the x-axis, but we're still calculating the same area. Proving that using properties seems to have stumped me.
I prefer hints over solutions, thanks.

There is a small error in your first step, you didn't change the sign of the limits. Always keep in mind that after making any substitution, don't forget to change the limits. Since you have made a substitution $x=-t$, so change the limits accordingly. So corrected step 1 is $$-\int_{b}^af(x)dx$$
Now you can reverse the limits to get the correct answer i.e. 
$$\int_{a}^bf(x)dx$$
