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.

Let f(x) = x. then f(-x) = -x. Substituting -a and -b in the limits of the integration will lead to be f(a) = a then f(-a) = -(-a) = a.
it's simply this that you are multiplying the limits and the function by -1. if both are multiplied then they would get neutralized,
