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.
 A: Your first step is mistaken: it seems that you mistake $\int _a ^b (-f) (x) \ \Bbb d x$ for $\int _a ^b f (-x) \ \Bbb d x$; these two are completely different, and the homogeneity property applies only to the first formula, not to the second.
Just use the substitution $y = -x$, this will solve the problem in no time.
A: 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,
A: Your step 1 is wrong and you can realize the error by considering $a=0$, $b=1$, $f(x)=e^x$.
Then
$$
\int_{0}^{1}e^x\,dx=e-1,
\qquad
-\int_{-1}^0e^x\,dx=\frac{1}{e}-1
$$
which are quite different.
You can prove the statement by the definition, I'll use Riemann sums. A Riemann sum for $\int_{a}^{b}f(x)\,dx$ consists first in a choice $S$ of points
$$
a=x_0<x_1<x_2<\dots<x_{n-1}<x_n=b,
\qquad
c_i\in[x_{i-1},x_i],\ i=1,2,\dots,n
$$
and in considering
$$
\sigma(f;S)=\sum_{i=1}^n f(c_i)(x_i-x_{i-1})
$$
Define $\delta(S)=\max\{x_1-x_0,x_2-x_1,\dots,x_n-x_{n-1}\}$; then it's not much difficult to give a meaning to
$$
\lim_{\delta(S)\to0}\sigma(f;S)
$$
and, if this exists, it is called the integral.
Now note that for each Riemann sum for $f(x)$ over $[a,b]$ we can define a Riemann sum $\hat{S}$ for $g(x)=f(-x)$ over $[-b,-a]$ by simply taking the negative of each point (and renaming indices, if you prefer to make your life difficult). Then
$$
\sigma(g;\hat{S})=\sum_{i=1}^n g(-c_i)(-x_{i-1}-(-x_i))
=
\sum_{i=1}^n f(c_i)(x_i-x_{i-1})
=
\sigma(f;S)
$$
Thus the two limits are equal, because also each Riemann sum for $g$ over $[-b,-a]$ corresponds to a Riemann sum for $f$ over $[a,b]$, by the same construction.
Similarly if you define integrals with upper and lower sums.

If the function $f$ is continuous, you can use substitutions (through the fundamental theorem of calculus):
\begin{align}
\int_{-b}^{-a}f(-x)\,dx
&=\int_{b}^{a}f(t)\cdot(-1)\,dt && -x=t,\quad dx=-dt
\\
&=-\int_{b}^{a}f(t)\,dt
\\
&=\int_{a}^{b}f(t)\,dt
\\
&=\int_{a}^{b}f(x)\,dx && x=t,\quad dt=dx
\end{align}
A: 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. 
$$\int_{a}^bf(x)dx$$
