Here's a "picture proof":
Note that
The graph of $\color{maroon}{y=g(-x)}$ is the graph of $\color{darkgreen}{y=g(x)}$ reflected through the $y$-axis.
The graph of $\color{darkblue}{y=g\bigl(-(x-(a+b))\bigr)=g(a+b-x)}$ is the graph of $\color{maroon}{y=g(-x)}$ shifted to the right $a+b$ units.

Now think about the definite integral and it's relation to area.
Alternatively (this is just the substitution method):
If $G(x)$ is an antiderivative of $g(x)$, then $-G (a+b-x)$ is an antiderivative of $g(a+b-x)$; so $ \int_a^b g(a+b-x)\,dx =-G(a+b-x)\bigl|_a^b =-\bigl(G(a)-G(b) \bigr) =G(b)-G(a)=\int_a^bg(x)\,dx. $
But of course, since picture proofs aren't really proofs and since an integrable function does not necessarily have a primitive, the proper way to do this is to use Riemann sums. This would not be to difficult to do since any Riemann sum for
$g(x)$ over
$[a,b]$ corresponds to (by reversing the order of the summation) a Riemann sum for
$g(a+b-x)$.