Lets us define two functions $G1(x) = F(x)$ and $G2(x) = F(a+b-x)$
For any point $x$ in the range $x=a$ to $x=b$ we can define a variable scalar $P$ such that $x$ divides the interval $a...b$ in the ratio $(P)$:$(1-P)$ where $0 <= P <=1$.
Now we can define any point $x$ in two ways: $x1 = a + P(b-a)$ and $x2 = b - (1-P)(b-a)$
Now let us insert $x2$, $x1$ into the two different functions $G1$,$G2$
$G1(x) = F([x2]) = F([b - (1-P)(b-a)]) = F(a + P(b-a))$
$G2(x) = F(a+b-[x1]) = F(a + b - [a + P(b-a)]) = F(b - P(b-a))$
therefore $\int_0^1G1(x)\,\mathrm{d}P =\int_0^1G2(x)\,\mathrm{d}P$ because in the former integration we move across the interval from $x=a$ to $x=b$ and in the latter integration we move across the same interval from $x=b$ to $x=a$.
Therefore $\int_a^bF(x)$ = $\int_a^b F(a+b-x)$