Proof that integrating f(x) between a and b is the same as integrating f(a+b-x) over the same bounds?

Question

$$\int_a^bf(x)dx=\int_a^b f(a+b-x) dx$$

Is anyone able to show me a rigorous proof of this identity. I understand using a substitution of $u=a-x$ gives rise to the RHS in terms of $u$ but don’t know how to show this implies that it’s also true for $x$?

Answer 1

Let $a+b-x=z\implies dx=-dz$

When $x\to a,\quad z \to b \quad \text{and} \quad x \to b, \quad z \to a$

$\int_a^b f(a+b-x) dx=\int_b^af(z)(-dx)=-\int_b^af(z) dz=\int_a^bf(z)dz=\int_a^bf(x)dx$