How is the Integral of $\int_a^bf(x)~dx=\int_a^bf(a+b-x)~dx$

Can Some one tell me what this method is called and how it works With a detailed proof

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

I've been using this a lot in definite integration but haven't seemed to have realized why it is true. But whatever it is it always seems to work.

Basically a proof of how it is always true.

Here is a pictorial argument.

$\displaystyle \int_a^b f(x) dx$ is the area under the curve $y=f(x)$ in the interval $(a,b)$ when you integrate from left to right. $\displaystyle \int_a^b f(a+b-x) dx$ is the area under the curve $y=f(x)$ in the interval $(a,b)$ when you integrate from right to left. Hence, both are equal.

Change of variables: $a+b-x=t$, $dx = -dt$, and $$ \int_a^b f(a+b-x)\, dx = -\int_b^a f(t)\, dt = \int_a^b f(t)\, dt. $$

it is just substitution, if we let $u = a+b-x$, we have $du = -dx$ and hence (note that $u = b$ when $x= a$ and vice versa) \begin{align*} \int_a^b f(x)\,dx &= \int_a^b f(u)\, du\\ &= \int_b^a f(a+b-x)\bigl(-dx\bigr)\\ &= -\int_b^a f(a+b-x)\,dx\\ &= \int_a^b f(a+b-x)\, dx \end{align*}