If f (a + b – x) = f(x) then $\int\limits_a^bx\,f(x)\, dx$ is equal to

$(\frac{a+b}{2})\int\limits_a^bf(x)\, dx$

Let $I = \int\limits_a^bx\,f (x)\, dx = \int\limits_a^b(a+b-x)\, f (a +b-x)\, dx$

$= (a + b) \int\limits_a^bf(a +b –x)dx - \int\limits_a^bx\, f (a + b –x) dx$

$= (a + b) \int\limits_a^bf (x)\, dx -\int\limits_a^b x\, f (x)\, dx$.

Hence I = f(x) dx. Hence (B) is the correct answer.