If $f\left( {a + b - x} \right) = f\left( x \right)$ , then $\int\limits_a^b {xf\left( x \right)dx}$ is equal to

Option d is correct

Let $I = \int\limits_a^b {xf\left( x \right)dx}$

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

When $x = a,z = b$ and when $x = b,z = a$

$\therefore$ $I = - \int\limits_b^a {\left( {a + b - z} \right)f\left( z \right)dz} = \int\limits_a^b {\left( {a + b} \right)f\left( z \right)dz} - \int\limits_a^b {zf\left( z \right)dz}$

$= \left( {a + b} \right)\int\limits_a^b {f\left( x \right)dx} - \int\limits_a^b {x\,f\left( x \right)dx = \left( {a + b} \right)} \int\limits_a^b {f\left( x \right)dx} - I$

$\Rightarrow$ $2I = \left( {a + b} \right)\int\limits_a^b {f\left( x \right)dx}$

Hence, $I = \left( {\cfrac{{a + b}}{2}} \right)\int\limits_a^b {f\left( x \right)} dx$