Answer:
Option D
Explanation:
Let I= $\int_{a}^{b} x f(x) dx $
Let a+b-x= z $\Rightarrow$ -dx=dz
when x=a,z=b and when x=b,z=a
$\therefore$ $I= -\int_{b}^{a} (a+b-z)f(z) dz$
$I= (a+b)\int_{b}^{a} f(x) dx-\int_{a}^{b} x f(x) dx$
$I= (a+b)\int_{b}^{a} f(x) dx-I;$
$2I= (a+b)\int_{a}^{b} f(x) dx$
Hence , I= $\left(\frac{a+b}{2}\right)\int_{a}^{b} f(x) dx$