If $\mathrm{f}(\mathrm{a}+\mathrm{b}-\mathrm{x})=\mathrm{f}(\mathrm{x})$ then $\int_a^b…
If $\mathrm{f}(\mathrm{a}+\mathrm{b}-\mathrm{x})=\mathrm{f}(\mathrm{x})$ then $\int_a^b \mathrm{xf}(\mathrm{x}) \mathrm{dx}$ is equal to
$\frac{a+b}{2} \int_a^b f(a+b-x) d x$
$\frac{a+b}{2} \int_a^b f(b-x) d x$
$\frac{a+b}{2} \int_a^b f(x) d x$
$\frac{b-a}{2} \int_a^b f(x) d x$
Solution
$I=\int_a^b x f(x) d x=\int_a^b(a+b-x) f(a+b-x) d x$
$=(a+b) \int_a^b f(a+b-x) d x-\int_a^b x f(a+b-x) d x$
$=(a+b) \int_a^b f(a+b-x) d x-\int_a^b x f(x) d x$
$2 I=(a+b) \int_a^b f(x) d x$
$I=\frac{(a+b)}{2} \int_a^b f(x) d x ; I=\frac{(a+b)}{2} \int_a^b f(a+b-x) d x$