Show that

If $f(a+b-x)=f(x)$, then $\int_{a}^{b} x f(x) d x$ is equal to

A. $\frac{a+b}{2} \int_{a}^{b} f(b-x) d x$

B. $\frac{a+b}{2} \int_{a}^{b} f(b+x) d x$

C. $\frac{b-a}{2} \int_{a}^{b} f(x) d x$

D. $\frac{a+b}{2} \int_{a}^{b} f(x) d x$


Let $I=\int_{0}^{b} x f(x) d x$                     …(1)

$I=\int_{a}^{b}(a+b-x) f(a+b-x) d x$             $\left(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right)$

$\Rightarrow I=\int_{a}^{b}(a+b-x) f(x) d x$

$\Rightarrow I=(a+b) \int_{a}^{b} f(x) d x \quad-I$               $[$ Using $(1)]$

$\Rightarrow I+I=(a+b) \int_{a}^{b} f(x) d x$

$\Rightarrow 2 I=(a+b) \int_{a}^{b} f(x) d x$

$\Rightarrow I=\left(\frac{a+b}{2}\right) \int_{a}^{b} f(x) d x$

Hence, the correct answer is D.


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