Question

If $f(a+b-x)=f\left(x\right)$, then ${\int }_a^{b}xf\left(x\right)dx$ is equal to
A. $\frac{a+b}{2}{\int }_a^{b}f(b-x)dx$
B. $\frac{a+b}{2}{\int }_a^{b}f(b+x)dx$
C. $\frac{b-a}{2}{\int }_a^{b}f\left(x\right)dx$
D. $\frac{b+a}{2}{\int }_a^{b}f\left(x\right)dx$

Answer 1

Let $I={\int }_a^{b}xf\left(x\right)dx$ ...(1)
$\Rightarrow I={\int }_a^b(a+b-x)(a+b-x)dx$
$[\because {\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x\left)dx\right]$
$\Rightarrow I={\int }_a^b(a+b-x)f\left(x\right)dx$
$\{\because f(a+b-x)=f(x\left)\right\}$
$\Rightarrow I={\int }_a^b(a+b)f\left(x\right)dx-{\int }_a^{b}xf\left(x\right)dx$
$\Rightarrow I=(a+b){\int }_a^{b}f\left(x\right)dx-I$
$thereforeI=\frac{(a+b)}{2}{\int }_a^{b}f\left(x\right)dx$
Hence option (D) is correct.