Question

If $f(a+b-x)=f\left(x\right)$ then ${\int }_a^{b}xf\left(x\right)dx$ equals

• A. $\frac{a+b}{2}{\int }_a^{b}f\left(x\right)dx$
• B. $\frac{b-a}{2}{\int }_a^{b}f\left(x\right)dx$
• C. $\frac{a+b}{2}{\int }_a^{b}f(a+bx)dx$
• D. $\frac{a+b}{2}{\int }_a^{b}f(b-x)dx$

Answer 1

The correct option is A $\frac{a+b}{2}{\int }_a^{b}f\left(x\right)dx$

Let $I={\int }_a^{b}xf\left(x\right)dx$ ...(1)
Using property ${\int }_A^{B}f\left(x\right)dx={\int }_A^{B}f(A+B-x)dx$
$I={\int }_a^b(a+b-x)f(a+b-x)dx$ ...(2)
Adding (1) and (2) and using $f(a+b-x)=f\left(x\right)$
We get $2I=(a+b){\int }_a^{b}f\left(x\right)dx\Rightarrow I=\frac{(a+b)}{2}{\int }_a^{b}f\left(x\right)dx$