Question

If $f(a+b-x)=f\left(x\right),$ then the value of ${\int }_a^{b}xf\left(x\right)dx$ is

• A. $\frac{a+b}{2}{\int }_a^{b}f(b-x)dx$
• B. $\frac{a+b}{2}{\int }_a^{b}f\left(x\right)dx$
• C. $\frac{b-a}{2}{\int }_a^{b}f\left(x\right)dx$
• D. None of these

Answer 1

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

Since $I={\int }_a^{b}xf\left(x\right)dx={\int }_a^b(a+b-x)f(a+b-x)dx\text{Given}f(a+b-x)=f\left(x\right)I={\int }_a^b(a+b)f\left(x\right)dx-{\int }_a^{b}xf\left(x\right)dxI={\int }_a^b(a+b)f\left(x\right)dx-I2I=(a+b){\int }_a^{b}f\left(x\right)dxI={\int }_a^{b}xf\left(x\right)dx=\frac{a+b}{2}{\int }_a^{b}f\left(x\right)dx$