Question

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

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

Answer 1

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

As $f(a+b-x)=f\left(x\right)$

${\int }_a^{b}xf\left(x\right)dx={\int }_a^b(a+b-x)f(a+b-x)dx={\int }_a^b(a+b-x)f\left(x\right)dx={\int }_a^b(a+b)f\left(x\right)dx-{\int }_a^{b}xf\left(x\right)dx\Rightarrow 2{\int }_a^{b}xf\left(x\right)dx=(a+b){\int }_a^{b}f\left(x\right)dx\Rightarrow {\int }_a^{b}xf\left(x\right)dx=\frac{(a+b)}{2}{\int }_a^{b}f\left(x\right)dx$