Question

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

• A. $\frac{1}{2}(a+b)$
• B. $\frac{1}{2}(a-b)$
• C. $\frac{1}{2}(a^2+b^2)$
• D. $\frac{1}{2}(a^2-b^2)$

Answer 1

The correct option is A $\frac{1}{2}(a+b)$

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