Question

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

• A. $\left(\frac{a+b}{2}\right){\int }_a^{b}f\left(x\right)dx$
• B. ${\int }_a^{b}f\left(x\right)dx$
• C. $(a+b){\int }_a^{b}f\left(x\right)dx$
• D. $0$

Answer 1

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

$I={\int }_a^{b}xf\left(x\right)dx$

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

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

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

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

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