Question

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

• 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. $\frac{a+b}{2}{\int }_a^{b}f(a+b+x)dx$

Answer 1

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

$LettI={\int }_a^{b}xf\left(x\right)dx-------\left(1\right)$
$\Rightarrow I={\int }_a^b(a+b-x)f(a+b-x)dx$
$(\because {\int }_a^{b}f(x)dx={\int }_b^{b}f(a+b-x\left)dx\right)$
$\Rightarrow I={\int }_b^b(a+b-x)f\left(x\right)dx$
$(\because Given)$
$\Rightarrow I=(a+b){\int }_a^{b}f\left(x\right)dx-{\int }_a^{b}xf\left(x\right)d_x$
$\Rightarrow I=(a+b){\int }_a^{b}f\left(x\right)dx-I[fromEq.(1\left)\right]$
$\Rightarrow I=\left(\frac{a+b}{2}\right){\int }_a^{b}f\left(x\right)dx$