Question

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

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

Answer 1

Answer: (D)

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

Let $I={\int }_a^{b}xf\left(x\right)dx={\int }_a^{b}xf(a+b-x)dx$
Let $a+b-x=z\Rightarrow -dx=dz$
When $x=a,z=b$ and when $x=b,z=a$
$\therefore I=-{\int }_b^a(a+b-z)f\left(z\right)dz$

$⟹I=-{\int }_b^a(a+b)f\left(z\right)dx-{\int }_b^a(-z)f\left(z\right)dx$

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

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

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

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