Question

If $f(a+b-x)=f(x)$, then $\underset{a}{\overset{b}{\int }}xf(x)dx$ is equal to

• A. $\frac{a+b}{2}\underset{a}{\overset{b}{\int }}f(b-x)dx$
• B. $\frac{a+b}{2}\underset{a}{\overset{b}{\int }}f(x)dx$
• C. $\frac{b-a}{2}\underset{a}{\overset{b}{\int }}f(x)dx$
• D. $\frac{a+b}{2}\underset{a}{\overset{b}{\int }}f(b+x)dx$

Answer 1

Answer: (B)

Let $I=\underset{a}{\overset{b}{\int }}xf(x)dx$ ....(1)
Also, $I=\underset{a}{\overset{b}{\int }}(a+b-x)f(a+b-x)dx$ ....(2)
(Using King property)
$\therefore (1)+(2)\Rightarrow 2I=\underset{a}{\overset{b}{\int }}(a+b-x+x)f(x)dx$
So, $I=\left(\frac{a+b}{2}\right)\underset{a}{\overset{b}{\int }}f(x)dx$
[As $f(a+b-x)=f(x)$ (Given)]