Question

If $f(a+b+1-x)=f(x)$, for a ll x, where a and b are fixed positive real numbers, then $\frac{1}{a+b}{\int }_a^{b}x\left(f\left(x\right)+f\left(x+1\right)\right)dx$ is equal to :

• A. ${\int }_{a-1}^{b-1}f\left(x\right)dx$
• B. ${\int }_{a+1}^{b+1}f\left(x+1\right)dx$
• C. ${\int }_{a-1}^{b-1}f\left(x+1\right)dx$
• D. ${\int }_{a+1}^{b+1}f\left(x\right)dx$

Answer 1

Answer: (C)

$f\left(x+1\right)=f\left(a+b-x\right)$
$I=\frac{1}{\left(a+b\right)}{\int }_a^{b}xf\left(x\right)+f\left(x+1\right)dx.....\left(1\right)$
$I=\frac{1}{\left(a+b\right)}{\int }_a^b\left(a+b-x\right)\left(f\left(x+1\right)+f\left(x\right)\right)dx...\left(2\right)$
from $\left(1\right)$ and $\left(2\right)$
$2I={\int }_a^b\left(f\left(x\right)+f\left(x+1\right)\right)dx$
$2I={\int }_a^{b}f\left(a+b-x\right)dx+{\int }_a^{b}f\left(x+1\right)dx$
$2I={\int }_a^{b}f\left(x+1\right)dx\Rightarrow I={\int }_a^{b}f\left(x+1\right)dx$
$={\int }_{a+1}^{b+1}f\left(x\right)dx$
OR
$I=\frac{1}{\left(a+b\right)}{\int }_a^{b}xf\left(x\right)+f\left(x+1\right)dx.....\left(1\right)$
$I=\frac{1}{\left(a+b\right)}{\int }_a^b\left(a+b-x\right)\left(f\left(a+b-x\right)+f\left(a+b+1-x\right)\right)dx$
$I=\frac{1}{\left(a+b\right)}{\int }_a^b\left(a+b-x\right)\left(f\left(x+1\right)+f\left(x\right)\right)dx...\left(2\right)$
equation $\left(1\right)+\left(2\right)$
$2I=\frac{1}{\left(a+b\right)}{\int }_a^b\left(a+b\right)\left(f\left(x+1\right)+f\left(x\right)\right)dx$
$I=\frac{1}{2}\left[{\int }_a^{b}f\left(x+1\right)dx+{\int }_a^{b}f\left(x\right)dx\right]$
$=\frac{1}{2}\left[{\int }_a^{b}f\left(x\right)dx+{\int }_a^{b}f\left(x\right)dx\right]$
$I={\int }_a^{b}f\left(x\right)dx$
Let $x=T+1$
$={\int }_{a-1}^{b-1}f\left(x+1\right)dx$