Question

If $f(a+b+1-x)=f\left(x\right)$ $\forall x\in R,$ where $a$ and $b$ are fixed positive real numbers, then $\frac{1}{a+b}\underset{a}{\overset{b}{\int }}x\left[f\right(x)+f(x+1\left)\right]dx$ is equal to

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

Answer 1

The correct option is B $\underset{a-1}{\overset{b-1}{\int }}f(x+1)dx$

$I=\frac{1}{a+b}\underset{a}{\overset{b}{\int }}x\left[f\right(x)+f(x+1\left)\right]dx\cdots \left(i\right)$
using kings property, $x\to (a+b-x)$
$\Rightarrow I=\frac{1}{a+b}\underset{a}{\overset{b}{\int }}(a+b-x)\left[f\right(a+b-x)+f(a+b+1-x\left)\right]dxI=\frac{1}{a+b}\underset{a}{\overset{b}{\int }}(a+b-x)\left[f\right(x+1)+f(x\left)\right]dx\cdots \left(ii\right)[\because f(a+b+1-x)=f(x),f(a+b-x)=f(x+1\left)\right]$
Adding $\left(i\right)$ and $\left(ii\right):$
$\Rightarrow 2I=\underset{a}{\overset{b}{\int }}\left[f\right(x+1)+f(x\left)\right]dx\Rightarrow 2I=\underset{a}{\overset{b}{\int }}f(x+1)dx+\underset{a}{\overset{b}{\int }}f\left(x\right)dx$
Using property
$\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$2I=\underset{a}{\overset{b}{\int }}f(a+b+1-x)dx+\underset{a}{\overset{b}{\int }}f\left(x\right)dx\Rightarrow 2I=2\underset{a}{\overset{b}{\int }}f\left(x\right)dx\Rightarrow I=\underset{a}{\overset{b}{\int }}f\left(x\right)dx\therefore I=\underset{a-1}{\overset{b-1}{\int }}f(x+1)dx$
$\text{OR}$
$I=\underset{a+1}{\overset{b+1}{\int }}f(x-1)dx$