Question

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

Answer 1

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

$f(a+b+1-x)=f\left(x\right)...\left(1\right)$
$x\to x+1$
$f(a+b-x)=f(x+1)...\left(2\right)$
$I=\frac{1}{a+b}\underset{a}{\overset{b}{\int }}x\left(f\right(x)+f(x+1\left)\right)dx...\left(3\right)$

From $\left(1\right)$ and $\left(2\right)$
$I=\frac{1}{a+b}\underset{a}{\overset{b}{\int }}(a+b-x)\left(f\right(x+1)+f(x\left)\right)dx...\left(4\right)$

Adding $\left(3\right)$ and $\left(4\right)$
$2I=\underset{a}{\overset{b}{\int }}\left(f\right(x)+f(x+1\left)\right)dx\Rightarrow 2I=\underset{a}{\overset{b}{\int }}f(x+1)dx+\underset{a}{\overset{b}{\int }}f\left(x\right)dx\Rightarrow 2I=\underset{a}{\overset{b}{\int }}f(a+b-x+1)dx+\underset{a}{\overset{b}{\int }}f\left(x\right)dx\Rightarrow 2I=2\underset{a}{\overset{b}{\int }}f\left(x\right)dx\Rightarrow I=\underset{a}{\overset{b}{\int }}f\left(x\right)dx$
Let $x=t+1,dx=dt$
$\Rightarrow I=\underset{a-1}{\overset{b-1}{\int }}f(t+1)dt\Rightarrow I=\underset{a-1}{\overset{b-1}{\int }}f(x+1)dx$