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  4. If f(a + b +1 - x) = f (x), for a ll x, where a and b are fixed positive real numbers, then (1/a+b) ∫ba x(f(x)+f(x+1))dx is equal to :

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

JEE MainJEE Main 2020Integrals

Solution:

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