Definite Integration — Maths STD 12 Science — Question

Let $I=\int_a^b f(x) d x$
Put x= a + b - t
∴ dx = -dt
When x = a, t = b and when x = b, t = a
$\therefore I=\int_b^a f(a+b-t)(-d t)$
$\therefore I=-\int_b^a f(a+b-t) d t$
$\therefore I=\int_a^b f(a+b-t) d t \ldots\left[\because \int_a^b f(x) d x=-\int_b^a f(x) d x\right]$
$\therefore \int_a^b f(x) d x=\int_a^b f(a+b-x) d x \ldots\left[\because \int_a^b f(x) d x=\int_a^b f(t) d t\right]$
Let $I=\int_a^b \frac{f(x)}{f(x)+f(a+b-x)} d x \ldots(i)$
$\therefore I=\int_a^b \frac{f(a+b-x)}{f(a+b-x)+f(a+b-(a+b-x))} d x$
$\therefore I=\int_a^b \frac{f(a+b-x)}{f(a+b-x)+f(x)} d x \ldots(i i)$
Adding (i) and (ii) we get
$\begin{aligned} & 2 I=\int_a^b \frac{f(x)+f(a+b-x)}{f(x)+f(a+b-x)} d x \\ & \therefore 2 I=\int_a^b 1 d x \\ & \therefore 2 I=[x]_a^b \\ & \therefore I=\frac{b-a}{2}\end{aligned}$