The value of $\int\limits_{1 / n}^{(a n-1) / n} \frac{\sqrt{x}}{\sqrt{a-x}+\sqrt{x}} d x$, is
Answer & explanation
Correct answer: option 3
Let
$I=\int\limits_{1 / n}^{(a n-1) / n} \frac{\sqrt{x}}{\sqrt{a-x}+\sqrt{x}} d x$ .....(i)
Using $\int\limits_a^b f(x) d x=\int\limits_a^b f(a+b-x) d x$, we have
$I=\int\limits_{1 / n}^{(a n-1) / n} \frac{\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}} d x$ ....(ii)
Adding (i) and (ii), we get
$2 I=\int\limits_{1 / n}^{(a n-1) / n} 1 . d x=\frac{a n-2}{n} \Rightarrow I=\frac{a n-2}{2 n}$