Evaluate $\int\limits_{1}^{3} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{4 - x}} dx$.

1

The correct answer is Option (1) → 1

$I = \int\limits_{1}^{3} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{4 - x}} dx \dots(i)$

Using property $\int\limits_{a}^{b} f(x)dx = \int\limits_{a}^{b} f(a+b-x)dx$, we get

$I = \int\limits_{1}^{3} \frac{\sqrt{4 - x}}{\sqrt{4 - x} + \sqrt{x}} dx \dots(ii)$

On adding eqs. (i) and (ii), we get

$2I = \int\limits_{1}^{3} \frac{\sqrt{x} + \sqrt{4 - x}}{\sqrt{x} + \sqrt{4 - x}} dx$

$= \int\limits_{1}^{3} 1 dx = [x]_1^3$

$= 3 - 1 = 2$

$∴I = 1$