Question

By using the properties of definite integrals,evaluate the integrals: $\int^a_0 {\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a- x}} \, dx$

Answer 1

Answer:

$\frac{a}{2}$

Step-by-step explanation:

$concept:\\\\\int\limits^a_0{f(x)}\:dx=\int\limits^a_0{f(a-x)\:dx$

$Let,\\I=\int\limits^a_0{\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\:dx.............(1)$

By properties of definite integrals

$I=\int\limits^a_0{\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}\:dx$

$I=\int\limits^a_0{\frac{\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}\:dx...........(2)$

Adding (1) and (2)

$2I=\int\limits^a_0{\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\:dx+\int\limits^a_0{\frac{\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}\:dx$

$2I=\int\limits^a_0[\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}+\frac{\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}]\:dx$

$2I=\int\limits^a_0[\frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}\:dx$

$2I=\int\limits^a_0{dx}\\\\2I=[x]^a_0\\\\2I=a\\\\I=\frac{a}{2}$