Question

Solve :

$\underset{0}{\overset{a}{\int }}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx$

Answer 1

We have,

$I=\underset{0}{\overset{a}{\int }}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx$ $........\left(1\right)$

Apply property,

$I=\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$

Therefore,

$I=\underset{0}{\overset{a}{\int }}\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{a-a+x}}dx$

$I=\underset{0}{\overset{a}{\int }}\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}dx$ $.......\left(2\right)$

On adding both equations, we get

$2I=\underset{0}{\overset{a}{\int }}\frac{\sqrt{a-x}+\sqrt{x}}{\sqrt{a-x}+\sqrt{x}}dx$

$2I=\underset{0}{\overset{a}{\int }}1dx$

$2I=[x{]}_0^a$

$2I=a-0$

$2I=a$

$I=\frac{a}{2}$

Hence, this is the answer.