Question

$I={\int }_0^a\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx$

Then $I=?$

• A. $\frac{a}{2}$
• B. $\frac{a}{3}$
• C. $\frac{1}{2}$
• D. $\frac{a}{5}$

Answer 1

Answer: (A)

$\frac{a}{2}$

$I={\int }_0^a\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx$ ______ (1)

using queen rule, we can write $\Rightarrow$

$I={\int }_0^a\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}dx$ _______ (2)

$\left(1\right)+\left(2\right)\Rightarrow$

$2I={\int }_0^{a}1dx$

$=a$

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