Question

${\int }_{\frac{a}{4}}^{\frac{3a}{4}}\frac{\sqrt{x}}{\sqrt{a-x}+\sqrt{x}}dx$ is equal to

• A. $\frac{a}{4}$
• B. $a$
• C. $-a$
• D. none of these

Answer 1

The correct option is A $\frac{a}{4}$

Let $I={\int }_{\frac{a}{4}}^{\frac{3a}{4}}\frac{\sqrt{x}}{\sqrt{a-x}+\sqrt{x}}dx$ ....$\left(1\right)$

Replace $x\to a-x$

$={\int }_{\frac{a}{4}}^{\frac{3a}{4}}\frac{\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}dx$ ....$\left(2\right)$

Adding eqns$\left(1\right)$ and $\left(2\right)$ we get

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

$\Rightarrow 2I={\int }_{\frac{a}{4}}^{\frac{3a}{4}}\frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}dx$

$\Rightarrow 2I={\int }_{\frac{a}{4}}^{\frac{3a}{4}}1.dx$

$\Rightarrow 2I={\left[x\right]}_{\frac{a}{4}}^{\frac{3a}{4}}$

$\Rightarrow 2I=\frac{3a}{4}-\frac{a}{4}=\frac{2a}{4}$

$\therefore I=\frac{a}{4}$