Question

The value of ${\int }_{1/n}^{(an-1)/n}\frac{\sqrt{x}}{\sqrt{a-x}+\sqrt{x}}dx$ is equal to

• A. $\frac{a}{2}$
• B. $\frac{n\cdot a+2}{2n}$
• C. $\frac{n\cdot a-2}{2n}$
• D. $\frac{n\cdot a}{2}$

Answer 1

The correct option is C $\frac{n\cdot a-2}{2n}$

Let, $I={\int }_{1/2}^{(an-1)/n}\frac{\sqrt{x}}{\sqrt{a-x}+\sqrt{x}}dx$
$={\int }_{\frac{1}{n}}^{a-\frac{1}{n}}\frac{\sqrt{x}}{\sqrt{a-x}+\sqrt{x}}dx$ ....(i)
$={\int }_{\frac{1}{n}}^{a-\frac{1}{n}}\frac{\left(\sqrt{\frac{1}{n}+a-\frac{1}{n}-x}\right)}{\sqrt{a-(\frac{1}{n}+a-\frac{1}{n}-x)}+\sqrt{\frac{1}{n}+a-\frac{1}{n}-x}}dx$
$\Rightarrow I={\int }_{\frac{1}{n}}^{a-\frac{1}{n}}\frac{\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}dx$ .... (ii)
On adding Eqs. (i) and (ii), we get
$2I={\int }_{\frac{1}{n}}^{a-\frac{1}{n}}1dx=[x{]}_{\frac{1}{n}}^{a-\frac{1}{n}}$
$=a-\frac{1}{n}-\frac{1}{n}=\frac{na-2}{n}\Rightarrow I=\frac{na-2}{2n}$.