Question

What is the value of the definite integral, ${\int }_0^a\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx?$

• A. zero
• B. $a/2$
• C. $a$
• D. $2a$

Answer 1

The correct option is B $a/2$

$I={\int }_0^a\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx...\left(i\right)$
$I={\int }_0^a\left(\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{a-(a-x)}}\right)dx...\left(ii\right)$
$\left\{Using{\int }_a^{b}f\right(x)dx={\int }_a^{b}f(a+b-x\left)dx\right\}$
$\left(i\right)+\left(ii\right)$
$\Rightarrow 2I={\int }_0^a\left(\frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}\right)dx$
$\Rightarrow 2I={\int }_0^{a}1dx$
$\Rightarrow 2I=(a-0)$
$I=a/2$