Question

${\int }_1^3\frac{\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}}dx=$

• A. 0
• B. 1
• C. 3
• D. 2

Answer 1

Answer: (B)

Let $I={\int }_1^3\frac{\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}}dx...(i)$
$\Rightarrow I={\int }_1^3\frac{\sqrt{4-(4-x)}}{\sqrt{4-x}+\sqrt{4-(4-x)}}dx$
$\left[\because {\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx\right]$
$\Rightarrow I={\int }_1^3\frac{\sqrt{x}}{\sqrt{4-x}+\sqrt{x}}dx...(ii)$
On adding Eqs. (i) and (ii), we get
$2I={\int }_1^{3}1dx=[x{]}_1^3$
$I=\frac{2}{2}=1$