Question

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