Question

$\int \limits^{2017}_{2016} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{4033 - x}} dx$ is equal to

• A. 44565
• B. 44622
• C. 42767
• D. 44563

Answer 1

Answer: (D)

44563

Answer 2

Let $I=\int\limits_{2016}^{2017} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{4033-x}} d x\,\,\,...(i)$
$\therefore I=\int\limits_{2016}^{2017} \frac{\sqrt{4033-x}}{\sqrt{4033-X}+\sqrt{4033-(4033-x)}} d x$
$\left[\because \int\limits_{a}^{b} f(x) d x=\int_{b}^{a} f(a+b-x) d x\right]$
$\Rightarrow I=\int\limits_{2016}^{2017} \frac{\sqrt{4033-X}}{\sqrt{4033-X}+\sqrt{x}} d x \ldots$ (ii)
On adding Eqs. (i) and (ii), we get
$2 I=\int\limits_{2016}^{2017} d x$
$\Rightarrow 2\, I=[x]_{2016}^{2017}$
$\Rightarrow 2 \,I=1$
$\Rightarrow I=\frac{1}{2}$