$\int\limits_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{2 \pi-x}} d x$ is equal to
Answer & explanation
Correct answer: option 3
$I =\int\limits_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{2 \pi-x}} d x$ ....(1)
$ =\int\limits_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} \frac{\sqrt{2 \pi-x}}{\sqrt{2 \pi-x}+\sqrt{x}} d x$ ....(2)
as $\int\limits_a^b f(x) d x=\int\limits_a^b f(a+b-x) d x$
eq. (1) + eq. (2)
⇒ $2 I=\int\limits_{\pi / 2}^{3 \pi / 2} \frac{\sqrt{x}+\sqrt{2 \pi-x}}{\sqrt{x}+\sqrt{2 \pi-x}} d x$
$\Rightarrow 2 I=\int\limits_{\pi / 2}^{3 \pi / 2} 1 d x$
$\Rightarrow 2 I=\left[\frac{3 \pi}{2} - \frac{\pi}{2}\right]$
so $2 I=\pi$
so $I = \frac{\pi}{2}$