Practicing Success
$\int\limits_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{2 \pi-x}} d x$ is equal to |
$\frac{\pi}{4}$ $\pi$ $\frac{\pi}{2}$ 1 |
$\frac{\pi}{2}$ |
$I =\int\limits_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{2 \pi-x}} d x$ ....(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 $I = \frac{\pi}{2}$ |