The area of the region bounded by the Y-axis $y = \cos x$ and $y = \sin x$, where $0 \leq x \leq \frac{\pi}{2}$, is

$(\sqrt{2} - 1)$ sq units

The correct answer is Option (3) → $(\sqrt{2} - 1)$ sq units

We have, Y-axis i.e., $x = 0$, $y = \cos x$ and $y = \sin x$, where $0 \leq x \leq \frac{\pi}{2}$

$∴\text{Required area} = \int_{0}^{\pi/4} (\cos x - \sin x) dx$

$= [\sin x]_{0}^{\pi/4} + [\cos x]_{0}^{\pi/4}$

$= \left( \sin \frac{\pi}{4} - \sin 0 \right) + \left( \cos \frac{\pi}{4} - \cos 0 \right)$

$= \left( \frac{1}{\sqrt{2}} - 0 \right) + \left( \frac{1}{\sqrt{2}} - 1 \right)$

$\left[ ∵\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}, \sin 0 = 0, \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}, \cos 0 = 1 \right]$

$= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - 1 = -1 + \frac{2}{\sqrt{2}} = (-1 + \sqrt{2})$

$= (\sqrt{2} - 1) \text{ sq. units}$