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

• $\sqrt 2$ sq units

• $(\sqrt 2 + 1)$ sq units

• $(\sqrt 2 - 1)$ sq units

• $(2\sqrt 2 - 1)$ sq units

Correct Option (c)

We have $y = \cos x$ and $y = \sin x,$ where $0 \le x \le \frac{\pi }{2}$

Solving curve, we get $\sin x = \cos x$
$\therefore \quad x = \frac{\pi }{4}$
Graphs are as shown in the following figure.

From the figure, area of the shaded region,
$A = \int_0^{\pi /4} {(\cos x - \sin x)} dx$
$= [\sin x + \cos x]_0^{\pi /4}$
$= \left( {\sin \frac{\pi }{4} + \cos \frac{\pi }{4} - \sin 0 - \cos 0} \right)$
$= \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} - 1 = (\sqrt 2 - 1)$ sq. units