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

$2- \sqrt{2}$ sq.units

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

The curves $y = \sin x$ and $y = \cos x$ intersect at $x = \frac{\pi}{4}$.

• From $0$ to $\frac{\pi}{4}$, $\sin x < \cos x$, so the region is bounded by the x-axis and $\sin x$.
• From $\frac{\pi}{4}$ to $\frac{\pi}{2}$, $\cos x < \sin x$, so the region is bounded by the x-axis and $\cos x$.

Hence, total area:

$\text{Area} = \int_{0}^{\pi/4} \sin x \, dx + \int_{\pi/4}^{\pi/2} \cos x \, dx$

Evaluating:

• $\int_{0}^{\pi/4} \sin x \, dx = 1 - \frac{\sqrt{2}}{2}$
• $\int_{\pi/4}^{\pi/2} \cos x \, dx = 1 - \frac{\sqrt{2}}{2}$

Adding both:

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