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

$\sqrt{2}$ sq.units

Curves: $y = \cos x$ and $y = \sin x$, bounded by $x$-axis, $0 \le x \le \pi/2$

Intersection point: $\sin x = \cos x \Rightarrow x = \pi/4$

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

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

= $[\sin x]_0^{\pi/4} + [-\cos x]_{\pi/4}^{\pi/2}$

= $(\sin (\pi/4) - \sin 0) + (-\cos (\pi/2) + \cos (\pi/4))$

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

= $\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}$

Answer: $\sqrt{2}$