Find the area of the region bounded by (i) $y=\cos x$ and the x-axis in the interval $[0,2\pi ]$

$A={\int }_0^{\pi /2} \cos xdx+{\int }_{\pi /2}^{3\pi /2} -\cos xdx+{\int }_{3\pi /2}^{2\pi } \cos xdx$

$A=[\sin x{]}_0^{\pi /2}-[\sin x{]}_{\pi /2}^{\frac{3\pi }{2}}+[\sin x{]}_{\frac{3\pi }{2}}^{2\pi }$

$=[1-0]-[-1-1]+[0-(-1\left)\right]$

$A=1+2+1\therefore A=4\text{ sq. units }$

$\therefore$ The area of the region bounded by the curve

$y=\cos x\text{ between }[0,2\pi ]\text{ is }4\text{ sq. units }$