The area between x-axis and curve, , y= cos x when $0\le \mathrm{x}\le 2\mathrm{\pi }$ is

From the given question, $\mathrm{y}=\cos \mathrm{x},0\le \mathrm{x}\le 2\mathrm{\pi }$

$$\begin{gathered} \Rightarrow {\int }_0^{2\mathrm{\pi }} \mathrm{ydx}={\int }_0^{\frac{\mathrm{\pi }}{2}} \mathrm{ydx}-{\int }_{\frac{\mathrm{\pi }}{2}}^{\frac{3\mathrm{\pi }}{2}} \mathrm{ydx}+{\int }_{\frac{3\mathrm{\pi }}{2}}^{2\mathrm{\pi }} \mathrm{ydx} \\ \Rightarrow {\int }_0^{2\mathrm{\pi }} \mathrm{ydx}={\int }_0^{\frac{\mathrm{\pi }}{2}} \cos \mathrm{xdx}-{\int }_{\frac{1}{2}}^{\frac{3\mathrm{\pi }}{2}} \cos \mathrm{xdx}+{\int }_{\frac{3\mathrm{\pi }}{2}}^{2\mathrm{\pi }} \cos \mathrm{xdx} \\ \Rightarrow {\int }_{ }^{2\mathrm{\pi }} \mathrm{ydx}=[\sin \mathrm{x}{]}_0^{\frac{\mathrm{\pi }}{2}}-[\sin \mathrm{x}{]}_{\frac{\mathrm{\pi }}{2}}^{\frac{3\mathrm{\pi }}{2}}+[\sin \mathrm{x}{]}_{\frac{3\mathrm{\pi }}{2}}^{2\mathrm{\pi }} \\ {\int }_0^{2\mathrm{\pi }} y \mathrm{dx}=1-0-(-1-1)+(0+1)=4 \end{gathered}$$

Hence, correct option is (b).