Question

The area bounded by the curve $y=\cos x$, the $x-axis$ and the ordinates $x=0$and $x=2\mathrm{\pi }$

Answer 1

The area bounded by the curve $y=\cos x$, the $x-axis$ and the ordinates $x=0$and $x=2\mathrm{\pi }$

is given by-

Area bounded $=$${\int }_0^{2\mathrm{\pi }}\left|y\right|dx$

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

Hence, the area bounded is$4sq.units$