Question

Find the area bounded by the curve y = cos x, x-axis and the ordinates x = 0 and x = 2π.

Answer 1

$$\text{ The shaded region is the required area bound by the curve }y=\cos x,x\text{ axis and }x=0,x=2\pi$$
$$\text{ Consider a vertical strip of length }=\left|y\right|\text{ and width }=dx\text{ in the first quadrant }$$
$$\text{ Area of the approximating rectangle }=\left|y\right|dx$$
$$\text{ The approximating rectangle moves from }x=0\text{ to }x=2\pi$$
$$\text{ Now },0\le x\le \frac{\pi }{2}\text{ and }\frac{3\pi }{2}\le x\le 2\pi ,y>0\Rightarrow \left|y\right|=y$$
$$\frac{\pi }{2}\le x\le \frac{3\pi }{2},y<0\Rightarrow \left|y\right|=-y$$
$$\Rightarrow \text{ Area of the shaded region }={\int }_0^{2\pi }\left|y\right|dx$$
$$\Rightarrow A={\int }_0^{\frac{\pi }{2}}\left|y\right|dx+{\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\left|y\right|dx+{\int }_{\frac{3\pi }{2}}^{2\pi }\left|y\right|dx$$
$$\Rightarrow A={\int }_0^{\frac{\pi }{2}}ydx+{\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}-ydx+{\int }_{\frac{3\pi }{2}}^{2\pi }ydx$$
$$\Rightarrow A={\int }_0^{\frac{\pi }{2}}\cos xdx+{\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}-\cos xdx+{\int }_{\frac{3\pi }{2}}^{2\pi }\cos xdx$$
$$\Rightarrow A={[\sin x]}_0^{\frac{\pi }{2}}+{[-\sin x]}_{\frac{\pi }{2}}^{\frac{3\pi }{2}}+{[-\sin x]}_{\frac{3\pi }{2}}^{2\pi }$$
$$\Rightarrow A=1+(1+1)+(0-(-1))$$
$$\Rightarrow A=4\text{ sq . units }$$
$$\text{ Area bound by the curve }y=\cos x,x-\text{ axis and }x=0,x=2\pi =4\text{ sq . units }$$