Question

Show that the areas under the curves y = sin x and y = sin 2x between x = 0 and x = $\frac{\mathrm{\pi }}{3}$ are in the ratio 2 : 3.

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{shaded}\mathrm{area}{\mathrm{A}}_1\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{area}\mathrm{bound}\mathrm{by}\mathrm{the}\mathrm{curve}y=\sin x\mathrm{and}x=0,x=\frac{\pi }{3} \\ \mathrm{The}\mathrm{shaded}\mathrm{area}{\mathrm{A}}_2\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{area}\mathrm{bound}\mathrm{by}\mathrm{the}\mathrm{curve}y=\sin 2x\mathrm{and}x=0,x=\frac{\pi }{3} \\ \therefore {\mathrm{A}}_1={\int }_0^{\frac{\mathrm{\pi }}{3}}\left|y\right|dx \\ \Rightarrow {\mathrm{A}}_1={\int }_0^{\frac{\mathrm{\pi }}{3}}ydx\left[\mathrm{As},y>0,\left|y\right|=y\right] \\ \Rightarrow {\mathrm{A}}_1={\int }_0^{\frac{\mathrm{\pi }}{3}}\sin xdx \\ \Rightarrow {\mathrm{A}}_1={\left[-\cos x\right]}_0^{\frac{\mathrm{\pi }}{3}} \\ \Rightarrow {\mathrm{A}}_1=-\cos \frac{\mathrm{\pi }}{3}+\cos 0 \\ \Rightarrow {\mathrm{A}}_1=-\frac{1}{2}+1=\frac{1}{2}...\left(1\right) \\ \mathrm{And}, \\ {A}_2={\int }_0^{\frac{\mathrm{\pi }}{3}}\left|y\right|dx \\ \Rightarrow A_2={\int }_0^{\frac{\mathrm{\pi }}{3}}ydx\left[\mathrm{As},y>0,\left|y\right|=y\right] \\ \Rightarrow A_2={\int }_0^{\frac{\mathrm{\pi }}{3}}\sin 2xdx \\ \Rightarrow A_2={\int }_0^{\frac{\mathrm{\pi }}{3}}2\sin x\cos xdx \\ \Rightarrow A_2=2{\left[\frac{-1}{4}\cos 2x\right]}_0^{\frac{\mathrm{\pi }}{3}} \\ \Rightarrow A_2=\frac{1}{2}\left[-\cos \frac{2\mathrm{\pi }}{3}+\cos 0\right] \\ \Rightarrow A_2=\frac{1}{2}\left[1+\frac{1}{2}\right]=\frac{1}{2}\times \frac{3}{2}...\left(2\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right) \\ \frac{A_1}{A_2}=\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{3}{2}}}=\frac{2}{3} \\ \mathrm{Thus},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{curve}y=\sin x\mathrm{and}y=\sin 2x\mathrm{for}x=0\mathrm{and}x=\frac{\mathrm{\pi }}{3}\mathrm{are}\mathrm{in}\mathrm{the}\mathrm{ratio}2:3 \end{gathered}$$