Question

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

Answer 1

\[\text{ The shaded area }A_1\text{ is the required area bound by the curve }y = \sin x\text{ and }x = 0, x = \frac{\pi}{3}\]
\[\text{ The shaded area }A_2 \text{ is the required area bound by the curve }y = \sin 2x\text{ and }x = 0, x = \frac{\pi}{3}\]
\[ \therefore A_1 = \int_0^\frac{\pi}{3} \left| y \right| dx\]
\[ \Rightarrow A_1 = \int_0^\frac{\pi}{3} y dx ................\left[ As, y > 0 , \left| y \right| = y \right]\]
\[ \Rightarrow A_1 = \int_0^\frac{\pi}{3} \sin x dx\]
\[ \Rightarrow A_1 = \left[ - \cos x \right]_0^\frac{\pi}{3} \]
\[ \Rightarrow A_1 = - \cos \frac{\pi}{3} + \cos 0\]
\[ \Rightarrow A_1 = - \frac{1}{2} + 1 = \frac{1}{2} ................\left( 1 \right)\]
And,
\[ A_2 = \int_0^\frac{\pi}{3} \left| y \right| dx\]
\[ \Rightarrow A_2 = \int_0^\frac{\pi}{3} y dx ..............\left[ As, y > 0 , \left| y \right| = y \right]\]
\[ \Rightarrow A_2 = \int_0^\frac{\pi}{3} \sin 2x dx\]
\[ \Rightarrow A_2 = \int_0^\frac{\pi}{3} 2 \sin x \cos x dx\]
\[ \Rightarrow A_2 = 2 \left[ \frac{- 1}{4}\cos 2x \right]_0^\frac{\pi}{3} \]
\[ \Rightarrow A_2 = \frac{1}{2}\left[ - \cos \frac{2\pi}{3} + \cos 0 \right]\]
\[ \Rightarrow A_2 = \frac{1}{2}\left[ 1 + \frac{1}{2} \right] = \frac{1}{2} \times \frac{3}{2} ..............\left( 2 \right)\]
\[\text{ From }\left( 1 \right) \text{ and }\left( 2 \right)\]
\[\frac{A_1}{A_2} = \frac{\frac{1}{2}}{\frac{1}{2} \times \frac{3}{2}} = \frac{2}{3}\]
\[\text{ Thus, the area of curve }y = \sin x\text{ and }y = \sin 2x \text{ for }x = 0\text{ and }x = \frac{\pi}{3}\text{ are in the ratio }2 : 3 \]