The ratio of areas under the curves \(y = \sin x\) and \(y = \sin 2x\), from \(x = 0\) to \(x = \displaystyle\frac{\pi}{3}\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2 : 3
On \(\left[0, \frac{\pi}{3}\right]\) both \(\sin x\) and \(\sin 2x\) are non-negative (\(2x\) goes only up to \(\frac{2\pi}{3}\)), so each area under the curve is the definite integral.
\( \displaystyle A_1 = \int_{0}^{\pi/3}\sin x\,dx = \Big[-\cos x\Big]_{0}^{\pi/3} \)
\( \displaystyle = -\frac{1}{2} + 1 = \frac{1}{2} \)
\( \displaystyle A_2 = \int_{0}^{\pi/3}\sin 2x\,dx = \Big[-\frac{\cos 2x}{2}\Big]_{0}^{\pi/3} \)
\( \displaystyle = -\frac{1}{2}\left(-\frac{1}{2}\right) + \frac{1}{2} = \frac{3}{4} \)
So
\( \displaystyle A_1 : A_2 = \frac{1}{2} : \frac{3}{4} = 2 : 3 \)