Question

The ratio of areas under the curves \(y=sinx \) and \(y=sin2x\) ,from \(x=0\) to  \( x=\frac{\pi}{3}\) is

• A. \(3:2\)
• B. \(2:3\)
• C. \(\sqrt{3}:2\)
• D. \(1:\sqrt{3}\)

Answer 1

The Correct Option is B

To determine the ratio of the areas under the curves \(y=\sin x\) and \(y=\sin 2x\) from \(x=0\) to \(x=\frac{\pi}{3}\), we need to evaluate the definite integrals of these functions over the given interval.

The area under \(y=\sin x\) from \(x=0\) to \(x=\frac{\pi}{3}\) is given by the integral: \[\int_0^{\frac{\pi}{3}}\sin x \, dx\]

The antiderivative of \(\sin x\) is \(-\cos x\), so we have: \[-\cos x \bigg|_0^{\frac{\pi}{3}} = -\cos\left(\frac{\pi}{3}\right) + \cos(0)\]

This evaluates to: \[-\frac{1}{2} + 1 = \frac{1}{2}\]

Now, for the area under \(y=\sin 2x\), the integral is: \[\int_0^{\frac{\pi}{3}} \sin 2x \, dx\]

We use the substitution \(u=2x\), \(du=2\,dx\), hence \(dx=\frac{du}{2}\). The limits of integration change accordingly: when \(x=0\), \(u=0\); and when \(x=\frac{\pi}{3}\), \(u=\frac{2\pi}{3}\). This transforms the integral to: \[\frac{1}{2}\int_0^{\frac{2\pi}{3}} \sin u \, du\]

The antiderivative of \(\sin u\) is \(-\cos u\), so: \[-\frac{1}{2}\cos u \bigg|_0^{\frac{2\pi}{3}} = -\frac{1}{2}\left(\cos\left(\frac{2\pi}{3}\right)\right) + \frac{1}{2}\left(\cos(0)\right)\]

Evaluating this gives: \[-\frac{1}{2}\left(-\frac{1}{2}\right) + \frac{1}{2}\cdot1 = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}\]

Thus, the area under each curve is: \(\frac{1}{2}\) for \(y=\sin x\) and \(\frac{3}{4}\) for \(y=\sin 2x\).

The ratio of these areas is \(\frac{1/2}{3/4}=\frac{1}{2}\cdot\frac{4}{3}=\frac{2}{3}\).

Thus, the ratio of the areas is \(2:3\).