Find the area of the region enclosed by the curves \( y=\sin x \) and \( y=\sin 2 x \) for \( 0 \leq x \leq \pi \).

To find the area enclosed by the curves \( y = \sin x \) and \( y = \sin 2x \) from \( 0 \) to \( \pi \), we first need to find the points where these two curves intersect. Setting \( \sin x = \sin 2x \) gives us: \[ \sin x = 2 \sin x \cos x \] Rearranging this leads to: \[ \sin x (1 - 2 \cos x) = 0 \] This equation yields two cases: 1. \( \sin x = 0 \) -> which gives \( x = 0 \) and \( x = \pi \). 2. \( 1 - 2 \cos x = 0 \) -> which yields \( \cos x = \frac{1}{2} \) -> hence \( x = \frac{\pi}{3} \) (in the given interval). Now we have the points of intersection: \( x = 0 \), \( x = \frac{\pi}{3} \), and \( x = \pi \). Next, we determine which function is on top in the interval \( [0, \pi] \). We can analyze the functions at the midpoint \( x = \frac{\pi}{6} \): \[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}, \quad \sin\left(2 \cdot \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \] Thus, \( y = \sin 2x \) (above) is larger than \( y = \sin x \) from \( 0 \) to \( \frac{\pi}{3} \), and \( y = \sin x \) is above \( y = \sin 2x \) from \( \frac{\pi}{3} \) to \( \pi \). Now, we can compute the area \( A \) as: \[ A = \int_0^{\frac{\pi}{3}} (\sin 2x - \sin x) \, dx + \int_{\frac{\pi}{3}}^{\pi} (\sin x - \sin 2x) \, dx \] Calculating the first integral: \[ \int_0^{\frac{\pi}{3}} (\sin 2x - \sin x) \, dx = \left[-\frac{1}{2} \cos 2x + \cos x \right]_0^{\frac{\pi}{3}} \] Evaluating at the bounds: \[ = \left(-\frac{1}{2} \cos\left(\frac{2\pi}{3}\right) + \cos\left(\frac{\pi}{3}\right)\right) - \left(-\frac{1}{2} \cos(0) + \cos(0)\right) \] \[ = \left(-\frac{1}{2} \left(-\frac{1}{2}\right) + \frac{1}{2}\right) - \left(-\frac{1}{2} \cdot 1 + 1\right) \] \[ = \left(\frac{1}{4} + \frac{1}{2}\right) - \left(-\frac{1}{2} + 1\right) = \frac{3}{4} - \frac{1}{2} = \frac{1}{4} \] Now the second integral: \[ \int_{\frac{\pi}{3}}^{\pi} (\sin x - \sin 2x) \, dx = \left[-\cos x + \frac{1}{2} \cos 2x\right]_{\frac{\pi}{3}}^{\pi} \] Evaluating at the bounds: \[ = \left[-\cos(\pi) + \frac{1}{2} \cos(2\pi)\right] - \left[-\cos\left(\frac{\pi}{3}\right) + \frac{1}{2} \cos\left(\frac{2\pi}{3}\right)\right] \] \[ = \left[-(-1) + \frac{1}{2} \cdot 1\right] - \left[-\frac{1}{2} + \frac{1}{2}\left(-\frac{1}{2}\right)\right] \] \[ = \left[1 + \frac{1}{2}\right] - \left[-\frac{1}{2