If y = 2 sin x + sin 2x for 0 le x le 2 pi , | vedclass

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(B) The given curve is $y = 2 \sin x + \sin 2x$. For $0 \le x \le \pi$,$y \ge 0$,and for $\pi \le x \le 2 \pi$,$y \le 0$.The area $A$ is given by $\in

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$8$

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(B) The given curve is $y = 2 \sin x + \sin 2x$. For $0 \le x \le \pi$,$y \ge 0$,and for $\pi \le x \le 2 \pi$,$y \le 0$.The area $A$ is given by $\int_0^{2 \pi} |y| \, dx = \int_0^{\pi} (2 \sin x + \sin 2x) \, dx + \int_{\pi}^{2 \pi} -(2 \sin x + \sin 2x) \, dx$.Evaluating the first integral: $\int_0^{\pi} (2 \sin x + \sin 2x) \, dx = [-2 \cos x - \frac{1}{2} \cos 2x]_0^{\pi} = (-2(-1) - \frac{1}{2}(1)) - (-2(1) - \frac{1}{2}(1)) = (2 - 0.5) - (-2 - 0.5) = 1.5 + 2.5 = 4$.Evaluating the second integral: $\int_{\pi}^{2 \pi} -(2 \sin x + \sin 2x) \, dx = -[-2 \cos x - \frac{1}{2} \cos 2x]_{\pi}^{2 \pi} = -[(-2(1) - \frac{1}{2}(1)) - (-2(-1) - \frac{1}{2}(1))] = -[(-2.5) - (1.5)] = -[-4] = 4$.Total area $A = 4 + 4 = 8$.

If $y = 2 \sin x + \sin 2x$ for $0 \le x \le 2 \pi ,$ then the area enclosed by the curve and the $x-$axis is

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