The area of the region bounded by the curve y | vedclass

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(B) The area $A$ is given by the integral of the absolute value of the function: $A = \int_{-\pi}^{\frac{3\pi}{2}} |\sin x| dx$ We split the integral

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$5 	ext{ (unit)}^2$

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(B) The area $A$ is given by the integral of the absolute value of the function: $A = \int_{-\pi}^{\frac{3\pi}{2}} |\sin x| dx$ We split the integral based on the sign of $\sin x$: $A = \int_{-\pi}^{0} |\sin x| dx + \int_{0}^{\pi} |\sin x| dx + \int_{\pi}^{\frac{3\pi}{2}} |\sin x| dx$ Since $\sin x \le 0$ for $x \in [-\pi, 0]$ and $x \in [\pi, \frac{3\pi}{2}]$,and $\sin x \ge 0$ for $x \in [0, \pi]$: $A = \int_{-\pi}^{0} (-\sin x) dx + \int_{0}^{\pi} (\sin x) dx + \int_{\pi}^{\frac{3\pi}{2}} (-\sin x) dx$ $A = [\cos x]_{-\pi}^{0} + [-\cos x]_{0}^{\pi} + [\cos x]_{\pi}^{\frac{3\pi}{2}}$ $A = (\cos 0 - \cos(-\pi)) + (-\cos \pi + \cos 0) + (\cos(\frac{3\pi}{2}) - \cos \pi)$ $A = (1 - (-1)) + (-(-1) + 1) + (0 - (-1))$ $A = (1 + 1) + (1 + 1) + (0 + 1) = 2 + 2 + 1 = 5 	ext{ (unit)}^2$

The area of the region bounded by the curve $y=\sin x$ between $x=-\pi$ and $x=\frac{3\pi}{2}$ is

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