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 $y = \sin 2x$ from $x = 0$ to $x = \pi$.$A = \int_{0}^{\pi} |\sin 2x|

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$2$

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(B) The area $A$ is given by the integral of the absolute value of the function $y = \sin 2x$ from $x = 0$ to $x = \pi$.$A = \int_{0}^{\pi} |\sin 2x| \, dx$Since $\sin 2x$ changes sign at $x = \frac{\pi}{2}$,we split the integral:$A = \int_{0}^{\pi/2} \sin 2x \, dx + \int_{\pi/2}^{\pi} -\sin 2x \, dx$Evaluating the first part:$\int_{0}^{\pi/2} \sin 2x \, dx = [-\frac{\cos 2x}{2}]_{0}^{\pi/2} = -\frac{1}{2}(\cos \pi - \cos 0) = -\frac{1}{2}(-1 - 1) = 1$Evaluating the second part:$\int_{\pi/2}^{\pi} -\sin 2x \, dx = [\frac{\cos 2x}{2}]_{\pi/2}^{\pi} = \frac{1}{2}(\cos 2\pi - \cos \pi) = \frac{1}{2}(1 - (-1)) = 1$Total Area $A = 1 + 1 = 2$ sq. units.

The area of the region bounded by the curve $y = \sin 2x$,the $x$-axis,and the lines $x = 0$ and $x = \pi$ is . . . . . . sq. units.

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