The area (in square units) of the region boun | vedclass

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(D) The required area $A$ is given by the integral $A = \int_0^{2\pi} |\sin 2x| \, dx$. Since the function $f(x) = |\sin 2x|$ is periodic with period

Vedclass · Accepted Answer

$4$

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(D) The required area $A$ is given by the integral $A = \int_0^{2\pi} |\sin 2x| \, dx$. Since the function $f(x) = |\sin 2x|$ is periodic with period $\frac{\pi}{2}$,the area over $[0, 2\pi]$ consists of $4$ identical humps,each over an interval of length $\frac{\pi}{2}$. Thus,$A = 4 \int_0^{\pi/2} \sin 2x \, dx$. Evaluating the integral: $A = 4 \left[ -\frac{\cos 2x}{2} \right]_0^{\pi/2}$. $A = 4 \left( -\frac{1}{2} (\cos \pi - \cos 0) \right) = 4 \left( -\frac{1}{2} (-1 - 1) \right) = 4 \left( -\frac{1}{2} (-2) \right) = 4(1) = 4$. Therefore,the area is $4$ square units. Hence,option $D$ is correct.

The area (in square units) of the region bounded by the curve $y = |\sin 2x|$ and the $X$-axis in the interval $[0, 2\pi]$ is:

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