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

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(C) The function is $y = |\sin 2x|$. We need to find the area bounded by this curve and the $X$-axis in the interval $[0, 2\pi]$.Since the period of $

Vedclass · Accepted Answer

$4$

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(C) The function is $y = |\sin 2x|$. We need to find the area bounded by this curve and the $X$-axis in the interval $[0, 2\pi]$.Since the period of $|\sin 2x|$ is $\frac{\pi}{2}$, the function repeats its shape every $\frac{\pi}{2}$ units.In the interval $[0, 2\pi]$, there are $4$ such identical humps (from $0$ to $\frac{\pi}{2}$, $\frac{\pi}{2}$ to $\pi$, $\pi$ to $\frac{3\pi}{2}$, and $\frac{3\pi}{2}$ to $2\pi$).Thus, the total area $A$ is given by:$A = \int_0^{2\pi} |\sin 2x| \, dx = 4 \int_0^{\frac{\pi}{2}} \sin 2x \, dx$$A = 4 \left[ \frac{-\cos 2x}{2} \right]_0^{\frac{\pi}{2}}$$A = -2 [\cos(2 \cdot \frac{\pi}{2}) - \cos(0)]$$A = -2 [\cos(\pi) - \cos(0)]$$A = -2 [-1 - 1] = -2 [-2] = 4$Therefore, the area is $4$ square units.

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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