The area (in sq. units) of the region bounded by the curve $y = \sin x, -2\pi \leq x \leq 2\pi$ and $x - axis$ is equal to

The curve $y = \sin x$ oscillates above and below the $x$-axis. Area is always positive, so the absolute value of the function must be used where it is negative.

For the interval $-2\pi \leq x \leq 2\pi$:

$\sin x \geq 0$ on $[-\pi, 0]$ and $[0, \pi]$

$\sin x < 0$ on $[-2\pi, -\pi]$ and $[\pi, 2\pi]$

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For $0 \leq x \leq \pi$, where $\sin x \geq 0$:

Area $= \int_0^\pi \sin x \, dx$

$= [-\cos x]_0^\pi$

$= -\cos \pi - (-\cos 0)$

$= -(-1) + 1$

$= 2$

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For $\pi \leq x \leq 2\pi$, where $\sin x < 0$:

Area $= \int_\pi^{2\pi} |\sin x| \, dx$

$= \int_\pi^{2\pi} (-\sin x) \, dx$

$= [\cos x]_\pi^{2\pi}$

$= \cos 2\pi - \cos \pi$

$= 1 - (-1)$

$= 2$

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For $-\pi \leq x \leq 0$, where $\sin x \geq 0$:

Area $= \int_{-\pi}^0 \sin x \, dx$

$= [-\cos x]_{-\pi}^0$

$= -\cos 0 - (-\cos(-\pi))$

$= -1 + (-1)$

$= -2$

Taking the absolute value: Area $= 2$

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For $-2\pi \leq x \leq -\pi$, where $\sin x < 0$:

Area $= \int_{-2\pi}^{-\pi} |\sin x| \, dx$

$= \int_{-2\pi}^{-\pi} (-\sin x) \, dx$

$= [\cos x]_{-2\pi}^{-\pi}$

$= \cos(-\pi) - \cos(-2\pi)$

$= -1 - 1$

$= -2$

Taking the absolute value: Area $= 2$

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Total Area $= 2 + 2 + 2 + 2$

$= 8$

Therefore, the area of the region bounded by the curve and the $x$-axis is $8$ square units.