The area (in sq. units) of the region bounded by the curve $y = \sqrt{16-x^2}$ and x-axis is

Given curve: $y = \sqrt{16-x^2}$

$y^2 = 16-x^2$

$x^2 + y^2 = 16$

This is the equation of a circle with center at origin $(0, 0)$ and radius $= \sqrt{16} = 4$ units.

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The original equation is $y = \sqrt{16-x^2}$ with the square root.

Since square root gives only positive values, $y \geq 0$

This represents only the upper half of the circle (above the x-axis).

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When $y = 0$:

$0 = \sqrt{16-x^2}$

$0 = 16-x^2$

$x^2 = 16$

$x = \pm 4$

The curve touches the x-axis at $x = -4$ and $x = 4$.

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The region bounded by the curve and x-axis is a semicircle (upper half).

Area of semicircle $= \frac{1}{2} \times \pi r^2$

Where radius $r = 4$:

Area $= \frac{1}{2} \times \pi \times (4)^2$

Area $= \frac{1}{2} \times \pi \times 16$

Area $= 8\pi$ sq. units

Therefore, the area of the region is $8\pi$ square units.