Find the area of the region bounded by and axis using integration.

The curve $y = \sqrt{4 - x^2}$ represents the upper semicircle of $x^2 + y^2 = 4$ for $y \ge 0$, bounded by $x = -2$ and $x = 2$. Required area bounded by the curve and the $x$-axis: $A = \int_{-2}^{2} \sqrt{4 - x^2} \, dx = 2 \int_{0}^{2} \sqrt{2^2 - x^2} \, dx$. Using $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$: $A = 2 \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2} = 2 \left[ 0 + 2\sin^{-1}(1) - 0 \right] = 2 \times 2\left(\frac{\pi}{2}\right) = 2\pi\text{ sq units}$.