Area under curve

By Calculus, we first substitute: $x=\sin{\theta} \quad \Rightarrow dx = \cos{\theta}\space d\theta$ , then:

$\displaystyle I = \int_0^{\frac{1}{2}} {\sqrt{1-x^2)}\space dx} = \int_0^{\frac{\pi}{6}} {\cos^2 {\theta}\space dx} = \frac {1}{2} \int_0^{\frac{\pi}{6}} {(\cos {2\theta} + 1) \space dx}$

$\displaystyle = \frac {1}{2} \left[ \frac {\sin{2\theta}}{2}+\theta \right] _0^{\frac{\pi}{6}} = \frac {1}{2} \left[ \frac {\sqrt{3}}{4}+\frac{\pi}{6} \right] = \frac {\sqrt{3}}{8}+\frac{\pi}{12} \approx \boxed{0.478}$

By Geometry; the integral $I$ is the area of the shaded region in the first quadrant of the unit-radius circle (see figure below). It is the area of the triangle plus the area of the circle segment.

Therefore, we have:

$\displaystyle I = \frac {1}{2}\dot{}\frac{1}{2}\dot{}\frac{\sqrt{3}}{3} + \frac {\frac{\pi}{6}}{2\pi} \pi(1^2)= \frac {\sqrt{3}}{8}+\frac{\pi}{12} \approx \boxed{0.478}$

This is the integral for finding the area inside the unit circle above the $x$ -axis for $0 \le x \le \frac{1}{2}$ . This region can be split up into two smaller regions: a right triangle with base $\frac{1}{2}$ and height $\frac{\sqrt{3}}{2}$ and a circular sector with radius $1$ and angle measure $\frac{\pi}{6}$ radians. Thus:

$\int_0^{0.5} \sqrt{1 - x^2} \: dx = A_{triangle} + A_{sector} \\ = \frac{1}{2}\cdot\frac{1}{2}\cdot\frac{\sqrt{3}}{2} + \frac{1}{2}\cdot\frac{\pi}{6}\cdot 1^2 = \frac{\sqrt{3}}{8} + \frac{\pi}{12} \approx \boxed{0.478}$