1. We are asked to evaluate the definite integral $$\int_0^2 \sqrt{4 - x^2} \, dx$$. 2. This integral represents the area under the curve of the function $y = \sqrt{4 - x^2}$ from $x=0$ to $x=2$. 3. Notice that $y = \sqrt{4 - x^2}$ is the upper half of a circle centered at the origin with radius 2, since $x^2 + y^2 = 4$. 4. The integral from 0 to 2 corresponds to a quarter of the circle's area (the first quadrant). 5. The area of a full circle with radius 2 is $$\pi \times 2^2 = 4\pi$$. 6. Therefore, the area for the quarter circle is $$\frac{1}{4} \times 4\pi = \pi$$. 7. Hence, $$\int_0^2 \sqrt{4 - x^2} \, dx = \pi$$. 8. For verification, we can use trigonometric substitution: let $x = 2\sin \theta$, then $dx = 2\cos \theta \, d\theta$. 9. When $x=0$, $\theta=0$; when $x=2$, $\theta=\frac{\pi}{2}$. 10. Substitute into the integral: $$\int_0^{\pi/2} \sqrt{4 - 4\sin^2 \theta} \times 2\cos \theta \, d\theta = \int_0^{\pi/2} \sqrt{4(1 - \sin^2 \theta)} \times 2\cos \theta \, d\theta$$ 11. Simplify inside the square root: $$\sqrt{4\cos^2 \theta} = 2|\cos \theta|$$ 12. Since $\theta$ is in $[0, \pi/2]$, $\cos \theta \geq 0$, so $|\cos \theta| = \cos \theta$. 13. The integral becomes: $$\int_0^{\pi/2} 2\cos \theta \times 2\cos \theta \, d\theta = \int_0^{\pi/2} 4\cos^2 \theta \, d\theta$$ 14. Use the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$: $$\int_0^{\pi/2} 4 \times \frac{1 + \cos 2\theta}{2} \, d\theta = \int_0^{\pi/2} 2(1 + \cos 2\theta) \, d\theta$$ 15. Integrate term by term: $$2 \int_0^{\pi/2} 1 \, d\theta + 2 \int_0^{\pi/2} \cos 2\theta \, d\theta = 2 \left[ \theta \right]_0^{\pi/2} + 2 \left[ \frac{\sin 2\theta}{2} \right]_0^{\pi/2}$$ 16. Evaluate: $$2 \times \frac{\pi}{2} + 2 \times \frac{\sin \pi - \sin 0}{2} = \pi + 0 = \pi$$ 17. Final answer: $$\boxed{\pi}$$