Problem 147

Points O(0, 0), A(2, 0), B(1, $-2\sqrt{3}$), and P, a moving point on the curve $y = \sqrt{1 - \frac{x^2}{4}}$ are given.

Let P be represented as $(2\cos{\theta}, \sin{\theta})$, where $\theta \in [0, \pi]$.

Then, $\overrightarrow{OP} \cdot \overrightarrow{BA} = 2\cos{\theta} + 2\sqrt{3}\sin{\theta} = 4\sin{(\theta + \frac{\pi}{6})} \in [-2, 4]$.

Therefore, the answer is $\boxed{[-2, 4]}$.

This problem can be solved by setting up P's coordinates using the given conditions, applying the dot product of vectors, and utilizing trigonometric functions of the sum and difference of angles. Accurately setting up P's coordinates is the key to solving this problem. It tests the application of the dot product of vectors, the parametric equation of an ellipse, and trigonometric functions of the sum and difference of angles.