1. **State the problem:** Find the average value of the function $f(x) = \sqrt{4 - x^2}$ on the interval $[-2, 2]$. 2. **Formula for average value:** The average value $f_{avg}$ of a continuous function $f(x)$ on $[a,b]$ is given by $$f_{avg} = \frac{1}{b - a} \int_a^b f(x) \, dx$$ 3. **Apply the formula:** Here, $a = -2$ and $b = 2$, so $$f_{avg} = \frac{1}{2 - (-2)} \int_{-2}^2 \sqrt{4 - x^2} \, dx = \frac{1}{4} \int_{-2}^2 \sqrt{4 - x^2} \, dx$$ 4. **Interpret the integral:** The function $f(x) = \sqrt{4 - x^2}$ represents the upper half of a circle with radius 2 centered at the origin. 5. **Calculate the integral:** The integral $$\int_{-2}^2 \sqrt{4 - x^2} \, dx$$ is the area of the upper semicircle of radius 2. 6. **Area of semicircle:** The area of a full circle is $\pi r^2 = \pi \times 2^2 = 4\pi$. The area of the upper semicircle is half of that: $$\frac{1}{2} \times 4\pi = 2\pi$$ 7. **Compute average value:** Substitute the integral value back: $$f_{avg} = \frac{1}{4} \times 2\pi = \frac{2\pi}{4} = \frac{\pi}{2}$$ **Final answer:** The average value of $f(x)$ on $[-2, 2]$ is $$\boxed{\frac{\pi}{2}}$$