1. **State the problem:** Find the area of Region S bounded by the curves $f(x) = 2 - \frac{1}{4}x^2$ and $g(x) = -\cos\left(\frac{\pi}{3}x\right)$ from $x=0$ to $x=2$. 2. **Formula for area between curves:** $$\text{Area} = \int_0^2 \bigl(f(x) - g(x)\bigr) \, dx$$ where $f(x)$ is the upper curve and $g(x)$ is the lower curve. 3. **Set up the integral:** $$\int_0^2 \left(2 - \frac{1}{4}x^2 - \left(-\cos\left(\frac{\pi}{3}x\right)\right)\right) dx = \int_0^2 \left(2 - \frac{1}{4}x^2 + \cos\left(\frac{\pi}{3}x\right)\right) dx$$ 4. **Evaluate the integral step-by-step:** - Integral of $2$ from 0 to 2 is $2x \big|_0^2 = 4$ - Integral of $-\frac{1}{4}x^2$ is $-\frac{1}{4} \cdot \frac{x^3}{3} = -\frac{x^3}{12}$, evaluated from 0 to 2: $$-\frac{2^3}{12} = -\frac{8}{12} = -\frac{2}{3}$$ - Integral of $\cos\left(\frac{\pi}{3}x\right)$: Use substitution $u = \frac{\pi}{3}x$, so $du = \frac{\pi}{3} dx$, $dx = \frac{3}{\pi} du$. $$\int_0^2 \cos\left(\frac{\pi}{3}x\right) dx = \int_0^{\frac{2\pi}{3}} \cos u \cdot \frac{3}{\pi} du = \frac{3}{\pi} \sin u \bigg|_0^{\frac{2\pi}{3}} = \frac{3}{\pi} \left(\sin \frac{2\pi}{3} - 0\right) = \frac{3}{\pi} \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2\pi}$$ 5. **Sum all parts:** $$4 - \frac{2}{3} + \frac{3\sqrt{3}}{2\pi} \approx 4 - 0.6667 + 0.8279 = 4.1612$$ **Final answer:** $$\boxed{\text{Area of Region S} \approx 4.16}$$