1. **State the problem:** We need to find the area of the region enclosed by the curves $y = \cos x$ and $y = 2 - 2\cos x$ over the interval $0 < x < 2$. 2. **Identify the curves and interval:** The curves are: - $y_1 = \cos x$ - $y_2 = 2 - 2\cos x$ The interval is $0 < x < 2$. 3. **Find points of intersection:** Set $\cos x = 2 - 2\cos x$ to find where the curves intersect. $$\cos x = 2 - 2\cos x$$ $$3\cos x = 2$$ $$\cos x = \frac{2}{3}$$ Find $x$ such that $\cos x = \frac{2}{3}$ in $0 < x < 2$: $$x = \arccos\left(\frac{2}{3}\right) \approx 0.8411$$ 4. **Determine which curve is on top:** For $x=1$ (between 0 and 2), evaluate: - $y_1 = \cos 1 \approx 0.5403$ - $y_2 = 2 - 2\cos 1 = 2 - 2(0.5403) = 2 - 1.0806 = 0.9194$ Since $y_2 > y_1$, $y_2$ is the upper curve and $y_1$ is the lower curve on the interval. 5. **Set up the integral for the area:** The area $A$ is given by: $$A = \int_0^2 \left( y_2 - y_1 \right) dx = \int_0^2 \left( 2 - 2\cos x - \cos x \right) dx = \int_0^2 \left( 2 - 3\cos x \right) dx$$ 6. **Calculate the integral:** $$\int_0^2 2 dx = 2x \Big|_0^2 = 4$$ $$\int_0^2 3\cos x dx = 3 \sin x \Big|_0^2 = 3(\sin 2 - \sin 0) = 3 \sin 2$$ So, $$A = 4 - 3 \sin 2$$ 7. **Final answer:** The area enclosed by the curves is: $$\boxed{4 - 3 \sin 2}$$ This represents the exact area of the region bounded by the given curves on $0 < x < 2$.