1. **State the problem:** Find the area bounded by the polar curve given by $$r = 4 - 3\cos\theta$$. 2. **Formula for area in polar coordinates:** The area enclosed by a polar curve $$r(\theta)$$ from $$\theta = a$$ to $$\theta = b$$ is given by $$\text{Area} = \frac{1}{2} \int_a^b r(\theta)^2 \, d\theta$$ 3. **Apply the formula:** Here, $$r(\theta) = 4 - 3\cos\theta$$, so $$r(\theta)^2 = (4 - 3\cos\theta)^2 = 16 - 24\cos\theta + 9\cos^2\theta$$ 4. **Integral setup:** We integrate from $$0$$ to $$2\pi$$ to cover the full curve: $$\text{Area} = \frac{1}{2} \int_0^{2\pi} (16 - 24\cos\theta + 9\cos^2\theta) \, d\theta$$ 5. **Simplify the integral:** Use the identity $$\cos^2\theta = \frac{1 + \cos 2\theta}{2}$$: $$9\cos^2\theta = 9 \times \frac{1 + \cos 2\theta}{2} = \frac{9}{2} + \frac{9}{2} \cos 2\theta$$ So the integral becomes $$\frac{1}{2} \int_0^{2\pi} \left(16 - 24\cos\theta + \frac{9}{2} + \frac{9}{2} \cos 2\theta \right) d\theta = \frac{1}{2} \int_0^{2\pi} \left(\frac{41}{2} - 24\cos\theta + \frac{9}{2} \cos 2\theta \right) d\theta$$ 6. **Integrate term-by-term:** - $$\int_0^{2\pi} \frac{41}{2} d\theta = \frac{41}{2} \times 2\pi = 41\pi$$ - $$\int_0^{2\pi} -24\cos\theta d\theta = -24 \times 0 = 0$$ (since $$\int_0^{2\pi} \cos\theta d\theta = 0$$) - $$\int_0^{2\pi} \frac{9}{2} \cos 2\theta d\theta = \frac{9}{2} \times 0 = 0$$ (since $$\int_0^{2\pi} \cos 2\theta d\theta = 0$$) 7. **Calculate the area:** $$\text{Area} = \frac{1}{2} \times 41\pi = \frac{41\pi}{2}$$ 8. **Final answer:** $$\boxed{\frac{41\pi}{2}}$$ This matches the integral evaluation and confirms the area bounded by the curve. **Check with graphing calculator:** The integral expression corresponds to the option $$(10.25\theta - 24 \sin \theta + 2.25 \sin 2\theta) \big|_0^{2\pi}$$ which evaluates to $$10.25 \times 2\pi = 20.5\pi$$, half of which is $$\frac{41\pi}{2}$$, confirming the result.