1. **State the problem:** Find the area of the region $R$ bounded by the curves $f(x) = 3x^2$ and $g(x) = 3x + 2$. 2. **Find the points of intersection:** Set $3x^2 = 3x + 2$ to find the limits of integration. $$3x^2 - 3x - 2 = 0$$ Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=3$, $b=-3$, $c=-2$: $$x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 3 \cdot (-2)}}{2 \cdot 3} = \frac{3 \pm \sqrt{9 + 24}}{6} = \frac{3 \pm \sqrt{33}}{6}$$ So the intersection points are: $$x_1 = \frac{3 - \sqrt{33}}{6}, \quad x_2 = \frac{3 + \sqrt{33}}{6}$$ 3. **Determine which function is on top:** For $x$ between $x_1$ and $x_2$, $g(x) = 3x + 2$ is above $f(x) = 3x^2$. 4. **Set up the integral for the area:** $$\text{Area} = \int_{x_1}^{x_2} [g(x) - f(x)] \, dx = \int_{x_1}^{x_2} [(3x + 2) - 3x^2] \, dx$$ 5. **Integrate:** $$\int [(3x + 2) - 3x^2] \, dx = \int (3x + 2) \, dx - \int 3x^2 \, dx = \left( \frac{3x^2}{2} + 2x \right) - x^3 + C$$ 6. **Evaluate the definite integral:** $$\text{Area} = \left[ \frac{3x^2}{2} + 2x - x^3 \right]_{x_1}^{x_2}$$ Calculate numerically: $$x_1 \approx \frac{3 - 5.7446}{6} = \frac{-2.7446}{6} \approx -0.4574$$ $$x_2 \approx \frac{3 + 5.7446}{6} = \frac{8.7446}{6} \approx 1.4574$$ Evaluate at $x_2$: $$\frac{3(1.4574)^2}{2} + 2(1.4574) - (1.4574)^3 \approx \frac{3(2.124)}{2} + 2.9148 - 3.096 = 3.186 + 2.9148 - 3.096 = 3.0048$$ Evaluate at $x_1$: $$\frac{3(-0.4574)^2}{2} + 2(-0.4574) - (-0.4574)^3 \approx \frac{3(0.2092)}{2} - 0.9148 + 0.0957 = 0.3138 - 0.9148 + 0.0957 = -0.5053$$ Subtract: $$3.0048 - (-0.5053) = 3.5101$$ 7. **Final answer:** The area of the region $R$ is approximately **3.510** (rounded to the nearest thousandth).