1. **State the problem:** Find the area of the shaded region bounded by the curves $$y = 8 - 3x^2$$ and $$y = x^3 - 4x^2 - 6x + 8$$ 2. **Find the points of intersection:** Set the two functions equal to find the limits of integration. $$8 - 3x^2 = x^3 - 4x^2 - 6x + 8$$ Simplify by subtracting 8 from both sides: $$-3x^2 = x^3 - 4x^2 - 6x$$ Bring all terms to one side: $$0 = x^3 - 4x^2 - 6x + 3x^2$$ $$0 = x^3 - x^2 - 6x$$ Factor out $x$: $$0 = x(x^2 - x - 6)$$ Factor the quadratic: $$0 = x(x - 3)(x + 2)$$ So the intersection points are at: $$x = 0, x = 3, x = -2$$ 3. **Determine which function is upper and which is lower between the intersection points:** Check at $x=1$: $$y_1 = 8 - 3(1)^2 = 8 - 3 = 5$$ $$y_2 = 1^3 - 4(1)^2 - 6(1) + 8 = 1 - 4 - 6 + 8 = -1$$ So $y = 8 - 3x^2$ is above $y = x^3 - 4x^2 - 6x + 8$ between $x=-2$ and $x=3$. 4. **Set up the integral for the area:** $$\text{Area} = \int_{-2}^{3} \left[(8 - 3x^2) - (x^3 - 4x^2 - 6x + 8)\right] dx$$ Simplify the integrand: $$= \int_{-2}^{3} \left(8 - 3x^2 - x^3 + 4x^2 + 6x - 8\right) dx$$ $$= \int_{-2}^{3} \left(-x^3 + ( -3x^2 + 4x^2 ) + 6x + (8 - 8)\right) dx$$ $$= \int_{-2}^{3} \left(-x^3 + x^2 + 6x\right) dx$$ 5. **Integrate term-by-term:** $$\int -x^3 dx = -\frac{x^4}{4}$$ $$\int x^2 dx = \frac{x^3}{3}$$ $$\int 6x dx = 3x^2$$ So, $$\int_{-2}^{3} (-x^3 + x^2 + 6x) dx = \left[-\frac{x^4}{4} + \frac{x^3}{3} + 3x^2\right]_{-2}^{3}$$ 6. **Evaluate at the bounds:** At $x=3$: $$-\frac{3^4}{4} + \frac{3^3}{3} + 3(3^2) = -\frac{81}{4} + \frac{27}{3} + 3(9) = -20.25 + 9 + 27 = 15.75$$ At $x=-2$: $$-\frac{(-2)^4}{4} + \frac{(-2)^3}{3} + 3(-2)^2 = -\frac{16}{4} + \frac{-8}{3} + 3(4) = -4 - \frac{8}{3} + 12 = 8 - \frac{8}{3} = \frac{24}{3} - \frac{8}{3} = \frac{16}{3} \approx 5.3333$$ 7. **Calculate the area:** $$15.75 - 5.3333 = 10.4167$$ **Final answer:** $$\boxed{\text{Area} = \frac{125}{12} \approx 10.42}$$