1. The problem is to find the area of the region bounded by the curves $f(x) = x^3 + 2x^2 - x - 2$ and $g(x) = 4x + 4$. 2. To find the area between two curves, we first find their points of intersection by solving $f(x) = g(x)$. 3. Set the equations equal: $$x^3 + 2x^2 - x - 2 = 4x + 4$$ 4. Rearrange to one side: $$x^3 + 2x^2 - x - 2 - 4x - 4 = 0$$ $$x^3 + 2x^2 - 5x - 6 = 0$$ 5. We try to find rational roots using the Rational Root Theorem. Possible roots are $\pm1, \pm2, \pm3, \pm6$. 6. Test $x=1$: $$1 + 2 - 5 - 6 = -8 \neq 0$$ 7. Test $x=2$: $$8 + 8 - 10 - 6 = 0$$ So $x=2$ is a root. 8. Factor out $x-2$ using polynomial division or synthetic division: $$x^3 + 2x^2 - 5x - 6 = (x - 2)(x^2 + 4x + 3)$$ 9. Factor the quadratic: $$x^2 + 4x + 3 = (x + 1)(x + 3)$$ 10. So the roots are $x = 2, -1, -3$. 11. These are the points where the curves intersect. The bounded area is between the smallest and largest roots, i.e., from $x = -3$ to $x = 2$. 12. To find the area between the curves, integrate the absolute difference: $$\text{Area} = \int_{-3}^{2} |f(x) - g(x)| \, dx$$ 13. Determine which function is on top in each interval. Check at $x=0$: $$f(0) = -2, \quad g(0) = 4$$ So $g(x)$ is above $f(x)$ at $x=0$. 14. Check at $x=-2$: $$f(-2) = (-8) + 8 + 2 - 2 = 0, \quad g(-2) = -8 + 4 = -4$$ So $f(x)$ is above $g(x)$ at $x=-2$. 15. The order of functions changes at the roots. So split the integral: $$\text{Area} = \int_{-3}^{-1} (f(x) - g(x)) \, dx + \int_{-1}^{2} (g(x) - f(x)) \, dx$$ 16. Compute $f(x) - g(x)$: $$f(x) - g(x) = (x^3 + 2x^2 - x - 2) - (4x + 4) = x^3 + 2x^2 - 5x - 6$$ 17. Compute $g(x) - f(x) = - (f(x) - g(x)) = -x^3 - 2x^2 + 5x + 6$. 18. Calculate the first integral: $$\int_{-3}^{-1} (x^3 + 2x^2 - 5x - 6) \, dx$$ 19. Integrate term by term: $$\int x^3 dx = \frac{x^4}{4}, \quad \int 2x^2 dx = \frac{2x^3}{3}, \quad \int -5x dx = -\frac{5x^2}{2}, \quad \int -6 dx = -6x$$ 20. So the antiderivative is: $$F(x) = \frac{x^4}{4} + \frac{2x^3}{3} - \frac{5x^2}{2} - 6x$$ 21. Evaluate from $-3$ to $-1)$: $$F(-1) = \frac{1}{4} - \frac{2}{3} - \frac{5}{2} + 6 = 0.25 - 0.6667 - 2.5 + 6 = 3.0833$$ $$F(-3) = \frac{81}{4} - 54 - \frac{45}{2} + 18 = 20.25 - 54 - 22.5 + 18 = -38.25$$ 22. So the first integral is: $$F(-1) - F(-3) = 3.0833 - (-38.25) = 41.3333$$ 23. Calculate the second integral: $$\int_{-1}^{2} (-x^3 - 2x^2 + 5x + 6) \, dx$$ 24. Antiderivative: $$G(x) = -\frac{x^4}{4} - \frac{2x^3}{3} + \frac{5x^2}{2} + 6x$$ 25. Evaluate from $-1$ to $2)$: $$G(2) = -4 - \frac{16}{3} + 10 + 12 = -4 - 5.3333 + 10 + 12 = 12.6667$$ $$G(-1) = -\frac{1}{4} + \frac{2}{3} + \frac{5}{2} - 6 = -0.25 + 0.6667 + 2.5 - 6 = -3.0833$$ 26. So the second integral is: $$G(2) - G(-1) = 12.6667 - (-3.0833) = 15.75$$ 27. Total area: $$41.3333 + 15.75 = 57.0833$$ 28. Final answer: $$\boxed{\text{Area} \approx 57.08}$$