1. **State the problem:** We are given the curve defined by the function $$y = x^3 - x^2 - 2x$$. We need to compute two things over the interval $$[-1, 2]$$: a) The definite integral $$\int_{-1}^2 (x^3 - x^2 - 2x) \, dx$$. b) The area between the graph and the x-axis over the same interval. 2. **Recall the formulas and rules:** - The definite integral $$\int_a^b f(x) \, dx$$ gives the net area between the curve and the x-axis from $$x=a$$ to $$x=b$$. Areas above the x-axis count as positive, below as negative. - The total area between the curve and the x-axis is the integral of the absolute value of the function: $$\int_a^b |f(x)| \, dx$$. 3. **Find the antiderivative:** $$\int (x^3 - x^2 - 2x) \, dx = \frac{x^4}{4} - \frac{x^3}{3} - x^2 + C$$ 4. **Compute the definite integral:** Evaluate at $$x=2$$: $$\frac{2^4}{4} - \frac{2^3}{3} - 2^2 = \frac{16}{4} - \frac{8}{3} - 4 = 4 - \frac{8}{3} - 4 = -\frac{8}{3}$$ Evaluate at $$x=-1$$: $$\frac{(-1)^4}{4} - \frac{(-1)^3}{3} - (-1)^2 = \frac{1}{4} - \left(-\frac{1}{3}\right) - 1 = \frac{1}{4} + \frac{1}{3} - 1 = \frac{3}{12} + \frac{4}{12} - \frac{12}{12} = -\frac{5}{12}$$ So the definite integral is: $$\int_{-1}^2 (x^3 - x^2 - 2x) \, dx = -\frac{8}{3} - \left(-\frac{5}{12}\right) = -\frac{8}{3} + \frac{5}{12} = -\frac{32}{12} + \frac{5}{12} = -\frac{27}{12} = -\frac{9}{4} = -2.25$$ 5. **Find where the function crosses the x-axis to compute area:** Solve $$x^3 - x^2 - 2x = 0$$: Factor out $$x$$: $$x(x^2 - x - 2) = 0$$ Solve $$x=0$$ or $$x^2 - x - 2=0$$. Quadratic factors as $$(x-2)(x+1)=0$$, so roots are $$x=2$$ and $$x=-1$$. The roots on the interval $$[-1, 2]$$ are exactly the endpoints and $$0$$. 6. **Determine sign of function on subintervals:** - For $$x \in (-1,0)$$, pick $$x=-0.5$$: $$(-0.5)^3 - (-0.5)^2 - 2(-0.5) = -0.125 - 0.25 + 1 = 0.625 > 0$$ - For $$x \in (0,2)$$, pick $$x=1$$: $$1 - 1 - 2 = -2 < 0$$ 7. **Compute area as sum of absolute integrals:** $$\text{Area} = \int_{-1}^0 (x^3 - x^2 - 2x) \, dx - \int_0^2 (x^3 - x^2 - 2x) \, dx$$ Calculate each integral: At $$x=0$$: $$\frac{0^4}{4} - \frac{0^3}{3} - 0^2 = 0$$ So, $$\int_{-1}^0 (x^3 - x^2 - 2x) \, dx = 0 - \left(-\frac{5}{12}\right) = \frac{5}{12}$$ $$\int_0^2 (x^3 - x^2 - 2x) \, dx = -\frac{8}{3} - 0 = -\frac{8}{3}$$ Taking absolute value for the second part: $$\text{Area} = \frac{5}{12} + \frac{8}{3} = \frac{5}{12} + \frac{32}{12} = \frac{37}{12} \approx 3.0833$$ **Final answers:** - Definite integral over $$[-1, 2]$$ is $$-\frac{9}{4} = -2.25$$. - Area between the graph and x-axis over $$[-1, 2]$$ is $$\frac{37}{12} \approx 3.0833$$.