1. **State the problem:** We have the curve given by the function $$y = x^3 - x^2 - 2x$$ and we want to compute two things over the interval $$[-1, 2]$$: a) The definite integral using Riemann sums. b) The area between the graph and the x-axis. 2. **Recall the formula for the definite integral:** The definite integral of a function $$f(x)$$ from $$a$$ to $$b$$ is given by $$\int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x$$ where $$\Delta x = \frac{b-a}{n}$$ and $$x_i^*$$ is a sample point in the $$i^{th}$$ subinterval. 3. **Compute the definite integral analytically (equivalent to the limit of Riemann sums):** First, find the antiderivative of $$f(x) = x^3 - x^2 - 2x$$: $$F(x) = \frac{x^4}{4} - \frac{x^3}{3} - x^2 + C$$ 4. **Evaluate the definite integral:** $$\int_{-1}^2 (x^3 - x^2 - 2x) \, dx = F(2) - F(-1)$$ Calculate each term: $$F(2) = \frac{2^4}{4} - \frac{2^3}{3} - 2^2 = \frac{16}{4} - \frac{8}{3} - 4 = 4 - \frac{8}{3} - 4 = -\frac{8}{3}$$ $$F(-1) = \frac{(-1)^4}{4} - \frac{(-1)^3}{3} - (-1)^2 = \frac{1}{4} - \left(-\frac{1}{3}\right) - 1 = \frac{1}{4} + \frac{1}{3} - 1 = \frac{3}{12} + \frac{4}{12} - \frac{12}{12} = -\frac{5}{12}$$ So, $$\int_{-1}^2 (x^3 - x^2 - 2x) \, dx = -\frac{8}{3} - \left(-\frac{5}{12}\right) = -\frac{8}{3} + \frac{5}{12} = -\frac{32}{12} + \frac{5}{12} = -\frac{27}{12} = -\frac{9}{4} = -2.25$$ 5. **Interpretation:** The definite integral is $$-2.25$$, which means the net signed area between the curve and the x-axis over $$[-1, 2]$$ is negative. 6. **Find the area between the graph and the x-axis:** The area is the integral of the absolute value of the function over the interval. We need to find where the function crosses the x-axis in $$[-1, 2]$$. 7. **Find roots of $$y = x^3 - x^2 - 2x = x(x^2 - x - 2)$$:** Factor the quadratic: $$x^2 - x - 2 = (x - 2)(x + 1)$$ So roots are $$x = 0, 2, -1$$. 8. **Split the interval $$[-1, 2]$$ at roots: $$[-1, 0], [0, 2]$$** On each subinterval, determine the sign of $$f(x)$$: - For $$x \in (-1, 0)$$, pick $$x = -0.5$$: $$f(-0.5) = (-0.5)^3 - (-0.5)^2 - 2(-0.5) = -0.125 - 0.25 + 1 = 0.625 > 0$$ - For $$x \in (0, 2)$$, pick $$x = 1$$: $$f(1) = 1 - 1 - 2 = -2 < 0$$ 9. **Calculate area as sum of absolute integrals:** $$\text{Area} = \int_{-1}^0 f(x) \, dx - \int_0^2 f(x) \, dx$$ Calculate each integral: $$\int_{-1}^0 f(x) \, dx = F(0) - F(-1)$$ $$F(0) = 0$$ From step 4, $$F(-1) = -\frac{5}{12}$$ So, $$\int_{-1}^0 f(x) \, dx = 0 - \left(-\frac{5}{12}\right) = \frac{5}{12}$$ $$\int_0^2 f(x) \, dx = F(2) - F(0) = -\frac{8}{3} - 0 = -\frac{8}{3}$$ Taking absolute value: $$\text{Area} = \frac{5}{12} + \frac{8}{3} = \frac{5}{12} + \frac{32}{12} = \frac{37}{12} \approx 3.0833$$ **Final answers:** - Definite integral over $$[-1, 2]$$ is $$-\frac{9}{4} = -2.25$$. - Area between the graph and the x-axis over $$[-1, 2]$$ is $$\frac{37}{12} \approx 3.0833$$.