1. **State the problem:** Find the area of the region bounded by the curves $$y = x^2 - 2$$ and $$y = x$$ between the points $(-1, -1)$ and $(2, 2)$. 2. **Find the points of intersection:** Given points are $(-1, -1)$ and $(2, 2)$, which satisfy both equations. 3. **Set up the integral for the area:** The area between two curves $$y = f(x)$$ and $$y = g(x)$$ from $$x=a$$ to $$x=b$$ is $$\int_a^b |f(x) - g(x)| \, dx$$. Here, the parabola $$y = x^2 - 2$$ is above the line $$y = x$$ in the interval $$[-1, 2]$$. So, area $$A = \int_{-1}^2 ((x^2 - 2) - x) \, dx$$. 4. **Simplify the integrand:** $$x^2 - 2 - x = x^2 - x - 2$$. 5. **Compute the integral:** $$A = \int_{-1}^2 (x^2 - x - 2) \, dx = \left[ \frac{x^3}{3} - \frac{x^2}{2} - 2x \right]_{-1}^2$$. 6. **Evaluate at the bounds:** At $$x=2$$: $$\frac{2^3}{3} - \frac{2^2}{2} - 2(2) = \frac{8}{3} - 2 - 4 = \frac{8}{3} - 6 = \frac{8 - 18}{3} = -\frac{10}{3}$$. At $$x=-1$$: $$\frac{(-1)^3}{3} - \frac{(-1)^2}{2} - 2(-1) = -\frac{1}{3} - \frac{1}{2} + 2 = -\frac{1}{3} - \frac{1}{2} + 2$$. Find common denominator 6: $$-\frac{2}{6} - \frac{3}{6} + \frac{12}{6} = \frac{7}{6}$$. 7. **Calculate the area:** $$A = \left(-\frac{10}{3}\right) - \frac{7}{6} = -\frac{20}{6} - \frac{7}{6} = -\frac{27}{6} = -\frac{9}{2}$$. Since area is positive, take absolute value: $$A = \frac{9}{2}$$. **Final answer:** The area of the region bounded by the curves is $$\boxed{\frac{9}{2}}$$, which corresponds to option c.