1. Stating the problem: We need to evaluate the definite integral $$\int_{-1}^2 (x^2 - 5x - 4)\,dx$$. 2. Find the antiderivative: The integral of $$x^2$$ is $$\frac{x^3}{3}$$, the integral of $$-5x$$ is $$-\frac{5x^2}{2}$$, and the integral of $$-4$$ is $$-4x$$. So the antiderivative is: $$F(x) = \frac{x^3}{3} - \frac{5x^2}{2} - 4x$$. 3. Evaluate the antiderivative at the upper and lower limits: $$F(2) = \frac{2^3}{3} - \frac{5 \cdot 2^2}{2} - 4 \cdot 2 = \frac{8}{3} - \frac{20}{2} - 8 = \frac{8}{3} - 10 - 8 = \frac{8}{3} - 18 = -\frac{46}{3}$$ $$F(-1) = \frac{(-1)^3}{3} - \frac{5 \cdot (-1)^2}{2} - 4 \cdot (-1) = -\frac{1}{3} - \frac{5}{2} + 4 = -\frac{1}{3} - \frac{5}{2} + \frac{12}{3} = -\frac{1}{3} + \frac{12}{3} - \frac{5}{2} = \frac{11}{3} - \frac{5}{2} = \frac{22}{6} - \frac{15}{6} = \frac{7}{6}$$. 4. Subtract to find the definite integral: $$\int_{-1}^2 (x^2 -5x -4)\,dx = F(2) - F(-1) = -\frac{46}{3} - \frac{7}{6} = -\frac{92}{6} - \frac{7}{6} = -\frac{99}{6} = -\frac{33}{2}$$. Final answer: $$-\frac{33}{2}$$.