1. Problem: Evaluate the integral $$- \int_0^1 x^2 (1 - x)^3 \, dx$$. 2. Formula: Use the binomial expansion for $(1-x)^3 = 1 - 3x + 3x^2 - x^3$. 3. Substitute and expand: $$- \int_0^1 x^2 (1 - 3x + 3x^2 - x^3) \, dx = - \int_0^1 (x^2 - 3x^3 + 3x^4 - x^5) \, dx$$. 4. Integrate term-by-term: $$- \left[ \frac{x^3}{3} - \frac{3x^4}{4} + \frac{3x^5}{5} - \frac{x^6}{6} \right]_0^1 = - \left( \frac{1}{3} - \frac{3}{4} + \frac{3}{5} - \frac{1}{6} \right)$$. 5. Simplify inside the parentheses: $$\frac{1}{3} - \frac{3}{4} + \frac{3}{5} - \frac{1}{6} = \frac{20}{60} - \frac{45}{60} + \frac{36}{60} - \frac{10}{60} = \frac{1}{60}$$. 6. Final answer: $$- \frac{1}{60} = -0.0167$$. This completes the evaluation of the integral.