1. **State the problem:** We want to evaluate the definite integral $$\int_{-3}^{-1} (x^2 - 4x) \, dx$$ using the definition of the integral as a limit of Riemann sums. 2. **Recall the definition:** $$\int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x,$$ where $$\Delta x = \frac{b - a}{n}$$ and $$x_i = a + i \Delta x.$$ 3. **Identify parameters:** Here, $$a = -3$$ and $$b = -1$$, so $$\Delta x = \frac{-1 - (-3)}{n} = \frac{2}{n}.$$ 4. **Express $$x_i$$:** $$x_i = -3 + i \cdot \frac{2}{n} = -3 + \frac{2i}{n}.$$ 5. **Write the sum:** $$\sum_{i=1}^n f(x_i) \Delta x = \sum_{i=1}^n \left( x_i^2 - 4x_i \right) \Delta x = \sum_{i=1}^n \left( \left(-3 + \frac{2i}{n}\right)^2 - 4\left(-3 + \frac{2i}{n}\right) \right) \cdot \frac{2}{n}.$$ 6. **Expand inside the sum:** $$\left(-3 + \frac{2i}{n}\right)^2 = 9 - \frac{12i}{n} + \frac{4i^2}{n^2},$$ $$-4\left(-3 + \frac{2i}{n}\right) = 12 - \frac{8i}{n}.$$ So, $$f(x_i) = 9 - \frac{12i}{n} + \frac{4i^2}{n^2} + 12 - \frac{8i}{n} = 21 - \frac{20i}{n} + \frac{4i^2}{n^2}.$$ 7. **Sum becomes:** $$\sum_{i=1}^n \left(21 - \frac{20i}{n} + \frac{4i^2}{n^2}\right) \cdot \frac{2}{n} = \frac{2}{n} \sum_{i=1}^n \left(21 - \frac{20i}{n} + \frac{4i^2}{n^2}\right).$$ 8. **Separate the sum:** $$= \frac{2}{n} \left( \sum_{i=1}^n 21 - \sum_{i=1}^n \frac{20i}{n} + \sum_{i=1}^n \frac{4i^2}{n^2} \right) = \frac{2}{n} \left( 21n - \frac{20}{n} \sum_{i=1}^n i + \frac{4}{n^2} \sum_{i=1}^n i^2 \right).$$ 9. **Use formulas for sums:** $$\sum_{i=1}^n i = \frac{n(n+1)}{2},$$ $$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}.$$ 10. **Substitute sums:** $$= \frac{2}{n} \left( 21n - \frac{20}{n} \cdot \frac{n(n+1)}{2} + \frac{4}{n^2} \cdot \frac{n(n+1)(2n+1)}{6} \right) = \frac{2}{n} \left( 21n - 10(n+1) + \frac{4(n+1)(2n+1)}{6n} \right).$$ 11. **Simplify inside parentheses:** $$21n - 10n - 10 + \frac{4(n+1)(2n+1)}{6n} = 11n - 10 + \frac{4(n+1)(2n+1)}{6n}.$$ 12. **Multiply by $$\frac{2}{n}$$:** $$\frac{2}{n} \left( 11n - 10 + \frac{4(n+1)(2n+1)}{6n} \right) = \frac{2}{n} (11n - 10) + \frac{2}{n} \cdot \frac{4(n+1)(2n+1)}{6n} = 22 - \frac{20}{n} + \frac{8(n+1)(2n+1)}{6n^2}.$$ 13. **Evaluate the limit as $$n \to \infty$$:** As $$n \to \infty$$, $$\frac{20}{n} \to 0,$$ $$\frac{8(n+1)(2n+1)}{6n^2} = \frac{8(2n^2 + 3n + 1)}{6n^2} = \frac{16n^2 + 24n + 8}{6n^2} = \frac{16}{6} + \frac{24}{6n} + \frac{8}{6n^2} \to \frac{16}{6} = \frac{8}{3}.$$ 14. **Final value:** $$\lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x = 22 + \frac{8}{3} = \frac{66}{3} + \frac{8}{3} = \frac{74}{3}.$$ **Answer:** $$\int_{-3}^{-1} (x^2 - 4x) \, dx = \frac{74}{3}.$$