1. **Problem:** Calculate the definite integral $$\int_2^4 (-6x^5 + 3x^4 - 2x^3) \, dx$$. 2. **Formula:** The definite integral of a polynomial function $$f(x)$$ from $$a$$ to $$b$$ is $$\int_a^b f(x) \, dx = F(b) - F(a)$$ where $$F(x)$$ is the antiderivative of $$f(x)$$. 3. **Find the antiderivative:** $$\int (-6x^5 + 3x^4 - 2x^3) \, dx = -6 \int x^5 \, dx + 3 \int x^4 \, dx - 2 \int x^3 \, dx$$ Calculate each term: $$-6 \cdot \frac{x^{6}}{6} + 3 \cdot \frac{x^{5}}{5} - 2 \cdot \frac{x^{4}}{4} + C = -x^{6} + \frac{3}{5}x^{5} - \frac{1}{2}x^{4} + C$$ 4. **Evaluate at limits:** $$F(4) = -4^{6} + \frac{3}{5}4^{5} - \frac{1}{2}4^{4} = -4096 + \frac{3}{5} \times 1024 - \frac{1}{2} \times 256$$ Calculate each term: $$\frac{3}{5} \times 1024 = \frac{3072}{5} = 614.4$$ $$\frac{1}{2} \times 256 = 128$$ So, $$F(4) = -4096 + 614.4 - 128 = -3609.6$$ Similarly, $$F(2) = -2^{6} + \frac{3}{5}2^{5} - \frac{1}{2}2^{4} = -64 + \frac{3}{5} \times 32 - 8$$ Calculate: $$\frac{3}{5} \times 32 = \frac{96}{5} = 19.2$$ So, $$F(2) = -64 + 19.2 - 8 = -52.8$$ 5. **Calculate definite integral:** $$\int_2^4 (-6x^5 + 3x^4 - 2x^3) \, dx = F(4) - F(2) = -3609.6 - (-52.8) = -3609.6 + 52.8 = -3556.8$$ **Final answer:** $$\boxed{-3556.8}$$