1. **State the problem:** We need to evaluate the definite integral $$\int_{-2}^{3} (4x^2 + 2x - 5) \, dx$$. 2. **Recall the formula:** The integral of a polynomial $$ax^n$$ is $$\frac{a}{n+1}x^{n+1}$$ plus a constant of integration for indefinite integrals. For definite integrals, we evaluate the antiderivative at the upper and lower limits and subtract. 3. **Find the antiderivative:** $$\int (4x^2 + 2x - 5) \, dx = \int 4x^2 \, dx + \int 2x \, dx - \int 5 \, dx$$ $$= 4 \cdot \frac{x^3}{3} + 2 \cdot \frac{x^2}{2} - 5x + C = \frac{4}{3}x^3 + x^2 - 5x + C$$ 4. **Evaluate at the bounds:** $$F(3) = \frac{4}{3} \cdot 3^3 + 3^2 - 5 \cdot 3 = \frac{4}{3} \cdot 27 + 9 - 15 = 36 + 9 - 15 = 30$$ $$F(-2) = \frac{4}{3} \cdot (-2)^3 + (-2)^2 - 5 \cdot (-2) = \frac{4}{3} \cdot (-8) + 4 + 10 = -\frac{32}{3} + 14 = \frac{-32 + 42}{3} = \frac{10}{3}$$ 5. **Calculate the definite integral:** $$\int_{-2}^{3} (4x^2 + 2x - 5) \, dx = F(3) - F(-2) = 30 - \frac{10}{3} = \frac{90}{3} - \frac{10}{3} = \frac{80}{3}$$ **Final answer:** $$\boxed{\frac{80}{3}}$$