1. **State the problem:** We need to evaluate the double integral $$\int_{1}^{3} \int_{-2}^{4} (x^2 - 2xy + 3y^2 - 5) \, dy \, dx$$. 2. **Integrate with respect to $y$ first:** Treat $x$ as a constant and integrate the inner integral: $$\int_{-2}^{4} (x^2 - 2xy + 3y^2 - 5) \, dy = \int_{-2}^{4} x^2 \, dy - \int_{-2}^{4} 2xy \, dy + \int_{-2}^{4} 3y^2 \, dy - \int_{-2}^{4} 5 \, dy$$ 3. **Calculate each integral separately:** - $$\int_{-2}^{4} x^2 \, dy = x^2 y \Big|_{-2}^{4} = x^2 (4 - (-2)) = 6x^2$$ - $$\int_{-2}^{4} 2xy \, dy = 2x \int_{-2}^{4} y \, dy = 2x \left( \frac{y^2}{2} \Big|_{-2}^{4} \right) = 2x \left( \frac{16 - 4}{2} \right) = 2x \times 6 = 12x$$ - $$\int_{-2}^{4} 3y^2 \, dy = 3 \left( \frac{y^3}{3} \Big|_{-2}^{4} \right) = (64 - (-8)) = 72$$ - $$\int_{-2}^{4} 5 \, dy = 5y \Big|_{-2}^{4} = 5(4 - (-2)) = 30$$ 4. **Combine the results:** $$6x^2 - 12x + 72 - 30 = 6x^2 - 12x + 42$$ 5. **Now integrate with respect to $x$ from 1 to 3:** $$\int_{1}^{3} (6x^2 - 12x + 42) \, dx = \int_{1}^{3} 6x^2 \, dx - \int_{1}^{3} 12x \, dx + \int_{1}^{3} 42 \, dx$$ 6. **Calculate each integral:** - $$\int_{1}^{3} 6x^2 \, dx = 6 \left( \frac{x^3}{3} \Big|_{1}^{3} \right) = 6 \left( \frac{27 - 1}{3} \right) = 6 \times \frac{26}{3} = 52$$ - $$\int_{1}^{3} 12x \, dx = 12 \left( \frac{x^2}{2} \Big|_{1}^{3} \right) = 12 \times \frac{9 - 1}{2} = 12 \times 4 = 48$$ - $$\int_{1}^{3} 42 \, dx = 42(x) \Big|_{1}^{3} = 42(3 - 1) = 84$$ 7. **Combine these results:** $$52 - 48 + 84 = 88$$ **Final answer:** $$\boxed{88}$$