1. The problem is to evaluate the integral $$\int_0^1 (3x^3 - 2x^2 + x - 4) x y^2 \sqrt{x^2 - 3x + 24} \, dx$$ with respect to $x$ from 0 to 1, treating $y$ as a constant. 2. First, simplify the integrand by combining like terms: $$ (3x^3 - 2x^2 + x - 4) x = 3x^4 - 2x^3 + x^2 - 4x $$ 3. The integral becomes $$ \int_0^1 (3x^4 - 2x^3 + x^2 - 4x) y^2 \sqrt{x^2 - 3x + 24} \, dx $$ 4. Since $y^2$ is constant with respect to $x$, factor it out: $$ y^2 \int_0^1 (3x^4 - 2x^3 + x^2 - 4x) \sqrt{x^2 - 3x + 24} \, dx $$ 5. The integral involves a polynomial times a square root of a quadratic expression. This integral does not have a simple elementary antiderivative. 6. To evaluate it exactly, one would typically use numerical methods or special functions. 7. Therefore, the integral is expressed as $$ y^2 \int_0^1 (3x^4 - 2x^3 + x^2 - 4x) \sqrt{x^2 - 3x + 24} \, dx $$ and can be approximated numerically if needed. Final answer: $$ y^2 \int_0^1 (3x^4 - 2x^3 + x^2 - 4x) \sqrt{x^2 - 3x + 24} \, dx $$