1. **State the problem:** We need to evaluate the definite integral of the polynomial function $$\int_2^3 x^2(5 + 2x^3) + 8 \, dx$$ from $x=2$ to $x=3$. 2. **Rewrite the integrand:** Distribute $x^2$ inside the parentheses: $$x^2(5 + 2x^3) + 8 = 5x^2 + 2x^5 + 8$$ 3. **Set up the integral:** $$\int_2^3 (5x^2 + 2x^5 + 8) \, dx$$ 4. **Integrate term-by-term:** - Integral of $5x^2$ is $5 \cdot \frac{x^3}{3} = \frac{5x^3}{3}$ - Integral of $2x^5$ is $2 \cdot \frac{x^6}{6} = \frac{2x^6}{6} = \frac{x^6}{3}$ - Integral of $8$ is $8x$ So, $$\int (5x^2 + 2x^5 + 8) \, dx = \frac{5x^3}{3} + \frac{x^6}{3} + 8x + C$$ 5. **Evaluate the definite integral:** $$\left[ \frac{5x^3}{3} + \frac{x^6}{3} + 8x \right]_2^3 = \left( \frac{5 \cdot 3^3}{3} + \frac{3^6}{3} + 8 \cdot 3 \right) - \left( \frac{5 \cdot 2^3}{3} + \frac{2^6}{3} + 8 \cdot 2 \right)$$ Calculate each part: - For $x=3$: $$\frac{5 \cdot 27}{3} + \frac{729}{3} + 24 = 45 + 243 + 24 = 312$$ - For $x=2$: $$\frac{5 \cdot 8}{3} + \frac{64}{3} + 16 = \frac{40}{3} + \frac{64}{3} + 16 = \frac{104}{3} + 16 = \frac{104}{3} + \frac{48}{3} = \frac{152}{3}$$ 6. **Subtract to find the result:** $$312 - \frac{152}{3} = \frac{936}{3} - \frac{152}{3} = \frac{784}{3}$$ **Final answer:** $$\boxed{\frac{784}{3}}$$