calculate \( \int_{4}^{8}\left(3 x+x^{2}\right) d x \) given the following: \( \int_{4}^{8} x^{2} d x=\frac{448}{3} \int_{8}^{9} x^{2} d x \frac{2 \pi}{3} \int_{4}^{8} x d x=24 \)

To calculate the integral \( \int_{4}^{8}\left(3 x+x^{2}\right) d x \), we can split the integral into two parts: \[ \int_{4}^{8} (3x + x^{2}) \, dx = \int_{4}^{8} 3x \, dx + \int_{4}^{8} x^{2} \, dx \] Using the information given: 1. To calculate \( \int_{4}^{8} 3x \, dx \): \[ \int 3x \, dx = \frac{3}{2} x^{2} \] Evaluating from 4 to 8: \[ \left[ \frac{3}{2} (8^{2}) - \frac{3}{2} (4^{2}) \right] = \frac{3}{2} (64 - 16) = \frac{3}{2} \cdot 48 = 72 \] 2. We know that \( \int_{4}^{8} x^{2} \, dx = \frac{448}{3} \). Putting it all together: \[ \int_{4}^{8} (3x + x^{2}) \, dx = 72 + \frac{448}{3} = \frac{216}{3} + \frac{448}{3} = \frac{664}{3} \] Thus, the value of the integral \( \int_{4}^{8} (3x + x^{2}) \, dx \) is \( \frac{664}{3} \).