13. Evaluate the definite integral. \[ \int_{6}^{7} x \sqrt{x-6} d x \] \( \square \)

To evaluate the definite integral \( \int_{6}^{7} x \sqrt{x-6} \, dx \), we can use the substitution method. Let \( u = x - 6 \), which means \( du = dx \) and when \( x = 6 \), \( u = 0 \) and when \( x = 7 \), \( u = 1 \). Substituting, we get: \[ x = u + 6 \quad \text{so} \quad \int_{0}^{1} (u+6) \sqrt{u} \, du \] Now, distribute \( (u + 6) \sqrt{u} \): \[ \int_{0}^{1} (u^{3/2} + 6u^{1/2}) \, du \] Evaluating the integral term by term: 1. \( \int u^{3/2} \, du = \frac{u^{5/2}}{5/2} = \frac{2}{5} u^{5/2} \) 2. \( \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2} \) Therefore, the definite integral becomes: \[ \left[ \frac{2}{5} u^{5/2} + 6 \cdot \frac{2}{3} u^{3/2} \right]_{0}^{1} = \left( \frac{2}{5} \cdot 1^{5/2} + 4 \cdot 1^{3/2} \right) - \left( 0 + 0 \right) = \frac{2}{5} + 4 = \frac{2}{5} + \frac{20}{5} = \frac{22}{5} \] Thus, the value of the definite integral \( \int_{6}^{7} x \sqrt{x-6} \, dx \) is \(\frac{22}{5}\).