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

To evaluate the integral \( \int_{6}^{7} x \sqrt{x-6} \, dx \), we can use substitution. Let \( u = x - 6 \), which means \( x = u + 6 \) and \( dx = du \). The limits of integration change accordingly: when \( x = 6, u = 0 \) and when \( x = 7, u = 1 \). Thus, the integral becomes: \[ \int_{0}^{1} (u + 6) \sqrt{u} \, du \] We can expand this: \[ = \int_{0}^{1} (u^{3/2} + 6\sqrt{u}) \, du \] Now, we can integrate each term separately: 1. \( \int u^{3/2} \, du = \frac{u^{5/2}}{5/2} = \frac{2}{5} u^{5/2} \) 2. \( \int 6\sqrt{u} \, du = 6 \cdot \frac{u^{3/2}}{3/2} = 4u^{3/2} \) Thus, we have: \[ \int_{0}^{1} (u^{3/2} + 6\sqrt{u}) \, du = \left[ \frac{2}{5} u^{5/2} + 4u^{3/2} \right]_{0}^{1} \] Calculating the upper limit (at \( u = 1 \)): \[ = \left( \frac{2}{5} \cdot 1^{5/2} + 4 \cdot 1^{3/2} \right) = \frac{2}{5} + 4 = \frac{2}{5} + \frac{20}{5} = \frac{22}{5} \] Calculating the lower limit (at \( u = 0 \)): \[ = \left( \frac{2}{5} \cdot 0^{5/2} + 4 \cdot 0^{3/2} \right) = 0 \] Thus, the value of the definite integral is: \[ \int_{6}^{7} x \sqrt{x-6} \, dx = \frac{22}{5} \]