1. **State the problem:** We want to evaluate the definite integral $$\int_2^6 x \sqrt{2x - 3} \, dx$$ using u-substitution. 2. **Choose the substitution:** Let $$u = 2x - 3$$ because the square root contains this expression. 3. **Compute differential:** Then $$du = 2 \, dx \implies dx = \frac{du}{2}$$. 4. **Express x in terms of u:** From $$u = 2x - 3$$, we get $$x = \frac{u + 3}{2}$$. 5. **Change the limits:** When $$x=2$$, $$u = 2(2) - 3 = 1$$. When $$x=6$$, $$u = 2(6) - 3 = 9$$. 6. **Rewrite the integral:** $$\int_2^6 x \sqrt{2x - 3} \, dx = \int_1^9 \frac{u + 3}{2} \sqrt{u} \cdot \frac{du}{2} = \int_1^9 \frac{u + 3}{4} u^{1/2} \, du$$ 7. **Simplify the integrand:** $$\frac{u + 3}{4} u^{1/2} = \frac{1}{4} (u^{3/2} + 3u^{1/2})$$ 8. **Split the integral:** $$\int_1^9 \frac{1}{4} (u^{3/2} + 3u^{1/2}) \, du = \frac{1}{4} \int_1^9 u^{3/2} \, du + \frac{3}{4} \int_1^9 u^{1/2} \, du$$ 9. **Integrate each term:** - $$\int u^{3/2} \, du = \frac{u^{5/2}}{\frac{5}{2}} = \frac{2}{5} u^{5/2}$$ - $$\int u^{1/2} \, du = \frac{u^{3/2}}{\frac{3}{2}} = \frac{2}{3} u^{3/2}$$ 10. **Substitute back:** $$\frac{1}{4} \cdot \frac{2}{5} u^{5/2} + \frac{3}{4} \cdot \frac{2}{3} u^{3/2} = \frac{1}{10} u^{5/2} + \frac{1}{2} u^{3/2}$$ 11. **Evaluate definite integral from 1 to 9:** $$\left[ \frac{1}{10} u^{5/2} + \frac{1}{2} u^{3/2} \right]_1^9 = \left( \frac{1}{10} \cdot 9^{5/2} + \frac{1}{2} \cdot 9^{3/2} \right) - \left( \frac{1}{10} \cdot 1^{5/2} + \frac{1}{2} \cdot 1^{3/2} \right)$$ 12. **Calculate powers:** - $$9^{1/2} = 3$$ - $$9^{3/2} = 9^{1} \cdot 9^{1/2} = 9 \cdot 3 = 27$$ - $$9^{5/2} = 9^{2} \cdot 9^{1/2} = 81 \cdot 3 = 243$$ 13. **Plug in values:** $$\left( \frac{1}{10} \cdot 243 + \frac{1}{2} \cdot 27 \right) - \left( \frac{1}{10} + \frac{1}{2} \right) = \left( 24.3 + 13.5 \right) - \left( 0.1 + 0.5 \right) = 37.8 - 0.6 = 37.2$$ **Final answer:** $$\int_2^6 x \sqrt{2x - 3} \, dx = 37.2$$