1. **State the problem:** Evaluate the definite integral $$\int_0^2 2x \sqrt{x+2} \, dx$$ and verify that it equals $$\frac{32}{15}(2 + \sqrt{2})$$. 2. **Set up the integral:** We have the integral of the function $$f(x) = 2x \sqrt{x+2}$$ from 0 to 2. 3. **Use substitution:** Let $$u = x + 2$$, so that $$du = dx$$ and when $$x=0$$, $$u=2$$; when $$x=2$$, $$u=4$$. 4. **Rewrite the integral in terms of $$u$$:** $$\int_0^2 2x \sqrt{x+2} \, dx = \int_2^4 2(u-2) \sqrt{u} \, du$$ 5. **Expand the integrand:** $$2(u-2) \sqrt{u} = 2u \sqrt{u} - 4 \sqrt{u} = 2u^{3/2} - 4u^{1/2}$$ 6. **Rewrite the integral:** $$\int_2^4 (2u^{3/2} - 4u^{1/2}) \, du = 2 \int_2^4 u^{3/2} \, du - 4 \int_2^4 u^{1/2} \, du$$ 7. **Integrate each term:** - $$\int u^{3/2} \, du = \frac{u^{5/2}}{\frac{5}{2}} = \frac{2}{5} u^{5/2}$$ - $$\int u^{1/2} \, du = \frac{u^{3/2}}{\frac{3}{2}} = \frac{2}{3} u^{3/2}$$ 8. **Substitute back into the integral:** $$2 \times \frac{2}{5} u^{5/2} - 4 \times \frac{2}{3} u^{3/2} = \frac{4}{5} u^{5/2} - \frac{8}{3} u^{3/2}$$ 9. **Evaluate from $$u=2$$ to $$u=4$$:** $$\left[ \frac{4}{5} u^{5/2} - \frac{8}{3} u^{3/2} \right]_2^4 = \left( \frac{4}{5} 4^{5/2} - \frac{8}{3} 4^{3/2} \right) - \left( \frac{4}{5} 2^{5/2} - \frac{8}{3} 2^{3/2} \right)$$ 10. **Calculate powers:** - $$4^{5/2} = (4^{1/2})^5 = 2^5 = 32$$ - $$4^{3/2} = (4^{1/2})^3 = 2^3 = 8$$ - $$2^{5/2} = (2^{1/2})^5 = (\sqrt{2})^5 = 4 \sqrt{2}$$ - $$2^{3/2} = (2^{1/2})^3 = 2 \sqrt{2}$$ 11. **Substitute values:** $$\left( \frac{4}{5} \times 32 - \frac{8}{3} \times 8 \right) - \left( \frac{4}{5} \times 4 \sqrt{2} - \frac{8}{3} \times 2 \sqrt{2} \right) = \left( \frac{128}{5} - \frac{64}{3} \right) - \left( \frac{16}{5} \sqrt{2} - \frac{16}{3} \sqrt{2} \right)$$ 12. **Simplify each group:** - $$\frac{128}{5} - \frac{64}{3} = \frac{128 \times 3}{15} - \frac{64 \times 5}{15} = \frac{384 - 320}{15} = \frac{64}{15}$$ - $$\frac{16}{5} \sqrt{2} - \frac{16}{3} \sqrt{2} = \left( \frac{16 \times 3}{15} - \frac{16 \times 5}{15} \right) \sqrt{2} = \frac{48 - 80}{15} \sqrt{2} = -\frac{32}{15} \sqrt{2}$$ 13. **Combine:** $$\frac{64}{15} - \left(-\frac{32}{15} \sqrt{2} \right) = \frac{64}{15} + \frac{32}{15} \sqrt{2} = \frac{32}{15} (2 + \sqrt{2})$$ 14. **Conclusion:** The value of the integral is $$\boxed{\frac{32}{15} (2 + \sqrt{2})}$$, which matches the given expression. This confirms the correctness of the integral evaluation.