Evaluate the definite integral \( \int_{0}^{2} x \sqrt{2-x} \, dx \).

Using property \( \int_{0}^{a} f(x)dx = \int_{0}^{a} f(a-x)dx \), the integral becomes \( \int_{0}^{2} (2-x)\sqrt{x} dx = \int_{0}^{2} (2x^{1/2} - x^{3/2})dx = [\frac{4}{3}x^{3/2} - \frac{2}{5}x^{5/2}]_0^2 = \frac{16\sqrt{2}}{15} \).