1. **State the problem:** Evaluate the double integral $$\int_0^1 \int_x^{\sqrt{x}} y(x + y) \, dy \, dx$$. 2. **Understand the integral:** The integral is over $y$ from $x$ to $\sqrt{x}$ and then over $x$ from 0 to 1. 3. **Integrate with respect to $y$ first:** $$\int_x^{\sqrt{x}} y(x + y) \, dy = \int_x^{\sqrt{x}} (xy + y^2) \, dy = x \int_x^{\sqrt{x}} y \, dy + \int_x^{\sqrt{x}} y^2 \, dy$$ 4. **Calculate each integral:** $$x \int_x^{\sqrt{x}} y \, dy = x \left[ \frac{y^2}{2} \right]_x^{\sqrt{x}} = x \left( \frac{(\sqrt{x})^2}{2} - \frac{x^2}{2} \right) = x \left( \frac{x}{2} - \frac{x^2}{2} \right) = \frac{x^2}{2} - \frac{x^3}{2}$$ $$\int_x^{\sqrt{x}} y^2 \, dy = \left[ \frac{y^3}{3} \right]_x^{\sqrt{x}} = \frac{(\sqrt{x})^3}{3} - \frac{x^3}{3} = \frac{x^{3/2}}{3} - \frac{x^3}{3}$$ 5. **Sum the results:** $$\int_x^{\sqrt{x}} y(x + y) \, dy = \frac{x^2}{2} - \frac{x^3}{2} + \frac{x^{3/2}}{3} - \frac{x^3}{3} = \frac{x^2}{2} + \frac{x^{3/2}}{3} - \left( \frac{x^3}{2} + \frac{x^3}{3} \right)$$ 6. **Simplify the $x^3$ terms:** $$\frac{x^3}{2} + \frac{x^3}{3} = \frac{3x^3}{6} + \frac{2x^3}{6} = \frac{5x^3}{6}$$ So the expression becomes: $$\frac{x^2}{2} + \frac{x^{3/2}}{3} - \frac{5x^3}{6}$$ 7. **Now integrate with respect to $x$ from 0 to 1:** $$\int_0^1 \left( \frac{x^2}{2} + \frac{x^{3/2}}{3} - \frac{5x^3}{6} \right) dx = \frac{1}{2} \int_0^1 x^2 dx + \frac{1}{3} \int_0^1 x^{3/2} dx - \frac{5}{6} \int_0^1 x^3 dx$$ 8. **Calculate each integral:** $$\int_0^1 x^2 dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3}$$ $$\int_0^1 x^{3/2} dx = \left[ \frac{x^{5/2}}{5/2} \right]_0^1 = \frac{2}{5}$$ $$\int_0^1 x^3 dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4}$$ 9. **Substitute back:** $$\frac{1}{2} \times \frac{1}{3} + \frac{1}{3} \times \frac{2}{5} - \frac{5}{6} \times \frac{1}{4} = \frac{1}{6} + \frac{2}{15} - \frac{5}{24}$$ 10. **Find common denominator (120):** $$\frac{1}{6} = \frac{20}{120}, \quad \frac{2}{15} = \frac{16}{120}, \quad \frac{5}{24} = \frac{25}{120}$$ 11. **Sum:** $$\frac{20}{120} + \frac{16}{120} - \frac{25}{120} = \frac{11}{120}$$ **Final answer:** $$\boxed{\frac{11}{120}}$$