1. **State the problem:** We need to evaluate the double integral $$\iint_R xy \, dy \, dx$$ where the region $R$ is bounded by the curves $y = x^2$ and $x + y = 2$ (or $y = 2 - x$). 2. **Understand the region $R$:** The region is between the parabola $y = x^2$ and the line $y = 2 - x$. 3. **Find the intersection points:** Set $x^2 = 2 - x$ to find the limits for $x$. $$x^2 + x - 2 = 0$$ Factor: $$(x + 2)(x - 1) = 0$$ So, $x = -2$ or $x = 1$. 4. **Set up the integral:** For each $x$ in $[-2,1]$, $y$ goes from the lower curve $y = x^2$ to the upper curve $y = 2 - x$. Thus, $$\iint_R xy \, dy \, dx = \int_{x=-2}^{1} \int_{y=x^2}^{2 - x} xy \, dy \, dx$$ 5. **Integrate with respect to $y$ first:** $$\int_{y=x^2}^{2 - x} xy \, dy = x \int_{y=x^2}^{2 - x} y \, dy = x \left[ \frac{y^2}{2} \right]_{y=x^2}^{2 - x} = x \left( \frac{(2 - x)^2}{2} - \frac{(x^2)^2}{2} \right) = \frac{x}{2} \left( (2 - x)^2 - x^4 \right)$$ 6. **Simplify the expression inside the integral:** $$(2 - x)^2 = 4 - 4x + x^2$$ So, $$\frac{x}{2} \left(4 - 4x + x^2 - x^4 \right) = \frac{x}{2} (4 - 4x + x^2 - x^4)$$ 7. **Set up the single integral:** $$\int_{-2}^{1} \frac{x}{2} (4 - 4x + x^2 - x^4) \, dx = \frac{1}{2} \int_{-2}^{1} x (4 - 4x + x^2 - x^4) \, dx$$ 8. **Distribute $x$ inside the integral:** $$\frac{1}{2} \int_{-2}^{1} (4x - 4x^2 + x^3 - x^5) \, dx$$ 9. **Integrate term by term:** $$\frac{1}{2} \left[ 2x^2 - \frac{4}{3} x^3 + \frac{x^4}{4} - \frac{x^6}{6} \right]_{-2}^{1}$$ 10. **Evaluate at the bounds:** At $x=1$: $$2(1)^2 - \frac{4}{3}(1)^3 + \frac{1^4}{4} - \frac{1^6}{6} = 2 - \frac{4}{3} + \frac{1}{4} - \frac{1}{6}$$ At $x=-2$: $$2(-2)^2 - \frac{4}{3}(-2)^3 + \frac{(-2)^4}{4} - \frac{(-2)^6}{6} = 2(4) - \frac{4}{3}(-8) + \frac{16}{4} - \frac{64}{6} = 8 + \frac{32}{3} + 4 - \frac{64}{6}$$ 11. **Calculate each value:** At $x=1$: $$2 - \frac{4}{3} + \frac{1}{4} - \frac{1}{6} = \frac{24}{12} - \frac{16}{12} + \frac{3}{12} - \frac{2}{12} = \frac{9}{12} = \frac{3}{4}$$ At $x=-2$: $$8 + \frac{32}{3} + 4 - \frac{64}{6} = 12 + \frac{32}{3} - \frac{64}{6}$$ Convert to common denominator 6: $$12 = \frac{72}{6}, \quad \frac{32}{3} = \frac{64}{6}$$ So, $$\frac{72}{6} + \frac{64}{6} - \frac{64}{6} = \frac{72}{6} = 12$$ 12. **Subtract:** $$\left( \frac{3}{4} \right) - 12 = \frac{3}{4} - \frac{48}{4} = -\frac{45}{4}$$ 13. **Multiply by $\frac{1}{2}$:** $$\frac{1}{2} \times -\frac{45}{4} = -\frac{45}{8}$$ **Final answer:** $$\boxed{-\frac{45}{8}}$$