1. **State the problem:** Evaluate the double integral $$\int_0^1 \int_{\frac{y}{2}}^2 (1 - y^2) \, dx \, dy$$. 2. **Understand the integral:** The integrand is $$1 - y^2$$, which depends only on $$y$$, and the limits for $$x$$ are from $$\frac{y}{2}$$ to $$2$$, and for $$y$$ from $$0$$ to $$1$$. 3. **Integrate with respect to $$x$$ first:** Since $$1 - y^2$$ is constant with respect to $$x$$, $$\int_{\frac{y}{2}}^2 (1 - y^2) \, dx = (1 - y^2) \int_{\frac{y}{2}}^2 dx = (1 - y^2) \left(2 - \frac{y}{2}\right)$$. 4. **Simplify the expression inside the parentheses:** $$2 - \frac{y}{2} = \frac{4 - y}{2}$$. So the integral becomes $$\int_0^1 (1 - y^2) \cdot \frac{4 - y}{2} \, dy = \frac{1}{2} \int_0^1 (1 - y^2)(4 - y) \, dy$$. 5. **Expand the integrand:** $$(1 - y^2)(4 - y) = 4 - y - 4y^2 + y^3$$. 6. **Rewrite the integral:** $$\frac{1}{2} \int_0^1 (4 - y - 4y^2 + y^3) \, dy$$. 7. **Integrate term-by-term:** $$\int_0^1 4 \, dy = 4y \Big|_0^1 = 4$$ $$\int_0^1 (-y) \, dy = -\frac{y^2}{2} \Big|_0^1 = -\frac{1}{2}$$ $$\int_0^1 (-4y^2) \, dy = -4 \cdot \frac{y^3}{3} \Big|_0^1 = -\frac{4}{3}$$ $$\int_0^1 y^3 \, dy = \frac{y^4}{4} \Big|_0^1 = \frac{1}{4}$$ 8. **Sum the results:** $$4 - \frac{1}{2} - \frac{4}{3} + \frac{1}{4} = \frac{48}{12} - \frac{6}{12} - \frac{16}{12} + \frac{3}{12} = \frac{29}{12}$$. 9. **Multiply by $$\frac{1}{2}$$:** $$\frac{1}{2} \times \frac{29}{12} = \frac{29}{24}$$. **Final answer:** $$\boxed{\frac{29}{24}}$$