1. **State the problem:** Evaluate the integral $$\int x^2 \sqrt{1-x} \, dx$$ using substitution. 2. **Choose substitution:** Let $$u = 1 - x$$. Then, $$du = -dx$$ or $$dx = -du$$. 3. **Rewrite the integral in terms of $$u$$:** Since $$x = 1 - u$$, then $$x^2 = (1 - u)^2$$ and $$\sqrt{1-x} = \sqrt{u}$$. 4. Substitute into the integral: $$\int x^2 \sqrt{1-x} \, dx = \int (1-u)^2 \sqrt{u} (-du) = -\int (1-u)^2 u^{1/2} \, du$$ 5. **Expand the integrand:** $$(1-u)^2 = 1 - 2u + u^2$$ So, $$-\int (1 - 2u + u^2) u^{1/2} \, du = -\int (u^{1/2} - 2u^{3/2} + u^{5/2}) \, du$$ 6. **Integrate term-by-term:** $$-\left( \int u^{1/2} du - 2 \int u^{3/2} du + \int u^{5/2} du \right)$$ Calculate each integral: $$\int u^{1/2} du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2}$$ $$\int u^{3/2} du = \frac{u^{5/2}}{5/2} = \frac{2}{5} u^{5/2}$$ $$\int u^{5/2} du = \frac{u^{7/2}}{7/2} = \frac{2}{7} u^{7/2}$$ 7. Substitute back: $$-\left( \frac{2}{3} u^{3/2} - 2 \cdot \frac{2}{5} u^{5/2} + \frac{2}{7} u^{7/2} \right) + C = -\left( \frac{2}{3} u^{3/2} - \frac{4}{5} u^{5/2} + \frac{2}{7} u^{7/2} \right) + C$$ 8. **Simplify:** $$= -\frac{2}{3} u^{3/2} + \frac{4}{5} u^{5/2} - \frac{2}{7} u^{7/2} + C$$ 9. **Back-substitute $$u = 1 - x$$:** $$= -\frac{2}{3} (1-x)^{3/2} + \frac{4}{5} (1-x)^{5/2} - \frac{2}{7} (1-x)^{7/2} + C$$ **Final answer:** $$\int x^2 \sqrt{1-x} \, dx = -\frac{2}{3} (1-x)^{3/2} + \frac{4}{5} (1-x)^{5/2} - \frac{2}{7} (1-x)^{7/2} + C$$