1. We are asked to evaluate two definite integrals: (2) $$\int_0^1 (x^4 - x) \, dx$$ (3) $$\int_0^{\frac{1}{2}} \frac{x}{\sqrt{1 - x^2}} \, dx$$ --- 2. For integral (2), recall the power rule for integration: $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$ where $n \neq -1$. 3. Compute the integral (2): $$\int_0^1 (x^4 - x) \, dx = \int_0^1 x^4 \, dx - \int_0^1 x \, dx$$ Calculate each separately: $$\int_0^1 x^4 \, dx = \left[ \frac{x^5}{5} \right]_0^1 = \frac{1^5}{5} - 0 = \frac{1}{5}$$ $$\int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1^2}{2} - 0 = \frac{1}{2}$$ So, $$\int_0^1 (x^4 - x) \, dx = \frac{1}{5} - \frac{1}{2} = \frac{2}{10} - \frac{5}{10} = -\frac{3}{10}$$ --- 4. For integral (3), use substitution or recognize the integral form. The integral is: $$\int_0^{\frac{1}{2}} \frac{x}{\sqrt{1 - x^2}} \, dx$$ Let $u = 1 - x^2$, then $du = -2x \, dx$ or $-\frac{1}{2} du = x \, dx$. Change limits: When $x=0$, $u=1-0=1$. When $x=\frac{1}{2}$, $u=1 - \left(\frac{1}{2}\right)^2 = 1 - \frac{1}{4} = \frac{3}{4}$. Rewrite the integral: $$\int_{u=1}^{\frac{3}{4}} \frac{x}{\sqrt{u}} \, dx = \int_1^{\frac{3}{4}} \frac{-\frac{1}{2} du}{\sqrt{u}} = -\frac{1}{2} \int_1^{\frac{3}{4}} u^{-\frac{1}{2}} \, du$$ Integrate: $$-\frac{1}{2} \left[ 2 u^{\frac{1}{2}} \right]_1^{\frac{3}{4}} = -\frac{1}{2} \times 2 \left( \sqrt{\frac{3}{4}} - \sqrt{1} \right) = - \left( \frac{\sqrt{3}}{2} - 1 \right) = 1 - \frac{\sqrt{3}}{2}$$ --- 5. Final answers: (2) $$-\frac{3}{10}$$ (3) $$1 - \frac{\sqrt{3}}{2}$$