1. **Problem Statement:** Evaluate the integrals: (a) $\int (y^2 + y^{-2})\, dy$ (b) $\int_1^2 (y^2 + y^{-2})\, dy$ (c) $\int_{-1}^2 (y^2 + y^{-2})\, dy$ 2. **Step (a): Indefinite integral** Write the integral: $$\int (y^2 + y^{-2})\, dy = \int y^2\, dy + \int y^{-2}\, dy$$ For each term, use the power rule for integration: $$\int y^n\, dy = \frac{y^{n+1}}{n+1} + C \quad \text{for } n \neq -1$$ Apply to each: $$\int y^2\, dy = \frac{y^{3}}{3} + C$$ $$\int y^{-2}\, dy = \frac{y^{-1}}{-1} + C = -\frac{1}{y} + C$$ Combine: $$\int (y^2 + y^{-2})\, dy = \frac{y^{3}}{3} - \frac{1}{y} + C$$ 3. **Step (b): Definite integral from 1 to 2** Evaluate: $$\int_1^2 (y^2 + y^{-2})\, dy = \left[ \frac{y^3}{3} - \frac{1}{y} \right]_1^2 = \left( \frac{2^3}{3} - \frac{1}{2} \right) - \left( \frac{1^3}{3} - 1 \right)$$ Calculate each term: $$\frac{2^3}{3} = \frac{8}{3}, \quad -\frac{1}{2} = -0.5$$ $$\frac{1^3}{3} = \frac{1}{3}, \quad -1 = -1$$ Substitute: $$\left( \frac{8}{3} - \frac{1}{2} \right) - \left( \frac{1}{3} - 1 \right) = \left( \frac{8}{3} - 0.5 \right) - \left( \frac{1}{3} - 1 \right)$$ Calculate inside each parenthesis: $$\frac{8}{3} - 0.5 = \frac{8}{3} - \frac{1}{2} = \frac{16}{6} - \frac{3}{6} = \frac{13}{6}$$ $$\frac{1}{3} - 1 = \frac{1}{3} - \frac{3}{3} = -\frac{2}{3}$$ So the expression becomes: $$\frac{13}{6} - \left(-\frac{2}{3}\right) = \frac{13}{6} + \frac{2}{3} = \frac{13}{6} + \frac{4}{6} = \frac{17}{6}$$ Thus, the value is $\frac{17}{6}$. 4. **Step (c): Definite integral from -1 to 2** The integrand $y^{-2} = \frac{1}{y^2}$ is not defined at $y=0$, so the integral is improper with a vertical asymptote at 0. We split the integral at 0: $$\int_{-1}^2 (y^2 + y^{-2})\, dy = \int_{-1}^0 (y^2 + y^{-2})\, dy + \int_0^2 (y^2 + y^{-2})\, dy$$ The term $\int y^{-2} dy$ diverges near 0 since $\lim_{y \to 0} -1/y$ goes to infinity. Hence, the integral does not converge (it diverges) due to the discontinuity at 0. **Final result:** The integral is divergent and thus undefined. **Summary:** (a) $\int (y^2 + y^{-2})\, dy = \frac{y^3}{3} - \frac{1}{y} + C$ (b) $\int_1^2 (y^2 + y^{-2})\, dy = \frac{17}{6}$ (c) $\int_{-1}^2 (y^2 + y^{-2})\, dy$ diverges and is undefined.