1. **State the problem:** We need to find the variance of the power expended by the student's game console per year. 2. **Recall the power function:** The power is given by $$P = 44X^2 + 22$$ where $X$ is the random variable with the given PDF. 3. **Variance formula:** $$\mathrm{Var}(P) = E[P^2] - (E[P])^2$$ 4. **Calculate $E[P]$ (mean power):** From part (a), $$E[P] = 102.667$$ (rounded to 3 decimals). 5. **Calculate $E[P^2]$:** $$P^2 = (44X^2 + 22)^2 = 1936X^4 + 1936X^2 + 484$$ So, $$E[P^2] = E[1936X^4 + 1936X^2 + 484] = 1936E[X^4] + 1936E[X^2] + 484$$ 6. **Find $E[X^2]$ and $E[X^4]$ using the PDF:** The PDF is piecewise: $$f(x) = \begin{cases} x & 0 < x < 1 \\ 2 - x & 1 \leq x < 2 \\ 0 & \text{otherwise} \end{cases}$$ Calculate $E[X^2]$: $$E[X^2] = \int_0^1 x^2 \cdot x \, dx + \int_1^2 x^2 (2 - x) \, dx = \int_0^1 x^3 \, dx + \int_1^2 (2x^2 - x^3) \, dx$$ Calculate each integral: $$\int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4}$$ $$\int_1^2 2x^2 \, dx = 2 \left[ \frac{x^3}{3} \right]_1^2 = 2 \left( \frac{8}{3} - \frac{1}{3} \right) = 2 \cdot \frac{7}{3} = \frac{14}{3}$$ $$\int_1^2 x^3 \, dx = \left[ \frac{x^4}{4} \right]_1^2 = \frac{16}{4} - \frac{1}{4} = \frac{15}{4}$$ So, $$\int_1^2 (2x^2 - x^3) \, dx = \frac{14}{3} - \frac{15}{4} = \frac{56}{12} - \frac{45}{12} = \frac{11}{12}$$ Therefore, $$E[X^2] = \frac{1}{4} + \frac{11}{12} = \frac{3}{12} + \frac{11}{12} = \frac{14}{12} = \frac{7}{6} \approx 1.167$$ 7. **Calculate $E[X^4]$ similarly:** $$E[X^4] = \int_0^1 x^4 \cdot x \, dx + \int_1^2 x^4 (2 - x) \, dx = \int_0^1 x^5 \, dx + \int_1^2 (2x^4 - x^5) \, dx$$ Calculate each integral: $$\int_0^1 x^5 \, dx = \left[ \frac{x^6}{6} \right]_0^1 = \frac{1}{6}$$ $$\int_1^2 2x^4 \, dx = 2 \left[ \frac{x^5}{5} \right]_1^2 = 2 \left( \frac{32}{5} - \frac{1}{5} \right) = 2 \cdot \frac{31}{5} = \frac{62}{5}$$ $$\int_1^2 x^5 \, dx = \left[ \frac{x^6}{6} \right]_1^2 = \frac{64}{6} - \frac{1}{6} = \frac{63}{6} = 10.5$$ So, $$\int_1^2 (2x^4 - x^5) \, dx = \frac{62}{5} - 10.5 = 12.4 - 10.5 = 1.9$$ Therefore, $$E[X^4] = \frac{1}{6} + 1.9 = 0.1667 + 1.9 = 2.067$$ 8. **Calculate $E[P^2]$:** $$E[P^2] = 1936 \times 2.067 + 1936 \times 1.167 + 484 = 4003.712 + 2259.312 + 484 = 6747.024$$ 9. **Calculate variance:** $$\mathrm{Var}(P) = E[P^2] - (E[P])^2 = 6747.024 - (102.667)^2 = 6747.024 - 10540.89 = -3793.866$$ Since variance cannot be negative, re-check calculations for $E[P^2]$ and $E[P]$. 10. **Recalculate $E[P]$ for verification:** $$E[P] = E[44X^2 + 22] = 44E[X^2] + 22 = 44 \times 1.167 + 22 = 51.333 + 22 = 73.333$$ This differs from the given 102.667, so use this corrected value. 11. **Recalculate variance:** $$\mathrm{Var}(P) = 6747.024 - (73.333)^2 = 6747.024 - 5377.778 = 1369.246$$ 12. **Final answer:** The variance of the power expended by the student's game console per year is approximately $$\boxed{1369.246}$$ (to 3 decimals).