1. **Problem statement:** Given the density function of a random variable $X$ as $f(x) = kx(2-x)$, find the constant $k$, the $r$th moment, the mean, and the variance. 2. **Step 1: Find $k$ using the property of probability density functions (PDFs):** The total area under the PDF must be 1, so $$\int_0^2 kx(2-x) \, dx = 1$$ 3. **Step 2: Calculate the integral:** $$\int_0^2 x(2-x) \, dx = \int_0^2 (2x - x^2) \, dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = (4 - \frac{8}{3}) - 0 = \frac{12}{3} - \frac{8}{3} = \frac{4}{3}$$ 4. **Step 3: Solve for $k$:** $$k \times \frac{4}{3} = 1 \implies k = \frac{3}{4}$$ 5. **Step 4: Find the $r$th moment $E[X^r]$:** $$E[X^r] = \int_0^2 x^r f(x) \, dx = \int_0^2 x^r \cdot \frac{3}{4} x (2-x) \, dx = \frac{3}{4} \int_0^2 x^{r+1} (2-x) \, dx$$ 6. **Step 5: Expand and integrate:** $$\frac{3}{4} \int_0^2 (2x^{r+1} - x^{r+2}) \, dx = \frac{3}{4} \left[ 2 \frac{x^{r+2}}{r+2} - \frac{x^{r+3}}{r+3} \right]_0^2 = \frac{3}{4} \left( \frac{2 \cdot 2^{r+2}}{r+2} - \frac{2^{r+3}}{r+3} \right)$$ 7. **Step 6: Simplify the $r$th moment:** $$E[X^r] = \frac{3}{4} \cdot 2^{r+2} \left( \frac{1}{r+2} - \frac{1}{2(r+3)} \right) = 3 \cdot 2^r \left( \frac{1}{r+2} - \frac{1}{2(r+3)} \right)$$ 8. **Step 7: Calculate the mean ($r=1$):** $$E[X] = 3 \cdot 2^1 \left( \frac{1}{3} - \frac{1}{2 \cdot 4} \right) = 6 \left( \frac{1}{3} - \frac{1}{8} \right) = 6 \left( \frac{8}{24} - \frac{3}{24} \right) = 6 \times \frac{5}{24} = \frac{5}{4} = 1.25$$ 9. **Step 8: Calculate the second moment ($r=2$):** $$E[X^2] = 3 \cdot 2^2 \left( \frac{1}{4} - \frac{1}{2 \cdot 5} \right) = 12 \left( \frac{1}{4} - \frac{1}{10} \right) = 12 \left( \frac{5}{20} - \frac{2}{20} \right) = 12 \times \frac{3}{20} = \frac{36}{20} = 1.8$$ 10. **Step 9: Calculate the variance:** $$\text{Var}(X) = E[X^2] - (E[X])^2 = 1.8 - (1.25)^2 = 1.8 - 1.5625 = 0.2375$$ **Final answers:** - $k = \frac{3}{4}$ - $r$th moment: $$E[X^r] = 3 \cdot 2^r \left( \frac{1}{r+2} - \frac{1}{2(r+3)} \right)$$ - Mean: $1.25$ - Variance: $0.2375$