1. **Problem statement:** Given the piecewise probability density function $$f(x) = \begin{cases} c(x + 3) & 0 < x < 2 \\ c(7 - x) & 2 < x < 4 \\ 0 & \text{elsewhere} \end{cases}$$ (i) We need to sketch the graph and find the constant $c$. (ii) Find the expectation $E(x)$. (iii) Find the probability $P(1 \leq x < 3)$. --- **Step 1 (Find $c$):** Since $f(x)$ is a probability density function, the total area under the curve must be 1: $$\int_{0}^{2} c(x+3) \, dx + \int_{2}^{4} c(7 - x) \, dx = 1$$ Calculate each integral: $$\int_{0}^{2} (x+3) \, dx = \left[\frac{x^{2}}{2} + 3x\right]_0^{2} = \left(\frac{4}{2} + 6 \right) - 0 = 2 + 6 = 8$$ $$\int_{2}^{4} (7-x) \, dx = \left[7x - \frac{x^{2}}{2}\right]_2^{4} = \left(28 - 8 \right) - \left(14 - 2\right) = 20 - 12 = 8$$ Total integral: $$c(8 + 8) = c \times 16 = 1 \implies c = \frac{1}{16}$$ --- **Step 2 (Expectation $E(x)$):** $$E(x) = \int_{0}^{2} x f(x) dx + \int_{2}^{4} x f(x) dx$$ Calculate each part: $$\int_{0}^{2} x c(x+3) dx = c \int_{0}^{2} (x^{2} + 3x) dx = c \left[ \frac{x^3}{3} + \frac{3x^{2}}{2} \right]_0^{2} = \frac{1}{16} \left( \frac{8}{3} + 6 \right) = \frac{1}{16} \times \frac{26}{3} = \frac{26}{48} = \frac{13}{24}$$ $$\int_{2}^{4} x c(7 - x) dx = c \int_{2}^{4} (7x - x^{2}) dx = c \left[ \frac{7x^2}{2} - \frac{x^3}{3} \right]_2^{4}$$ Calculate inside bracket: At 4: $\frac{7 \times 16}{2} - \frac{64}{3} = 56 - \frac{64}{3} = \frac{168 - 64}{3} = \frac{104}{3}$ At 2: $\frac{7 \times 4}{2} - \frac{8}{3} = 14 - \frac{8}{3} = \frac{42 - 8}{3} = \frac{34}{3}$ Difference: $$\frac{104}{3} - \frac{34}{3} = \frac{70}{3}$$ Multiply by $c$: $$\frac{1}{16} \times \frac{70}{3} = \frac{70}{48} = \frac{35}{24}$$ Sum expectation: $$E(x) = \frac{13}{24} + \frac{35}{24} = \frac{48}{24} = 2$$ --- **Step 3 (Find $P(1 \leq x < 3)$):** Split the integral since the function changes at $x=2$: $$P(1 \leq x < 2) = \int_1^{2} c(x+3) dx = c \left[ \frac{x^2}{2} + 3x \right]_1^{2} = \frac{1}{16} \left( (2 + 6) - \left(\frac{1}{2} + 3\right) \right) = \frac{1}{16} (8 - 3.5) = \frac{1}{16} \times 4.5 = \frac{9}{32}$$ And $$P(2 \leq x < 3) = \int_2^{3} c(7 - x) dx = c \left[ 7x - \frac{x^2}{2} \right]_2^{3} = \frac{1}{16} \left( (21 - 4.5) - (14 - 2) \right) = \frac{1}{16} (16.5 - 12) = \frac{1}{16} \times 4.5 = \frac{9}{32}$$ Add probabilities: $$P(1 \leq x < 3) = \frac{9}{32} + \frac{9}{32} = \frac{18}{32} = \frac{9}{16}$$ --- Final answers: - $c = \frac{1}{16}$ - $E(x) = 2$ - $P(1 \leq x < 3) = \frac{9}{16}$