Let's solve both parts step by step: --- ## **Q.4 (a): Composite Simpson's Rule** Given: \[ \int_{-2}^{2} x^3 e^x dx, \quad n = 4 \] ### **Step 1: Find \( h \)** \[ h = \frac{b-a}{n} = \frac{2 - (-2)}{4} = \frac{4}{4} = 1 \] ### **Step 2: Compute the x-values** \[ x_0 = -2 \\ x_1 = -1 \\ x_2 = 0 \\ x_3 = 1 \\ x_4 = 2 \] ### **Step 3: Compute f(x) at each point** Let \( f(x) = x^3 e^x \): - \( f(x_0) = (-2)^3 e^{-2} = -8 e^{-2} \) - \( f(x_1) = (-1)^3 e^{-1} = -1 e^{-1} \) - \( f(x_2) = (0)^3 e^{0} = 0 \) - \( f(x_3) = (1)^3 e^{1} = e \) - \( f(x_4) = (2)^3 e^{2} = 8 e^{2} \) ### **Step 4: Apply the Formula** The **Composite Simpson's Rule** for \( n = 4 \): \[ \int_{a}^{b} f(x)dx \approx \frac{h}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + f(x_4) \right] \] Plug in the computed values: \[ \int_{-2}^{2} x^3 e^x dx \approx \frac{1}{3} [f(-2) + 4f(-1) + 2f(0) + 4f(1) + f(2)] \] \[ = \frac{1}{3} \left[ -8e^{-2} + 4(-e^{-1}) + 2(0) + 4e + 8e^2 \right] \] \[ = \frac{1}{3} \left[ -8e^{-2} - 4e^{-1} + 4e + 8e^2 \right] \] ### **Step 5: Final Calculation** Let’s substitute approximate values: - \( e^{-2} \approx 0.1353 \) - \( e^{-1} \approx 0.3679 \) - \( e \approx 2.7183 \) - \( e^2 \approx 7.3891 \) \[ = \frac{1}{3} \left[ -8(0.1353) - 4(0.3679) + 4(2.7183) + 8(7.3891) \right] \] \[ = \frac{1}{3} \left[ -1.0824 - 1.4716 + 10.8732 + 59.1128 \right] \] \[ = \frac{1}{3} [67.43196] \] \[ = 22.4773 \] **Final Answer:** \[ \boxed{22.48} \] (Rounded to two decimal places) --- ## **Q.4 (b): Gauss-Seidel Iterative Method** Given system: \[ \begin{cases} 20x_1 + x_2 - 2x_3 = 17 \\ 3x_1 + 20x_2 - x_3 = -18 \\ 20x_1 - 3x_2 + 20x_3 = 25 \end{cases} \] Initial guess: \( x^{(0)} = (0, 0, 0)^T \) ### **Step 1: Rearranged Equations for Iterative Solution** 1. \( x_1 = \frac{1}{20}(17 - x_2 + 2x_3) \) 2. \( x_2 = \frac{1}{20}(-18 - 3x_1 + x_3) \) 3. \( x_3 = \frac{1}{20}(25 - 20x_1 + 3x_2) \) --- ### **Step 2: Iteration 1** Start with \( (x_1^{(0)}, x_2^{(0)}, x_3^{(0)}) = (0, 0, 0) \): - \( x_1^{(1)} = \frac{1}{20}[17 - 0 + 0] = \frac{17}{20} = 0.85 \) - \( x_2^{(1)} = \frac{1}{20}[-18 - 3(0.85) + 0] = \frac{1}{20}[-18 - 2.55] = \frac{-20.55}{20} = -1.0275 \) - \( x_3^{(1)} = \frac{1}{20}[25 - 20(0.85) + 3(-1.0275)] = \frac{1}{20}[25 - 17 + (-3.0825)] = \frac{1}{20}[4.9175] = 0.2459 \) --- ### **Step 3: Iteration 2** Now use previous results: - \( x_1^{(2)} = \frac{1}{20}[17 - (-1.0275) + 2(0.2459)] = \frac{1}{20}[17 + 1.0275 + 0.4918] = \frac{1}{20}[18.5193] = 0.9260 \) - \( x_2^{(2)} = \frac{1}{20}[-18 - 3(0.9260) + 0.2459] = \frac{1}{20}[-18 - 2.778 + 0.2459] = \frac{1}{20}[-20.5321] = -1.0266 \) - \( x_3^{(2)} = \frac{1}{20}[25 - 20(0.9260) + 3(-1.0266)] = \frac{1}{20}[25 - 18.52 - 3.0798] = \frac{1}{20}[3.4002] = 0.1700 \) --- ## **Summary Table** | Iteration | \(x_1\) | \(x_2\) | \(x_3\) | |-----------|----------|-----------|-----------| | 0 | 0 | 0 | 0 | | 1 | 0.85 | -1.0275 | 0.2459 | | 2 | 0.9260 | -1.0266 | 0.1700 | --- ### **Final Answer** After two iterations, the approximate solution is: \[ \boxed{ x_1 \approx 0.9260, \quad x_2 \approx -1.0266, \quad x_3 \approx 0.1700 } \] --- If you need more iterations or further details, let me know!

**Final Answers:** **Q.4 (a):** \(\boxed{22.48}\) **Q.4 (b):** \(\boxed{ x_1 \approx 0.9260,\quad x_2 \approx -1.0266,\quad x_3 \approx 0.1700 }\)