Let's solve **part (a): Composite Trapezoidal method with \(h = 0.2\)** step by step. We want to approximate the integral: \[ I = \int_{-1}^{1} x \cos(2x) \, dx \] ### **Step 1: Set up the Trapezoidal Rule** The interval \([-1, 1]\) is divided into subintervals of width \(h = 0.2\). - Number of subintervals: \( n = \frac{1 - (-1)}{0.2} = 10 \) - Points: \( x_0 = -1, \; x_1 = -0.8, \ldots, x_{10} = 1 \) - Trapezoidal Rule formula: \[ I \approx \frac{h}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n) \right] \] where \( f(x) = x \cos(2x) \) **Explanation:** We divide the integration range into 10 equal intervals of width 0.2 and set up the composite trapezoidal formula. --- ### **Step 2: Calculate the Function Values and Substitute** Calculate \( f(x) \) at all points: | \(i\) | \(x_i\) | \(f(x_i) = x_i\cos(2x_i)\) | |------|--------|-----------------------------| | 0 | -1.0 | \(-1.0 \cos(-2)\) | | 1 | -0.8 | \(-0.8 \cos(-1.6)\) | | 2 | -0.6 | \(-0.6 \cos(-1.2)\) | | 3 | -0.4 | \(-0.4 \cos(-0.8)\) | | 4 | -0.2 | \(-0.2 \cos(-0.4)\) | | 5 | 0 | \(0 \cos(0) = 0\) | | 6 | 0.2 | \(0.2 \cos(0.4)\) | | 7 | 0.4 | \(0.4 \cos(0.8)\) | | 8 | 0.6 | \(0.6 \cos(1.2)\) | | 9 | 0.8 | \(0.8 \cos(1.6)\) | | 10 | 1.0 | \(1.0 \cos(2)\) | Plugging these into the formula: \[ I \approx \frac{0.2}{2} \left[ f(-1) + 2f(-0.8) + 2f(-0.6) + \ldots + 2f(0.8) + f(1) \right] \] **Explanation:** We evaluate the function at all 11 points, apply the trapezoidal weights, and sum them up. --- ## **Final Answer (rounded to 4 decimal places):** Let's quickly calculate the actual values (using \(\cos(-x) = \cos(x)\)): - \(f(-1) = -1 \cos(2) \approx -1 \times -0.4161 = -0.4161\) - \(f(-0.8) = -0.8 \cos(1.6) \approx -0.8 \times -0.0290 = 0.0232\) - \(f(-0.6) = -0.6 \cos(1.2) \approx -0.6 \times 0.3624 = -0.2174\) - \(f(-0.4) = -0.4 \cos(0.8) \approx -0.4 \times 0.6967 = -0.2787\) - \(f(-0.2) = -0.2 \cos(0.4) \approx -0.2 \times 0.9211 = -0.1842\) - \(f(0) = 0\) - \(f(0.2) = 0.2 \times 0.9211 = 0.1842\) - \(f(0.4) = 0.4 \times 0.6967 = 0.2787\) - \(f(0.6) = 0.6 \times 0.3624 = 0.2174\) - \(f(0.8) = 0.8 \times -0.0290 = -0.0232\) - \(f(1) = 1 \times -0.4161 = -0.4161\) Sum: \[ I \approx 0.1 \left[ -0.4161 + 2(0.0232 - 0.2174 - 0.2787 - 0.1842 + 0.1842 + 0.2787 + 0.2174 - 0.0232) - 0.4161 \right] \] Calculate the sum inside the brackets: \[ 2(0.0232 - 0.2174 - 0.2787 - 0.1842 + 0.1842 + 0.2787 + 0.2174 - 0.0232) = 2(0) = 0 \] So the sum is: \[ -0.4161 + 0 - 0.4161 = -0.8322 \] Multiply by 0.1: \[ I \approx 0.1 \times -0.8322 = -0.0832 \] ### **Final Answer (Trapezoidal, \(h = 0.2\)):** \[ \boxed{-0.0832} \] --- Do you want to proceed with (b), (c), or (d)?

Sure! Here is the complete solution for all parts (a), (b), (c), and (d): --- ### **Part (a): Composite Trapezoidal Method with \(h=0.2\)** **Setup:** - Interval: \([-1, 1]\) - \(h = 0.2\) - Number of subintervals: \(n=10\) **Function evaluations:** | \(x_i\) | \(f(x_i) = x_i \cos(2x_i)\) | Approximate value | |---------|----------------------------|-------------------| | -1.0 | \(-1 \times \cos(-2)\) | \(-0.4161\) | | -0.8 | \(-0.8 \times \cos(-1.6)\) | \(0.0232\) | | -0.6 | \(-0.6 \times \cos(-1.2)\) | \(-0.2174\) | | -0.4 | \(-0.4 \times \cos(-0.8)\) | \(-0.2787\) | | -0.2 | \(-0.2 \times \cos(-0.4)\) | \(-0.1842\) | | 0 | 0 | 0 | | 0.2 | \(0.2 \times \cos(0.4)\) | 0.1842 | | 0.4 | \(0.4 \times \cos(0.8)\) | 0.2787 | | 0.6 | \(0.6 \times \cos(1.2)\) | 0.2174 | | 0.8 | \(0.8 \times \cos(1.6)\) | -0.0232 | | 1.0 | \(1 \times \cos(2)\) | -0.4161 | **Approximate integral:** \[ I \approx \frac{h}{2} \left[ f(-1) + 2 \sum_{i=1}^{n-1} f(x_i) + f(1) \right] \] Calculating: \[ I \approx 0.1 \times \left[ -0.4161 + 2(0.0232 - 0.2174 - 0.2787 - 0.1842 + 0.1842 + 0.2787 + 0.2174 - 0.0232) - 0.4161 \right] \] Sum of interior points: \[ 2 \times (0.0232 - 0.2174 - 0.2787 - 0.1842 + 0.1842 + 0.2787 + 0.2174 - 0.0232) = 0 \] Total sum: \[ -0.4161 + 0 - 0.4161 = -0.8322 \] Final approximation: \[ \boxed{-0.0832} \] --- ### **Part (b): Composite Simpson's Rule with \(n=5\)** **Setup:** - Interval: \([-1, 1]\) - \(n=5\), which is odd and suitable for Simpson's rule - \(h = \frac{2}{5} = 0.4\) **Points:** \[ x_i = -1 + i \times 0.4, \quad i=0,1,2,3,4,5 \] Values: | \(x_i\) | \(f(x_i)\) | |---------|------------| | -1.0 | \(-0.4161\) | | -0.6 | \(-0.2174\) | | -0.2 | \(-0.1842\) | | 0.2 | 0.1842 | | 0.6 | 0.2174 | | 1.0 | -0.4161 | **Simpson's rule formula:** \[ I \approx \frac{h}{3} \left[ f(x_0) + 4 \sum_{\text{odd }i} f(x_i) + 2 \sum_{\text{even }i} f(x_i), \text{excluding endpoints} \right] \] But more straightforwardly: \[ I \approx \frac{h}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + f(x_5) \right] \] Calculations: \[ I \approx \frac{0.4}{3} \left[ -0.4161 + 4( -0.2174 ) + 2( -0.1842 ) + 4( 0.1842 ) + 2( 0.2174 ) - 0.4161 \right] \] Compute sum inside: \[ -0.4161 + (-0.8696) - 0.3684 + 0.7368 + 0.4348 - 0.4161 = -0.8686 \] Then: \[ I \approx \frac{0.4}{3} \times (-0.8686) \approx 0.1333 \times (-0.8686) \approx -0.1158 \] **Final answer:** \[ \boxed{-0.1158} \] --- ### **Part (c): Simpson's 3/8 Rule** **Setup:** - Same interval: \([-1,1]\) - Number of subintervals: \(n=3\) (since Simpson's 3/8 rule requires n divisible by 3) - \(h = \frac{2}{3} \approx 0.6667\) Points: \[ x_i = -1 + i \times 0.6667, \quad i=0,1,2,3 \] Values: | \(x_i\) | \(f(x_i)\) | |---------|------------| | -1.0 | \(-0.4161\) | | -0.3333 | \(-0.1081\) | | 0.3333 | 0.1081 | | 1.0 | -0.4161 | **Formula:** \[ I \approx \frac{3h}{8} [f(x_0) + 3f(x_1) + 3f(x_2) + f(x_3)] \] Calculate: \[ I \approx \frac{3 \times 0.6667}{8} \times [ -0.4161 + 3(-0.1081) + 3(0.1081) - 0.4161 ] \] Sum inside brackets: \[ -0.4161 - 0.3243 + 0.3243 - 0.4161 = -0.8322 \] Coefficient: \[ \frac{2.0}{8} = 0.25 \] Final approximation: \[ I \approx 0.25 \times (-0.8322) = -0.2081 \] **Final answer:** \[ \boxed{-0.2081} \] --- ### **Part (d): Gaussian Quadrature with \(n=2\)** **Setup:** - Interval: \([-1,1]\) - Nodes and weights for 2-point Gaussian quadrature: \[ x_{1,2} = \pm \frac{1}{\sqrt{3}} \approx \pm 0.5774 \] Weights: both 1 **Transform nodes to the interval \([-1,1]\):** \[ x_i' = \frac{b - a}{2} x_i + \frac{a + b}{2} = 1 \times x_i + 0 \] since interval is symmetric about zero (from -1 to 1). Values: \[ f(-0.5774) = -0.5774 \times \cos(-1.1547) \approx -0.5774 \times 0.4049 \approx -0.2340 \] \[ f(0.5774) = 0.5774 \times \cos(1.1547) \approx 0.5774 \times 0.4049 \approx 0.2340 \] Gaussian quadrature approximation: \[ I \approx f(-0.5774) + f(0.5774) \approx -0.2340 + 0.2340 = 0 \] **Final answer:** \[ \boxed{0} \] --- **Summary:** | Part | Method | Approximate value | |---------|------------------------------|-------------------| | (a) | Composite Trapezoidal (\(h=0.2\)) | \(-0.0832\) | | (b) | Simpson's Rule (\(n=5\)) | \(-0.1158\) | | (c) | Simpson's 3/8 Rule (\(n=3\)) | \(-0.2081\) | | (d) | Gaussian Quadrature (\(n=2\)) | 0 | --- Let me know if you'd like detailed calculations or explanations for any specific part!