Let's solve part (a) step by step: --- ### (a) Composite Trapezoidal Method #### **Step 1: Determine the spacing and the nodes** Given the interval \([-3, 3]\), let's denote \( f(x) = x \cos 2x \). Suppose \( n \) subintervals and spacing \( h \) is not specified. Let's assume \( n \) subintervals (for demonstration, I'll use \( n = 6 \) as a reasonable example, giving \( h = 1 \)): - \( h = \frac{b-a}{n} = \frac{3-(-3)}{6} = 1 \) - Nodes: \( x_0 = -3, x_1 = -2, x_2 = -1, x_3 = 0, x_4 = 1, x_5 = 2, x_6 = 3 \) #### **Step 2: Apply the composite trapezoidal rule** The composite trapezoidal rule is: \[ I \approx \frac{h}{2} \left[ f(x_0) + 2\sum_{i=1}^{n-1} f(x_i) + f(x_n) \right] \] Let's evaluate \( f(x) \) at each node: - \( f(-3) = -3\cos(-6) = -3\cos(6) \) - \( f(-2) = -2\cos(-4) = -2\cos(4) \) - \( f(-1) = -1\cos(-2) = -1\cos(2) \) - \( f(0) = 0\cos(0) = 0 \) - \( f(1) = 1\cos(2) \) - \( f(2) = 2\cos(4) \) - \( f(3) = 3\cos(6) \) Plug into the trapezoidal formula: \[ I \approx \frac{1}{2} \Big[ -3\cos(6) + 2(-2\cos(4) - 1\cos(2) + 0 + 1\cos(2) + 2\cos(4)) + 3\cos(6) \Big] \] Calculate inside the bracket: The terms \( -2\cos(4) \) and \( 2\cos(4) \) cancel, The terms \( -1\cos(2) \) and \( 1\cos(2) \) cancel, The terms \( -3\cos(6) \) and \( 3\cos(6) \) cancel. So, everything cancels and the result is: \[ I \approx 0 \] **Explanation:** Due to the symmetry and the fact that \( x \cos 2x \) is an odd function over a symmetric interval about zero, the sum cancels out. **Final Answer for (a):** \[ \boxed{0} \] --- Let me know if you'd like to see the solution for part (b) next, or if you want a more detailed calculation!

Let's proceed to part (b). --- ### (b) Composite Simpson's Rule with \( n = 0.1 \) **Step 1: Determine the number of subintervals \( n \) and the step size \( h \).** Given the interval \([-3, 3]\) and \( n = 0.1 \), the total length is: \[ b - a = 3 - (-3) = 6 \] Number of subintervals: \[ N = \frac{\text{total length}}{h} = \frac{6}{0.1} = 60 \] Since Simpson's rule requires an even number of subintervals, \( N = 60 \) is suitable. Step size: \[ h = 0.1 \] **Step 2: Apply composite Simpson's rule** The formula: \[ I \approx \frac{h}{3} \left[ f(x_0) + 4 \sum_{i=1,3,5,\dots}^{N-1} f(x_i) + 2 \sum_{i=2,4,6,\dots}^{N-2} f(x_i) + f(x_N) \right] \] where: - \( x_0 = -3 \), - \( x_N = 3 \), - \( x_i = a + i h \). **Step 3: Compute the function at each node** The nodes are: \[ x_i = -3 + i \times 0.1,\quad i=0,1,2,\dots,60 \] And: \[ f(x) = x \cos 2x \] The key is to recognize that the function values can be computed and summed accordingly. **Step 4: Approximate the integral** Given the large number of points (60), manual calculation is tedious; however, the process is: - Calculate \( f(x_i) \) at each node. - Sum the terms with appropriate weights: - \( f(x_0) \) and \( f(x_{60}) \) with weight 1. - Odd indices (1,3,5,...,59) with weight 4. - Even indices (2,4,6,...,58) with weight 2. **Note:** Since the function \( f(x) = x \cos 2x \) is well-behaved, this approach provides a good approximation. --- ### **Final approximation:** Because manual computation for all nodes is impractical here, the approach is to: 1. Evaluate \( f(x_i) \) at each node. 2. Sum the weighted values as per Simpson's rule. 3. Multiply by \( h/3 = 0.1/3 \approx 0.03333 \). **Summary:** - The approximate value of the integral using composite Simpson's rule with \( h = 0.1 \) and \( N=60 \) is obtained by summing these weighted function values. **Final Answer (conceptual):** \[ \boxed{ I \approx \frac{0.1}{3} \left[ f(-3) + 4 \sum_{\text{odd } i} f(x_i) + 2 \sum_{\text{even } i} f(x_i) + f(3) \right] } \] --- **Note:** To get a numerical value, you'd compute all \( f(x_i) \) and perform the sums accordingly. --- Would you like me to proceed with part (c) or clarify any step?

Let's proceed to part (c). --- ### (c) Simpson's 3/8 Rule (Simpson's 2/3 rule extended) **Step 1: Determine the number of subintervals \( n \).** The Simpson's 3/8 rule applies when the number of subintervals \( n \) is a multiple of 3. Given the interval \([-3, 3]\): \[ b - a = 6 \] Suppose we choose \( n = 6 \) (since \( 6 \) is divisible by 3), then: \[ h = \frac{b - a}{n} = \frac{6}{6} = 1 \] Nodes: \[ x_i = -3 + i \times 1,\quad i=0,1,\dots,6 \] which are: \[ -3,\ -2,\ -1,\ 0,\ 1,\ 2,\ 3 \] **Step 2: Apply Simpson's 3/8 rule** The formula: \[ I \approx \frac{3h}{8} \left[ f(x_0) + 3 \sum_{i=1}^{n-1} f(x_i) \text{ with weights depending on position} + f(x_n) \right] \] But for uniform steps, the composite Simpson's 3/8 rule is: \[ I \approx \frac{3h}{8} \left[ f(x_0) + 3 \times \text{sum of certain points} + 2 \times \text{others} + f(x_n) \right] \] However, the standard composite Simpson's 3/8 rule for multiple segments is: \[ I \approx \frac{3h}{8} \left[ f(x_0) + 3 \sum_{i=1}^{n-1} f(x_i) \text{ where } i \equiv 1 \text{ mod } 3 \text{ or } 2 \text{ mod } 3 + 2 \sum_{i=1}^{n-1} f(x_i) \text{ where } i \equiv 2 \text{ mod } 3 + f(x_n) \right] \] But more straightforwardly, the composite Simpson's 3/8 rule is often implemented as: \[ I \approx \frac{3h}{8} \left[ f(x_0) + 3 \sum_{i=1,2,4,5,\dots}^{n-1} f(x_i) + 2 \sum_{i=3,6,9,\dots}^{n-3} f(x_i) + f(x_n) \right] \] Given the small \( n=6 \), we can explicitly write: \[ I \approx \frac{3 \times 1}{8} \left[ f(-3) + 3 \left( f(-2) + f(-1) + f(0) + f(1) + f(2) \right) + f(3) \right] \] But this is an approximation, and the standard way is: \[ I \approx \frac{3h}{8} \left[ f(x_0) + 3 \sum_{i \text{ where } i \equiv 1 \text{ or } 2 \mod 3} f(x_i) + 2 \sum_{i \text{ where } i \equiv 0 \mod 3, 1< i < n-1} f(x_i) + f(x_n) \right] \] Given the node points: - \( x_0 = -3 \) - \( x_1 = -2 \) - \( x_2 = -1 \) - \( x_3 = 0 \) - \( x_4 = 1 \) - \( x_5 = 2 \) - \( x_6 = 3 \) The weights are: - \( f(x_0) \) and \( f(x_6) \): weight 1. - \( f(x_1), f(x_2), f(x_4), f(x_5) \): weight 3. - \( f(x_3) \): weight 2. So, \[ I \approx \frac{3 \times 1}{8} \left[ f(-3) + 3(f(-2) + f(-1) + f(1) + f(2)) + 2 f(0) + f(3) \right] \] Calculating: \[ I \approx \frac{3}{8} \left[ f(-3) + 3 (f(-2)+f(-1)+f(1)+f(2)) + 2f(0) + f(3) \right] \] --- ### **Final step:** Compute \( f(x) = x \cos 2x \) at each node: - \( f(-3) = -3 \cos(-6) = -3 \cos 6 \) - \( f(-2) = -2 \cos(-4) = -2 \cos 4 \) - \( f(-1) = -1 \cos(-2) = -1 \cos 2 \) - \( f(0) = 0 \times \cos 0 = 0 \) - \( f(1) = 1 \cos 2 = \cos 2 \) - \( f(2) = 2 \cos 4 \) - \( f(3) = 3 \cos 6 \) Plug into the approximation: \[ I \approx \frac{3}{8} \left[ -3 \cos 6 + 3 \left( -2 \cos 4 - 1 \cos 2 + \cos 2 + 2 \cos 4 \right) + 2 \times 0 + 3 \cos 6 \right] \] Simplify inside the brackets: - \( -2 \cos 4 + 2 \cos 4 = 0 \), - \( -1 \cos 2 + \cos 2 = 0 \), so the sum reduces to: \[ -3 \cos 6 + 3 \times 0 + 0 + 3 \cos 6 = 0 \] Thus, \[ I \approx \frac{3}{8} \times 0 = 0 \] **Final answer:** \[ \boxed{0} \] --- ### **Summary:** - Due to symmetry and the nature of the function over the symmetric interval, the integral approximates to zero using this method as well. --- Would you like to proceed with part (d) or need further clarification?

Let's proceed to part (d): --- ### (d) Gaussian Quadrature with \( n=2 \) **Step 1: Understand Gaussian Quadrature with 2 points** - For \( n=2 \), the 2-point Gaussian quadrature approximates the integral as: \[ I \approx w_1 f(x_1) + w_2 f(x_2) \] - The nodes \( x_1, x_2 \) and weights \( w_1, w_2 \) are standard for \([-1,1]\): \[ x_{1,2} = \pm \frac{1}{\sqrt{3}} \] \[ w_{1,2} = 1 \] - To apply to the interval \([-3,3]\), we need to map these nodes from \([-1,1]\) to \([-3,3]\): \[ x_i^{*} = \frac{b - a}{2} x_i + \frac{a + b}{2} \] where \( a = -3 \), \( b=3 \): \[ x_i^{*} = 3 \times x_i \] So, \[ x_1^{*} = 3 \times \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \] \[ x_2^{*} = -\sqrt{3} \] and the weights need to be scaled: \[ w_i^{*} = \frac{b - a}{2} \times w_i = 3 \times 1 = 3 \] **Step 2: Write the approximation** \[ I \approx \sum_{i=1}^{2} w_i^{*} f(x_i^{*}) = 3 \left[ f(\sqrt{3}) + f(-\sqrt{3}) \right] \] Recall: \[ f(x) = x \cos 2x \] Compute: \[ f(\sqrt{3}) = \sqrt{3} \cos (2 \sqrt{3}) \] \[ f(-\sqrt{3}) = -\sqrt{3} \cos (-2 \sqrt{3}) = -\sqrt{3} \cos (2 \sqrt{3}) \] Thus, \[ f(\sqrt{3}) + f(-\sqrt{3}) = \sqrt{3} \cos (2 \sqrt{3}) - \sqrt{3} \cos (2 \sqrt{3}) = 0 \] Therefore, \[ I \approx 3 \times 0 = 0 \] **Final answer:** \[ \boxed{0} \] --- ### **Summary:** All methods, due to symmetry and the nature of the integrand, suggest the integral is approximately zero. --- **Final note:** The integral of an odd function \( f(x) = x \cos 2x \) over a symmetric interval \([-a, a]\) is exactly zero. Many of these numerical methods confirm this due to symmetry, especially when the function is odd. --- **Final Answer to the entire question:** \[ \boxed{0} \] The approximate value of the integral \(\int_{-3}^{3} x \cos 2x\, dx \) is **0**.

The integral \(\int_{-3}^{3} x \cos 2x\, dx\) involves an odd function over a symmetric interval, so its exact value is zero. **Numerical approximations:** - **(a) Composite Trapezoidal Method:** Due to symmetry and the odd nature of the integrand, the approximation cancels out, resulting in **0**. - **(b) Composite Simpson's Rule with \( n=0.1 \):** With a sufficiently fine partition over a symmetric interval, the approximation approaches **0**. - **(c) Simpson's 3/8 Rule (with \( n=6 \)):** Also yields **0** due to symmetry and the properties of the function. - **(d) Gaussian Quadrature with \( n=2 \):** Evaluates to **0**, since the sum of the function at symmetric nodes cancels out. **Final answer: \(\boxed{0}\)**