Let's break down the solution for each part step-by-step. ### **Given:** \[ T = \int_{0.25}^{1} \frac{dh}{10 - 20h^2} \] Let \( f(h) = \frac{1}{10 - 20h^2} \). #### **First, let's calculate the values of \( f(h) \) at the required points for all rules.** #### **Limits:** - Lower limit, \( a = 0.25 \) - Upper limit, \( b = 1 \) --- ## **A) Trapezoidal Rule** Let’s use \( n = 3 \) intervals (so \( h = \frac{b-a}{n} = \frac{1-0.25}{3} = 0.25 \)). Points: - \( h_0 = 0.25 \) - \( h_1 = 0.5 \) - \( h_2 = 0.75 \) - \( h_3 = 1 \) Calculate \( f(h) \) at these points: 1. \( f(0.25) = \frac{1}{10 - 20(0.25)^2} = \frac{1}{10 - 20 \times 0.0625} = \frac{1}{10 - 1.25} = \frac{1}{8.75} = 0.1142857 \) 2. \( f(0.5) = \frac{1}{10 - 20(0.5)^2} = \frac{1}{10 - 20 \times 0.25} = \frac{1}{10 - 5} = \frac{1}{5} = 0.2 \) 3. \( f(0.75) = \frac{1}{10 - 20(0.75)^2} = \frac{1}{10 - 20 \times 0.5625} = \frac{1}{10 - 11.25} = \frac{1}{-1.25} = -0.8 \) 4. \( f(1) = \frac{1}{10 - 20(1)^2} = \frac{1}{10 - 20} = \frac{1}{-10} = -0.1 \) Trapezoidal Rule formula: \[ T \approx \frac{h}{2} \left[f(h_0) + 2f(h_1) + 2f(h_2) + f(h_3)\right] \] \[ T \approx \frac{0.25}{2} \left[0.1142857 + 2(0.2) + 2(-0.8) + (-0.1)\right] \] \[ = 0.125 \left[0.1142857 + 0.4 - 1.6 - 0.1\right] \] \[ = 0.125 \left[-1.1857143\right] = -0.1482143 \] **Final answer (A):** \[ \boxed{T \approx -0.148} \] --- ## **B) Improved Simpson's 1/3 Rule (with \( h = 0.0625 \))** Number of intervals: \[ h = 0.0625, \quad n = \frac{1-0.25}{0.0625} = 12 \] So, there are 13 points: \( h_0, h_1, ..., h_{12} \) with \( h_k = 0.25 + 0.0625k \). Simpson's 1/3 Rule: \[ T \approx \frac{h}{3} \left[ f(h_0) + 4 \sum_{\text{odd } i} f(h_i) + 2 \sum_{\text{even } i, i\neq0,n} f(h_i) + f(h_{12}) \right] \] Let’s just set up the formula, as the calculations are lengthy for 13 points: Calculate \( f(h_i) \) for \( i = 0 \) to \( 12 \), sum them as per the rule above. **Formula setup:** \[ T \approx \frac{0.0625}{3} \left[ f(0.25) + 4(f(0.3125) + f(0.4375) + ... + f(0.9375)) + 2(f(0.375) + f(0.5) + ... + f(0.875)) + f(1) \right] \] Plug in the values for all 13 points and compute the sum. --- ## **C) Simpson's 3/8 Rule with 3 Segments** For Simpson’s 3/8 rule, n must be a multiple of 3. Here, \( n = 3 \), so \( h = \frac{1-0.25}{3} = 0.25 \) Simpson’s 3/8 Rule: \[ T \approx \frac{3h}{8} \left[ f(h_0) + 3f(h_1) + 3f(h_2) + f(h_3) \right] \] Values: - \( f(0.25) = 0.1142857 \) - \( f(0.5) = 0.2 \) - \( f(0.75) = -0.8 \) - \( f(1) = -0.1 \) \[ T \approx \frac{3 \times 0.25}{8} \left[ 0.1142857 + 3(0.2) + 3(-0.8) + (-0.1) \right] \] \[ = \frac{0.75}{8} \left[ 0.1142857 + 0.6 - 2.4 - 0.1 \right] \] \[ = 0.09375 \times (-1.7857143) = -0.1673661 \] **Final answer (C):** \[ \boxed{T \approx -0.167} \] --- ## **Summary Table** | Method | Approximate Value of T | |---------------------------------|:---------------------:| | Trapezoidal Rule | -0.148 | | Simpson's 1/3 Rule (h=0.0625) | (calculate sum) | | Simpson's 3/8 Rule (3 segments)| -0.167 | **Note:** For Simpson's 1/3 Rule, you need to sum the values as per the formula. If you want the exact numerical answer, let me know if you want me to evaluate the sum as well.

Certainly! Let's perform the detailed calculations for **Part B: Simpson's 1/3 Rule** with \( h = 0.0625 \). --- ### **Step 1: Determine the number of intervals** \[ n = \frac{b - a}{h} = \frac{1 - 0.25}{0.0625} = \frac{0.75}{0.0625} = 12 \] Since \( n = 12 \), the points are: \[ h_i = 0.25 + i \times 0.0625,\quad i=0,1,2,...,12 \] which gives: | \(i\) | \(h_i\) | Calculation | |--------|--------------|-------------------------------------| | 0 | 0.25 | initial point | | 1 | 0.3125 | 0.25 + 1×0.0625 | | 2 | 0.375 | 0.25 + 2×0.0625 | | 3 | 0.4375 | ... | | 4 | 0.5 | ... | | 5 | 0.5625 | ... | | 6 | 0.625 | ... | | 7 | 0.6875 | ... | | 8 | 0.75 | ... | | 9 | 0.8125 | ... | | 10 | 0.875 | ... | | 11 | 0.9375 | ... | | 12 | 1.0 | final point | --- ### **Step 2: Calculate \(f(h_i) = \frac{1}{10 - 20h_i^2}\) at each point** Let's compute each: - \( f(0.25) \): \[ f(0.25) = \frac{1}{10 - 20 \times (0.25)^2} = \frac{1}{10 - 20 \times 0.0625} = \frac{1}{10 - 1.25} = \frac{1}{8.75} \approx 0.1142857 \] - \( f(0.3125) \): \[ f(0.3125) = \frac{1}{10 - 20 \times (0.3125)^2} = \frac{1}{10 - 20 \times 0.09765625} = \frac{1}{10 - 1.953125} = \frac{1}{8.046875} \approx 0.124341 \] - \( f(0.375) \): \[ f(0.375) = \frac{1}{10 - 20 \times (0.375)^2} = \frac{1}{10 - 20 \times 0.140625} = \frac{1}{10 - 2.8125} = \frac{1}{7.1875} \approx 0.139229 \] - \( f(0.4375) \): \[ f(0.4375) = \frac{1}{10 - 20 \times (0.4375)^2} = \frac{1}{10 - 20 \times 0.19140625} = \frac{1}{10 - 3.828125} = \frac{1}{6.171875} \approx 0.162089 \] - \( f(0.5) \): \[ f(0.5) = \frac{1}{10 - 20 \times 0.25} = \frac{1}{10 - 5} = \frac{1}{5} = 0.2 \] - \( f(0.5625) \): \[ f(0.5625) = \frac{1}{10 - 20 \times 0.31640625} = \frac{1}{10 - 6.328125} = \frac{1}{3.671875} \approx 0.272453 \] - \( f(0.625) \): \[ f(0.625) = \frac{1}{10 - 20 \times 0.390625} = \frac{1}{10 - 7.8125} = \frac{1}{2.1875} \approx 0.457143 \] - \( f(0.6875) \): \[ f(0.6875) = \frac{1}{10 - 20 \times 0.47265625} = \frac{1}{10 - 9.453125} = \frac{1}{0.546875} \approx 1.826 \] - \( f(0.75) \): \[ f(0.75) = \frac{1}{10 - 20 \times 0.5625} = \frac{1}{10 - 11.25} = \frac{1}{-1.25} = -0.8 \] - \( f(0.8125) \): \[ f(0.8125) = \frac{1}{10 - 20 \times 0.66015625} = \frac{1}{10 - 13.203125} = \frac{1}{-3.203125} \approx -0.3123 \] - \( f(0.875) \): \[ f(0.875) = \frac{1}{10 - 20 \times 0.765625} = \frac{1}{10 - 15.3125} = \frac{1}{-5.3125} \approx -0.1882 \] - \( f(0.9375) \): \[ f(0.9375) = \frac{1}{10 - 20 \times 0.87890625} = \frac{1}{10 - 17.578125} = \frac{1}{-7.578125} \approx -0.1320 \] - \( f(1.0) \): \[ f(1.0) = \frac{1}{10 - 20 \times 1} = \frac{1}{10 - 20} = \frac{1}{-10} = -0.1 \] --- ### **Step 3: Apply Simpson's 1/3 Rule formula** \[ T \approx \frac{h}{3} \left[ f(h_0) + 4 \sum_{\text{odd } i} f(h_i) + 2 \sum_{\text{even } i, i \neq 0,n} f(h_i) + f(h_{12}) \right] \] - \( h = 0.0625 \) - Sum of \( f(h_i) \) at odd indices \( i=1,3,5,7,9,11 \): \[ f(0.3125) \approx 0.124341 \] \[ f(0.4375) \approx 0.162089 \] \[ f(0.5625) \approx 0.272453 \] \[ f(0.6875) \approx 1.826 \] \[ f(0.8125) \approx -0.3123 \] \[ f(0.9375) \approx -0.1320 \] Sum: \[ \text{Sum odd} = 0.124341 + 0.162089 + 0.272453 + 1.826 - 0.3123 - 0.1320 \] Calculations: \[ 0.124341 + 0.162089 = 0.28643 \] \[ 0.28643 + 0.272453 = 0.558883 \] \[ 0.558883 + 1.826 = 2.384883 \] \[ 2.384883 - 0.3123 = 2.072583 \] \[ 2.072583 - 0.1320 = 1.940583 \] - Sum of \( f(h_i) \) at even indices \( i=2,4,6,8,10 \): \[ f(0.375) \approx 0.139229 \] \[ f(0.5) = 0.2 \] \[ f(0.625) \approx 0.457143 \] \[ f(0.75) = -0.8 \] \[ f(0.875) \approx -0.1882 \] Sum: \[ 0.139229 + 0.2 + 0.457143 - 0.8 - 0.1882 \] Calculations: \[ 0.139229 + 0.2 = 0.339229 \] \[ 0.339229 + 0.457143 = 0.796372 \] \[ 0.796372 - 0.8 = -0.003628 \] \[ -0.003628 - 0.1882 = -0.191828 \] --- ### **Step 4: Final calculation** Putting all into the Simpson's 1/3 formula: \[ T \approx \frac{0.0625}{3} \left[ f(0.25) + 4 \times (1.940583) + 2 \times (-0.191828) + f(1.0) \right] \] Values: \[ f(0.25) \approx 0.1142857 \] \[ f(1.0) = -0.1 \] Compute: \[ T \approx 0.0208333 \times \left[ 0.1142857 + 4 \times 1.940583 + 2 \times (-0.191828) - 0.1 \right] \] Calculate inside the brackets: \[ 4 \times 1.940583 = 7.762332 \] \[ 2 \times (-0.191828) = -0.383656 \] Sum: \[ 0.1142857 + 7.762332 - 0.383656 - 0.1 = \] Step-by-step: \[ 0.1142857 + 7.762332 = 7.8766177 \] \[ 7.8766177 - 0.383656 = 7.492962 \] \[ 7.492962 - 0.1 = 7.392962 \] Finally: \[ T \approx 0.0208333 \times 7.392962 \approx 0.0208333 \times 7.392962 \approx 0.154 \] --- ### **### **Final Result for Part B:** \[ \boxed{ \boxed{ T \approx \mathbf{0.154} } } \] **Note:** The approximate value of the integral using Simpson's 1/3 rule with \( h=0.0625 \) is **about 0.154**. --- If you'd like further clarification or calculations for other methods, please let me know!

Final answers for each method: **A) Trapezoidal Rule (n=3):** \(\boxed{-0.148}\) **B) Simpson's 1/3 Rule (h=0.0625, n=12):** \(\boxed{0.154}\) **C) Simpson's 3/8 Rule (n=3):** \(\boxed{-0.167}\)