Let's break down the solution step by step for **each function**. We are asked to compute the **length of the curve** for each function over \( 0 \leq x \leq 1 \) using **Trapezoidal Rule** and **Simpson's Rule** for different \( n \) values. ### **General Formula** The length of the curve \( y = f(x) \) from \( x = a \) to \( x = b \) is: \[ I(f) = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} \, dx \] Let’s solve **(a)** in detail and outline the steps for (b) and (c). **You can apply the same process for (b) and (c)!** --- ## **(a) \( f(x) = \sin(\pi x) \)** ### 1. **Find \( f'(x) \)** \[ f'(x) = \pi \cos(\pi x) \] ### 2. **Plug into the formula** \[ I(f) = \int_{0}^{1} \sqrt{1 + [\pi \cos(\pi x)]^2} \, dx \] --- ### 3. **Numerical Integration** #### **Trapezoidal Rule:** For \( n \) subintervals, \( h = \frac{b-a}{n} \), the rule is: \[ I \approx \frac{h}{2} [f(x_0) + 2f(x_1) + 2f(x_2) + \ldots + 2f(x_{n-1}) + f(x_n)] \] where \( x_i = a + i h \). #### **Simpson's Rule** (\( n \) even): \[ I \approx \frac{h}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \ldots + 4f(x_{n-1}) + f(x_n) \right] \] --- ### 4. **Set up the function for integration:** Let \[ g(x) = \sqrt{1 + [\pi \cos(\pi x)]^2} \] --- ### 5. **Python (or Calculator) Implementation** Let's show how you would set this up in Python with `numpy` for \( n = 4 \): ```python import numpy as np def g(x): return np.sqrt(1 + (np.pi * np.cos(np.pi * x))**2) a, b = 0, 1 def trapezoidal_rule(f, a, b, n): h = (b - a) / n result = 0.5 * (f(a) + f(b)) for i in range(1, n): result += f(a + i * h) return result * h def simpsons_rule(f, a, b, n): if n % 2: raise ValueError("n must be even for Simpson's rule") h = (b - a) / n result = f(a) + f(b) for i in range(1, n, 2): result += 4 * f(a + i * h) for i in range(2, n-1, 2): result += 2 * f(a + i * h) return result * h / 3 n_values = [4, 8, 16, 512] for n in n_values: trap = trapezoidal_rule(g, a, b, n) simp = simpsons_rule(g, a, b, n) print(f"n={n}: Trapezoidal={trap:.6f}, Simpson={simp:.6f}") ``` --- ## **(b) \( f(x) = e^x \)** - \( f'(x) = e^x \) - Integrand: \( g(x) = \sqrt{1 + (e^x)^2} = \sqrt{1 + e^{2x}} \) - Use the same Trapezoidal and Simpson's Rule as above. ## **(c) \( f(x) = e^{x^2} \)** - \( f'(x) = 2x e^{x^2} \) - Integrand: \( g(x) = \sqrt{1 + [2x e^{x^2}]^2} = \sqrt{1 + 4x^2 e^{2x^2}} \) - Use the same Trapezoidal and Simpson's Rule as above. --- ### **Final Step: Compute for \( n = 4, 8, \ldots, 512 \)** - Plug each \( g(x) \) into the rules and compute as above. --- ## **Summary Table Structure (for your report):** | Function | n | Trapezoidal | Simpson's | |------------------|-----|-------------|-----------| | \( \sin(\pi x) \)| 4 | | | | | 8 | | | | ... | ... | | | | \( e^x \) | 4 | | | | ... | ... | | | | \( e^{x^2} \) | ... | | | --- ### **Remarks** - The larger the \( n \), the more accurate the result. - Simpson's rule usually converges faster. - For exact values, use `scipy.integrate.quad` to compare. --- If you need the actual computed values for each function and \( n \), let me know, and I can provide Python code output or walk you through the computation for each!

Certainly! Let's proceed step-by-step with detailed calculations for each function, focusing on the application of the trapezoidal rule and Simpson's rule for various \(n\) values. We will do this explicitly for the case \(f(x) = \sin(\pi x)\). Similar steps apply to the other functions, but due to complexity, here we'll focus on the first function in detail. --- ## Part (a): \( f(x) = \sin(\pi x) \) ### Step 1: Find the derivative \(f'(x)\) \[ f'(x) = \pi \cos(\pi x) \] ### Step 2: Express the integrand \[ g(x) = \sqrt{1 + [f'(x)]^2} = \sqrt{1 + (\pi \cos(\pi x))^2} = \sqrt{1 + \pi^2 \cos^2(\pi x)} \] Our goal: \[ I = \int_0^1 g(x) \, dx = \int_0^1 \sqrt{1 + \pi^2 \cos^2(\pi x)} \, dx \] --- ### Step 3: Choose \(n\) and compute \(h = \frac{b - a}{n}\) Let's pick specific \(n\) values: 4, 8, 16, 512. For each \(n\): \[ h = \frac{1 - 0}{n} = \frac{1}{n} \] --- ### Step 4: Compute the nodes \(x_i\) \[ x_i = a + i h = 0 + i \times h, \quad i = 0, 1, 2, ..., n \] --- ### Step 5: Evaluate \(g(x_i)\) For each node \(x_i\), compute: \[ g(x_i) = \sqrt{1 + \pi^2 \cos^2(\pi x_i)} \] --- ### Step 6: Apply the **Trapezoidal Rule** \[ I \approx T_n = \frac{h}{2} \left[ g(x_0) + 2 \sum_{i=1}^{n-1} g(x_i) + g(x_n) \right] \] --- ### Step 7: Apply **Simpson's Rule** Since Simpson's rule requires \(n\) even: \[ I \approx S_n = \frac{h}{3} \left[ g(x_0) + 4 \sum_{i=1,\, i \text{ odd}}^{n-1} g(x_i) + 2 \sum_{i=2,\, i \text{ even}}^{n-2} g(x_i) + g(x_n) \right] \] --- ## Numerical calculations for specific \(n\) ### For \(n=4\): - \(h = 1/4 = 0.25\) Nodes: \[ x_0 = 0, \quad x_1 = 0.25, \quad x_2= 0.5, \quad x_3=0.75, \quad x_4=1 \] Calculate \(g(x_i)\): 1. \(x=0\): \[ g(0) = \sqrt{1 + \pi^2 \cos^2(0)} = \sqrt{1 + \pi^2 \times 1} = \sqrt{1 + \pi^2} \approx \sqrt{1 + 9.8696} \approx \sqrt{10.8696} \approx 3.295 \] 2. \(x=0.25\): \[ g(0.25) = \sqrt{1 + \pi^2 \cos^2(\pi \times 0.25)} = \sqrt{1 + \pi^2 \cos^2(\pi/4)} \] \(\cos(\pi/4) = \frac{\sqrt{2}}{2} \approx 0.7071\) \[ g(0.25) = \sqrt{1 + \pi^2 \times 0.5} = \sqrt{1 + 9.8696 \times 0.5} = \sqrt{1 + 4.9348} = \sqrt{5.9348} \approx 2.436 \] 3. \(x=0.5\): \[ g(0.5) = \sqrt{1 + \pi^2 \cos^2(\pi \times 0.5)} = \sqrt{1 + \pi^2 \times 0} = \sqrt{1 + 0} = 1 \] 4. \(x=0.75\): \[ g(0.75) = \sqrt{1 + \pi^2 \cos^2(3\pi/4)} \] \(\cos(3\pi/4) = -\frac{\sqrt{2}}{2} \approx -0.7071\) \[ g(0.75) = \sqrt{1 + \pi^2 \times 0.5} \approx 2.436 \quad (\same as at 0.25) \] 5. \(x=1\): \[ g(1) = \sqrt{1 + \pi^2 \cos^2(\pi)} = \sqrt{1 + \pi^2 \times 1} \approx 3.295 \] --- ### Applying the trapezoidal rule: \[ T_4 = \frac{h}{2} [g(0) + 2(g(0.25)+ g(0.5)+ g(0.75)) + g(1)] \] \[ T_4 = \frac{0.25}{2} [3.295 + 2 (2.436 + 1 + 2.436) + 3.295] \] Calculate inside brackets: \[ = 0.125 [3.295 + 2 (2.436 + 1 + 2.436) + 3.295] \] Sum inside the parentheses: \[ 2 (2.436 + 1 + 2.436) = 2 (5.872) = 11.744 \] Now: \[ = 0.125 [3.295 + 11.744 + 3.295] = 0.125 [18.334] \] \[ T_4 \approx 0.125 \times 18.334 = 2.29175 \] --- ### Applying Simpson’s rule: \[ S_4 = \frac{h}{3} [g(0) + 4g(0.25) + 2g(0.5) + 4g(0.75) + g(1)] \] \[ S_4 = \frac{0.25}{3} [3.295 + 4 \times 2.436 + 2 \times 1 + 4 \times 2.436 + 3.295] \] Compute: \[ 4 \times 2.436 = 9.744 \] \[ 2 \times 1 = 2 \] \[ 4 \times 2.436 = 9.744 \] Sum: \[ 3.295 + 9.744 + 2 + 9.744 + 3.295 = (3.295 + 3.295) + (9.744 + 9.744) + 2 = 6.59 + 19.488 + 2 = 27.078 \] Now: \[ S_4 = \frac{0.25}{3} \times 27.078 \approx 0.0833 \times 27.078 \approx 2.256 \] --- ## **Summary for \(n=4\):** | Method | Approximate Length | |--------------------|--------------------| | Trapezoidal Rule | **~2.292** | | Simpson's Rule | **~2.256** | --- ## Extending to larger \(n\): - As \(n\) increases, the approximations become more accurate. - For \(n=8, 16, 512\), repeat the process: compute \(g(x_i)\) at each node and apply the formulas. --- ## **Note:** - Similar calculations can be performed for the other functions \(f(x) = e^x\) and \(f(x) = e^{x^2}\). - For these, you'd evaluate \(g(x) = \sqrt{1 + [f'(x)]^2}\) at each node and sum accordingly. --- ## **Final Remarks:** - The key steps involve computing the derivative, substituting into the integrand, evaluating at nodes, and applying the numerical rule. - Larger \(n\) increases accuracy. - The process outlined provides the detailed calculations needed for each approximation. --- Let me know if you'd like me to perform similar detailed calculations for the other functions!