Let's solve the problem step by step. ## Given: - Integral: \(\int_{-5}^{5} (5x^2 + 8) \, dx\) - Desired error: \(< 2 \times 10^{-4}\) - Find the minimum \(n\) (number of subintervals) for: - (a) **Trapezoidal Rule error estimate** - (b) **Simpson's Rule error estimate** --- ## **Step 1: Error Formulas** ### **a. Trapezoidal Rule Error Estimate** \[ |E_T| \leq \frac{(b-a)^3}{12 n^2} \max_{a \leq x \leq b} |f''(x)| \] --- ### **b. Simpson's Rule Error Estimate** \[ |E_S| \leq \frac{(b-a)^5}{180 n^4} \max_{a \leq x \leq b} |f^{(4)}(x)| \] --- ## **Step 2: Find Derivatives** Given \(f(x) = 5x^2 + 8\): - \(f'(x) = 10x\) - \(f''(x) = 10\) - \(f'''(x) = 0\) - \(f^{(4)}(x) = 0\) --- ## **Step 3: Plug into Error Formulas** ### **Interval**: \(a = -5\), \(b = 5\), so \(b-a = 10\) --- ### **a. Trapezoidal Rule** \[ |E_T| \leq \frac{(10)^3}{12 n^2} \cdot 10 = \frac{1000}{12 n^2} \cdot 10 = \frac{10000}{12 n^2} = \frac{2500}{3 n^2} \] Set error less than \(2 \times 10^{-4}\): \[ \frac{2500}{3 n^2} < 2 \times 10^{-4} \] \[ \frac{2500}{3 \times 2 \times 10^{-4}} < n^2 \] \[ \frac{2500}{6 \times 10^{-4}} < n^2 \] \[ \frac{2500}{0.0006} < n^2 \] \[ \frac{2500}{0.0006} = 4,166,666.67 \] \[ n^2 > 4,166,666.67 \] \[ n > \sqrt{4,166,666.67} \approx 2041.24 \] **Minimum integer \(n\):** \[ \boxed{2042} \] --- ### **b. Simpson's Rule** \[ |E_S| \leq \frac{(10)^5}{180 n^4} \cdot 0 = 0 \] Because the fourth derivative \(f^{(4)}(x) = 0\), the error is **zero**. This is because Simpson's Rule is **exact for polynomials of degree 3 or less** (here, degree 2). **So, any \(n\) (as long as it is even) will achieve zero error.** \[ \boxed{n = 2 \text{ (or any even number)}} \] --- ## **Summary Table** | Rule | Minimum \(n\) | |-------------------|--------------| | Trapezoidal Rule | **2042** | | Simpson's Rule | **2** | --- ### **Final Answers:** **a. Trapezoidal Rule:** \(\boxed{2042}\) **b. Simpson's Rule:** \(\boxed{2}\) (or any even \(n\))