Let's solve the problem step by step. --- ### Given: \[ \int_{0}^{3} (4t^3 + 6t) \, dt \] We want the error in the approximation to be less than \( 10^{-4} \). --- ## **a. Trapezoidal Rule Error Estimate** The error formula for the Trapezoidal Rule is: \[ |E_T| \leq \frac{(b-a)^3}{12n^2} \max_{a \leq x \leq b} |f''(x)| \] Given: \( f(t) = 4t^3 + 6t \), \( a = 0 \), \( b = 3 \) ### **Step 1: Compute the second derivative** \[ f'(t) = 12t^2 + 6 \] \[ f''(t) = 24t \] ### **Step 2: Find the maximum value of \( |f''(t)| \) on \([0, 3]\)** Since \( f''(t) = 24t \) is increasing on \([0, 3]\), the maximum occurs at \( t = 3 \): \[ |f''(t)|_{\text{max}} = 24 \times 3 = 72 \] ### **Step 3: Plug into the error formula** \[ |E_T| \leq \frac{(3-0)^3}{12n^2} \cdot 72 = \frac{27}{12n^2} \cdot 72 = \frac{27 \cdot 72}{12n^2} \] \[ \frac{27 \cdot 72}{12n^2} = \frac{1944}{12n^2} = \frac{162}{n^2} \] Set \( |E_T| < 10^{-4} \): \[ \frac{162}{n^2} < 10^{-4} \] \[ 162 \times 10^4 < n^2 \] \[ 1,620,000 < n^2 \] \[ n > \sqrt{1,620,000} \approx 1274.75 \] ### **Final Answer for (a):** \[ \boxed{n \geq 1275} \] --- ## **b. Simpson's Rule Error Estimate** The error formula for Simpson's Rule is: \[ |E_S| \leq \frac{(b-a)^5}{180n^4} \max_{a \leq x \leq b} |f^{(4)}(x)| \] ### **Step 1: Compute the fourth derivative** \[ f''(t) = 24t \implies f'''(t) = 24 \implies f^{(4)}(t) = 0 \] ### **Step 2: Simpson's Rule Error** Since the fourth derivative is zero, the error for Simpson's Rule is: \[ |E_S| \leq 0 \] So **any even \( n \)** will give exact result for this polynomial (degree ≤ 3). The minimum would be \( n = 2 \). ### **Final Answer for (b):** \[ \boxed{n = 2} \] --- ### **Summary Table** | Method | Minimum \( n \) | |------------------|:--------------:| | Trapezoidal Rule | \( 1275 \) | | Simpson's Rule | \( 2 \) | Let me know if you need further clarifications!