Let's solve each problem step by step. --- ## **1. \( \int_0^7 \sqrt{7x+3}\,dx \) with error less than \( 3 \times 10^{-4} \):** ### **a. Trapezoidal Rule Error Estimate** The error formula for the Trapezoidal Rule is: \[ |E_T| \leq \frac{(b-a)^3}{12n^2} \max_{a \leq x \leq b} |f''(x)| \] #### **Step 1: Find \( f(x) \) and its second derivative** \( f(x) = \sqrt{7x+3} \) First derivative: \[ f'(x) = \frac{1}{2}(7x+3)^{-1/2} \cdot 7 = \frac{7}{2\sqrt{7x+3}} \] Second derivative: \[ f''(x) = \frac{d}{dx} \left(\frac{7}{2\sqrt{7x+3}}\right) = -\frac{7}{4}(7x+3)^{-3/2} \cdot 7 = -\frac{49}{4}(7x+3)^{-3/2} \] #### **Step 2: Find the maximum of \( |f''(x)| \) on \( [0, 7] \)** \[ |f''(x)| = \frac{49}{4}(7x+3)^{-3/2} \] This is largest at \( x = 0 \): \[ |f''(0)| = \frac{49}{4} \cdot 3^{-3/2} = \frac{49}{4} \cdot \frac{1}{3\sqrt{3}} = \frac{49}{12\sqrt{3}} \] #### **Step 3: Plug into error formula and solve for \( n \):** \[ |E_T| \leq \frac{(7-0)^3}{12n^2} \cdot \frac{49}{12\sqrt{3}} < 3 \times 10^{-4} \] \[ \frac{343}{12n^2} \cdot \frac{49}{12\sqrt{3}} < 3 \times 10^{-4} \] \[ \frac{343 \cdot 49}{12 \cdot 12\sqrt{3}} \cdot \frac{1}{n^2} < 3 \times 10^{-4} \] \[ \frac{16807}{144\sqrt{3}} \cdot \frac{1}{n^2} < 3 \times 10^{-4} \] \[ \frac{16807}{144\sqrt{3} \cdot 3 \times 10^{-4}} < n^2 \] \[ n^2 > \frac{16807}{144\sqrt{3} \cdot 3 \times 10^{-4}} \] Calculate denominator: \[ 144 \cdot 3 = 432, \quad 432 \cdot 10^{-4} = 0.0432 \] \[ 0.0432 \cdot \sqrt{3} \approx 0.0432 \cdot 1.732 \approx 0.07488 \] So, \[ n^2 > \frac{16807}{0.07488} \approx 224419.9 \] \[ n > \sqrt{224419.9} \approx 474 \] So, **minimum \( n = 475 \) subintervals.** --- ### **b. Simpson's Rule Error Estimate** Error formula: \[ |E_S| \leq \frac{(b-a)^5}{180 n^4} \max_{a \leq x \leq b} |f^{(4)}(x)| \] #### **Step 1: Find fourth derivative \( f^{(4)}(x) \)** We already have: \[ f''(x) = -\frac{49}{4}(7x+3)^{-3/2} \] Third derivative: \[ f'''(x) = -\frac{49}{4} \cdot -\frac{3}{2}(7x+3)^{-5/2} \cdot 7 = \frac{49 \cdot 21}{8}(7x+3)^{-5/2} \] Fourth derivative: \[ f^{(4)}(x) = \frac{49 \cdot 21}{8} \cdot -\frac{5}{2}(7x+3)^{-7/2} \cdot 7 = -\frac{49 \cdot 21 \cdot 35}{16}(7x+3)^{-7/2} \] \[ 49 \cdot 21 = 1029,\ 1029 \cdot 35 = 36015 \] So: \[ f^{(4)}(x) = -\frac{36015}{16}(7x+3)^{-7/2} \] Maximum at \( x = 0 \): \[ |f^{(4)}(0)| = \frac{36015}{16} \cdot 3^{-7/2} = \frac{36015}{16} \cdot \frac{1}{3^3 \sqrt{3}} = \frac{36015}{16 \cdot 27 \sqrt{3}} = \frac{36015}{432\sqrt{3}} \] #### **Step 2: Plug into error formula and solve for \( n \):** \[ |E_S| \leq \frac{(7)^5}{180 n^4} \cdot \frac{36015}{432\sqrt{3}} < 3 \times 10^{-4} \] \[ 7^5 = 16807 \] \[ \frac{16807}{180 n^4} \cdot \frac{36015}{432\sqrt{3}} < 3 \times 10^{-4} \] \[ \frac{16807 \cdot 36015}{180 \cdot 432 \sqrt{3}} \cdot \frac{1}{n^4} < 3 \times 10^{-4} \] \[ 180 \cdot 432 = 77,760 \] \[ 16807 \cdot 36015 = 605,643,105 \] \[ \frac{605,643,105}{77,760\sqrt{3}} \cdot \frac{1}{n^4} < 3 \times 10^{-4} \] \[ 77,760 \cdot 1.732 = 134,653.92 \] \[ \frac{605,643,105}{134,653.92} = 4,497.5 \] \[ 4,497.5 \cdot \frac{1}{n^4} < 3 \times 10^{-4} \] \[ \frac{4,497.5}{3 \times 10^{-4}} < n^4 \] \[ \frac{4,497.5}{0.0003} = 14,991,667 \] \[ n^4 > 14,991,667 \implies n > (14,991,667)^{1/4} \approx 62.6 \] So, **minimum \( n = 64 \) (must be even for Simpson's Rule).** --- ## **2. \( \int_{-6}^{9} 8 \sin(x+9)\,dx \) with error less than \( 2 \times 10^{-4} \):** ### **a. Trapezoidal Rule Error Estimate** \[ |E_T| \leq \frac{(b-a)^3}{12n^2} \max_{a \leq x \leq b} |f''(x)| \] \( f(x) = 8\sin(x+9) \) \[ f''(x) = -8\sin(x+9) \] Maximum of \( |f''(x)| = 8 \) (since \( |\sin| \leq 1 \)). \( a = -6, b = 9 \implies b - a = 15 \) \[ |E_T| \leq \frac{15^3}{12n^2} \cdot 8 < 2 \times 10^{-4} \] \[ \frac{3375}{12n^2} \cdot 8 < 2 \times 10^{-4} \] \[ \frac{3375 \cdot 8}{12} \cdot \frac{1}{n^2} < 2 \times 10^{-4} \] \[ \frac{27,000}{12} = 2,250 \] \[ \frac{2,250}{n^2} < 2 \times 10^{-4} \] \[ \frac{2,250}{2 \times 10^{-4}} < n^2 \] \[ \frac{2,250}{0.0002} = 11,250,000 \] \[ n > \sqrt{11,250,000} \approx 3355 \] So, **minimum \( n = 3356 \) subintervals.** --- ### **b. Simpson's Rule Error Estimate** \[ |E_S| \leq \frac{(b-a)^5}{180 n^4} \max_{a \leq x \leq b} |f^{(4)}(x)| \] \( f^{(4)}(x) = 8\sin(x+9) \) Maximum of \( |f^{(4)}(x)| = 8 \). \[ |E_S| \leq \frac{15^5}{180 n^4} \cdot 8 < 2 \times 10^{-4} \] \( 15^5 = 759375 \) \[ \frac{759375}{180 n^4} \cdot 8 < 2 \times 10^{-4} \] \[ \frac{759375 \cdot 8}{180} = \frac{6,075,000}{180} = 33,750 \] \[ \frac{33,750}{n^4} < 2 \times 10^{-4} \] \[ \frac{33,750}{2 \times 10^{-4}} < n^4 \] \[ \frac{33,750}{0.0002} = 168,750,000 \] \[ n > (168,750,000)^{1/4} \approx 114.6 \] So, **minimum \( n = 116 \) (must be even for Simpson's Rule).** --- ## **Final Answers** ### **1. \( \int_0^7 \sqrt{7x+3}\,dx \), error \( < 3 \times 10^{-4} \):** - a. Trapezoidal Rule: **Minimum \( n = 475 \)** - b. Simpson's Rule: **Minimum \( n = 64 \)** ### **2. \( \int_{-6}^{9} 8\sin(x+9)\,dx \), error \( < 2 \times 10^{-4} \):** - a. Trapezoidal Rule: **Minimum \( n = 3356 \)** - b. Simpson's Rule: **Minimum \( n = 116 \)** Let me know if you want further simplification or any step explained in more detail!