Let's break down and solve the problem step-by-step: Given: \[ \int_{1}^{2} 35e^{1/x} \, dx \] You are to compute \( T_{10} \) (trapezoidal rule with \( n = 10 \)) and \( M_{10} \) (midpoint rule with \( n = 10 \)), estimate the errors, and then determine the \( n \) needed for 0.0001 accuracy. --- ## (a) **Approximations \( T_{10} \) and \( M_{10} \)** ### **Step 1: Set up the interval** - \( a = 1 \), \( b = 2 \) - \( n = 10 \) - \( h = \frac{b - a}{n} = \frac{2 - 1}{10} = 0.1 \) ### **Step 2: List the points** - For \( x_i = a + ih \), \( i = 0, 1, ..., 10 \) - For midpoints: \( x^*_{i} = a + \left(i - \frac{1}{2}\right)h \), \( i = 1, ..., 10 \) ### **Step 3: Write the formulas** For the function \( f(x) = 35e^{1/x} \): #### **Trapezoidal Rule:** \[ T_{10} = \frac{h}{2} \left[f(x_0) + 2f(x_1) + \cdots + 2f(x_9) + f(x_{10})\right] \] #### **Midpoint Rule:** \[ M_{10} = h \left[f(x^*_1) + f(x^*_2) + \cdots + f(x^*_{10})\right] \] ### **Step 4: Calculation** Let's compute the values: #### **Trapezoidal Rule Calculation** - \( x_0 = 1 \) - \( x_1 = 1.1 \) - \( x_2 = 1.2 \) - ... - \( x_{10} = 2 \) Calculate \( f(x_i) \) for \( i = 0, 1, ..., 10 \): \[ f(x) = 35 e^{1/x} \] Let's get the values: \[ \begin{align*} f(1.0) &= 35 e^{1/1} = 35 e^1 \approx 95.2292 \\ f(1.1) &= 35 e^{1/1.1} \approx 35 e^{0.9091} \approx 86.8845 \\ f(1.2) &= 35 e^{0.8333} \approx 80.4196 \\ f(1.3) &= 35 e^{0.7692} \approx 75.2238 \\ f(1.4) &= 35 e^{0.7143} \approx 70.9022 \\ f(1.5) &= 35 e^{0.6667} \approx 67.2099 \\ f(1.6) &= 35 e^{0.6250} \approx 63.9844 \\ f(1.7) &= 35 e^{0.5882} \approx 61.1123 \\ f(1.8) &= 35 e^{0.5556} \approx 58.5166 \\ f(1.9) &= 35 e^{0.5263} \approx 56.1438 \\ f(2.0) &= 35 e^{0.5} \approx 57.6409 \\ \end{align*} \] Now, sum them as per the formula: \[ T_{10} = \frac{0.1}{2} \left[95.2292 + 2(86.8845 + 80.4196 + 75.2238 + 70.9022 + 67.2099 + 63.9844 + 61.1123 + 58.5166 + 56.1438) + 57.6409\right] \] Sum the inside: - Sum of \( f(x_1) \) to \( f(x_9) \): \( 86.8845 + 80.4196 + 75.2238 + 70.9022 + 67.2099 + 63.9844 + 61.1123 + 58.5166 + 56.1438 = 620.3971 \) - \( 2 \times 620.3971 = 1240.7942 \) Now sum with the endpoints: - \( 95.2292 + 1240.7942 + 57.6409 = 1393.6643 \) Now multiply by \( \frac{0.1}{2} = 0.05 \): \[ T_{10} = 0.05 \times 1393.6643 = 69.6832 \] #### **Midpoint Rule Calculation** Midpoints: \( x^*_i = 1 + (i - 0.5) \times 0.1 \), \( i = 1, ..., 10 \): - \( x^*_1 = 1.05 \) - \( x^*_2 = 1.15 \) - \( x^*_3 = 1.25 \) - ... - \( x^*_{10} = 1.95 \) Now calculate \( f(x^*_i) \) for each: \[ \begin{align*} f(1.05) &\approx 35 e^{0.9524} \approx 35 \times 2.5913 = 90.6955 \\ f(1.15) &\approx 35 e^{0.8696} \approx 35 \times 2.3859 = 83.5056 \\ f(1.25) &\approx 35 e^{0.8} \approx 35 \times 2.2255 = 77.8925 \\ f(1.35) &\approx 35 e^{0.7407} \approx 35 \times 2.0975 = 73.4125 \\ f(1.45) &\approx 35 e^{0.6897} \approx 35 \times 1.9932 = 69.7619 \\ f(1.55) &\approx 35 e^{0.6452} \approx 35 \times 1.9065 = 66.7260 \\ f(1.65) &\approx 35 e^{0.6061} \approx 35 \times 1.8335 = 64.1725 \\ f(1.75) &\approx 35 e^{0.5714} \approx 35 \times 1.7718 = 62.0130 \\ f(1.85) &\approx 35 e^{0.5405} \approx 35 \times 1.7192 = 60.1732 \\ f(1.95) &\approx 35 e^{0.5128} \approx 35 \times 1.6735 = 58.5993 \\ \end{align*} \] Sum: \[ 90.6955 + 83.5056 + 77.8925 + 73.4125 + 69.7619 + 66.7260 + 64.1725 + 62.0130 + 60.1732 + 58.5993 = 706.9520 \] Multiply by \( h = 0.1 \): \[ M_{10} = 0.1 \times 706.9520 = 70.6952 \] --- **Final answers (rounded to 6 places):** \[ \boxed{T_{10} = 69.683215} \] \[ \boxed{M_{10} = 70.695199} \] --- ## (b) **Estimate the Errors** Error bounds for the Trapezoidal and Midpoint Rules: \[ |E_T| \leq \frac{(b-a)^3}{12 n^2} K \] \[ |E_M| \leq \frac{(b-a)^3}{24 n^2} K \] where \( K \) is the maximum of \( |f''(x)| \) on \([1,2]\). ### **Step 1: Compute \( f''(x) \) and find \( K \)** First, \[ f(x) = 35 e^{1/x} \] \[ f'(x) = 35 e^{1/x} \cdot \left(-\frac{1}{x^2}\right) = -\frac{35}{x^2} e^{1/x} \] \[ f''(x) = \frac{d}{dx} \left( -\frac{35}{x^2} e^{1/x} \right ) \] Use product rule: \[ f''(x) = -35 \frac{d}{dx}(x^{-2} e^{1/x}) = -35 \left( \frac{d}{dx}(x^{-2}) e^{1/x} + x^{-2} \frac{d}{dx}(e^{1/x}) \right ) \] \[ \frac{d}{dx}(x^{-2}) = -2 x^{-3} \] \[ \frac{d}{dx}(e^{1/x}) = e^{1/x} \cdot \left(-\frac{1}{x^2}\right) \] So: \[ f''(x) = -35 \left[ -2x^{-3} e^{1/x} + x^{-2} \cdot e^{1/x} \cdot \left(-\frac{1}{x^2}\right) \right ] \] \[ = -35 \left[ -2x^{-3} e^{1/x} - x^{-4} e^{1/x} \right ] \] \[ = 35 [ 2x^{-3} e^{1/x} + x^{-4} e^{1/x} ] \] \[ = 35 e^{1/x} \left( 2x^{-3} + x^{-4} \right ) \] Now, the maximum will occur at the left endpoint (since \( e^{1/x} \), \( x^{-3} \), and \( x^{-4} \) are decreasing in \( x \) for \( x > 0 \)), so at \( x = 1 \): \[ K = 35 e^{1/1} (2 \cdot 1^{-3} + 1^{-4}) = 35 e^1 (2 + 1) = 35 e^1 \cdot 3 \approx 35 \times 2.71828 \times 3 \approx 285.687 \] Use this value for \( K \). ### **Step 2: Plug into the error bounds** - \( (b-a) = 1 \) - \( n = 10 \) - \( (b-a)^3 = 1^3 = 1 \) - \( n^2 = 100 \) So, \[ |E_T| \leq \frac{1}{12 \times 100} K = \frac{K}{1200} \] \[ |E_M| \leq \frac{K}{2400} \] Plug \( K \approx 285.687 \): \[ |E_T| \leq \frac{285.687}{1200} = 0.238073 \] \[ |E_M| \leq \frac{285.687}{2400} = 0.119037 \] --- **Final answers (rounded to 6 decimal places):** \[ |E_T| \leq \boxed{0.238073} \] \[ |E_M| \leq \boxed{0.119037} \] --- ## (c) **Find \( n \) for error \( < 0.0001 \)** Set each error bound less than 0.0001 and solve for \( n \): ### **Trapezoidal rule:** \[ \frac{K}{12 n^2} < 0.0001 \implies n^2 > \frac{K}{12 \times 0.0001} \] \[ n > \sqrt{ \frac{K}{0.0012} } \] \[ n > \sqrt{ \frac{285.687}{0.0012} } \approx \sqrt{238072.5} \approx 487.91 \] So, \( n = 488 \) (rounded up) ### **Midpoint rule:** \[ \frac{K}{24 n^2} < 0.0001 \implies n^2 > \frac{K}{24 \times 0.0001} \] \[ n > \sqrt{ \frac{K}{0.0024} } \] \[ n > \sqrt{ \frac{285.687}{0.0024} } \approx \sqrt{119036.25} \approx 344.97 \] So, \( n = 345 \) (rounded up) --- **Final answers:** \[ \text{For } T_{n'}, \; n = \boxed{488} \] \[ \text{For } M_{n'}, \; n = \boxed{345} \] --- ### **Summary Table** | Part

**(a) Approximations:** \[ T_{10} \approx \boxed{69.683215} \] \[ M_{10} \approx \boxed{70.695199} \] **(b) Error bounds:** \[ |E_T| \leq \boxed{0.238073} \] \[ |E_M| \leq \boxed{0.119037} \] **(c) Number of subintervals for desired accuracy (0.0001):** \[ \text{Trapezoidal rule } n \approx \boxed{488} \] \[ \text{Midpoint rule } n \approx \boxed{345} \]