| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the arc length of the function \( y = x^{-1} \) over the interval \([1, 2]\) using Simpson's Rule \( S_8 \), we follow these steps: 1. **Determine the arc length formula**: The arc length \( L \) of a function \( y = f(x) \) over an interval \([a, b]\) is given by: \[ L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} \, dx \] 2. **Find the derivative of the function**: For \( y = x^{-1} \), the derivative \( f'(x) \) is: \[ f'(x) = -x^{-2} \] 3. **Square the derivative and add 1**: \[ 1 + [f'(x)]^2 = 1 + (-x^{-2})^2 = 1 + x^{-4} \] 4. **Formulate the integral for arc length**: \[ L = \int_{1}^{2} \sqrt{1 + x^{-4}} \, dx \] 5. **Apply Simpson's Rule \( S_8 \)**: - Divide the interval \([1, 2]\) into 8 equal subintervals. The width \( h \) of each subinterval is: \[ h = \frac{2 - 1}{8} = \frac{1}{8} \] - The nodes at which we evaluate the function are: \[ x_0 = 1, \, x_1 = 1.125, \, x_2 = 1.25, \, x_3 = 1.375, \, x_4 = 1.5, \, x_5 = 1.625, \, x_6 = 1.75, \, x_7 = 1.875, \, x_8 = 2 \] 6. **Evaluate the function at these nodes**: \[ f(x_i) = \sqrt{1 + x_i^{-4}} \] - Calculate \( f(x_i) \) for each \( x_i \): \[ f(1) = \sqrt{2}, \, f(1.125) \approx 1.041, \, f(1.25) \approx 1.075, \, f(1.375) \approx 1.103, \, f(1.5) \approx 1.125, \, f(1.625) \approx 1.143, \, f(1.75) \approx 1.158, \, f(1.875) \approx 1.171, \, f(2) \approx 1.031 \] 7. **Apply Simpson's Rule formula**: \[ S_8 = \frac{h}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + 4f(x_5) + 2f(x_6) + 4f(x_7) + f(x_8) \right] \] Substituting the values: \[ S_8 = \frac{1/8}{3} \left[ \sqrt{2} + 4(1.041) + 2(1.075) + 4(1.103) + 2(1.125) + 4(1.143) + 2(1.158) + 4(1.171) + 1.031 \right] \] \[ S_8 = \frac{1}{24} \left[ 1.414 + 4.164 + 2.15 + 4.412 + 2.25 + 4.572 + 2.316 + 4.684 + 1.031 \right] \] \[ S_8 = \frac{1}{24} \left[ 26.183 \right] \] \[ S_8 \approx 1.091 \] Thus, the approximate arc length of \( y = x^{-1} \) over the interval \([1, 2]\) using Simpson's Rule \( S_8 \) is approximately \( 1.091 \). |
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| deepseekr1 | To find the arc length of \( y = x^{-1} \) over the interval \([1, 2]\) using Simpson's Rule \( S_8 \): 1. **Arc Length Formula**: The arc length \( L \) of \( y = f(x) \) from \( x = a \) to \( x = b \) is given by: \[ L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx \] For \( f(x) = x^{-1} \), the derivative is \( f'(x) = -x^{-2} \). Thus, the integrand becomes: \[ \sqrt{1 + \left( \frac{1}{x^2} \right)^2} = \sqrt{1 + \frac{1}{x^4}} \] 2. **Simpson's Rule Setup**: Using \( n = 8 \) subintervals, the interval \([1, 2]\) is divided into 8 equal parts with \( \Delta x = \frac{2 - 1}{8} = 0.125 \). The points \( x_i \) are \( 1, 1.125, 1.25, 1.375, 1.5, 1.625, 1.75, 1.875, 2 \). 3. **Evaluate the Integrand**: Compute \( \sqrt{1 + \frac{1}{x_i^4}} \) for each \( x_i \): \[ \begin{aligned} f(1.0) & \approx 1.41421356 \\ f(1.125) & \approx 1.27467390 \\ f(1.25) & \approx 1.18743421 \\ f(1.375) & \approx 1.13124967 \\ f(1.5) & \approx 1.09434556 \\ f(1.625) & \approx 1.06931494 \\ f(1.75) & \approx 1.05196018 \\ f(1.875) & \approx 1.03966390 \\ f(2.0) & \approx 1.03077641 \\ \end{aligned} \] 4. **Apply Simpson's Rule**: The formula for Simpson's Rule \( S_8 \) is: \[ S_8 = \frac{\Delta x}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + 4f(x_5) + 2f(x_6) + 4f(x_7) + f(x_8) \right] \] Substituting the values: \[ S_8 = \frac{0.125}{3} \left[ 1.41421356 + 4(1.27467390) + 2(1.18743421) + 4(1.13124967) + 2(1.09434556) + 4(1.06931494) + 2(1.05196018) + 4(1.03966390) + 1.03077641 \right] \] Calculating the sum inside the brackets: \[ 1.41421356 + 5.09869562 + 2.37486842 + 4.52499867 + 2.18869112 + 4.27725975 + 2.10392035 + 4.15865558 + 1.03077641 \approx 27.17207948 \] Multiplying by \( \frac{0.125}{3} \): \[ S_8 \approx \frac{27.17207948}{24} \approx 1.1321699785 \] 5. **Final Answer**: The arc length is approximately \(\boxed{1.1322}\) using Simpson's Rule \( S_8 \). |
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| gpto1 | To find the arc length of the function \( y = x^{-1} \) over the interval \([1, 2]\) using Simpson's Rule with \( n = 8 \) subdivisions, we proceed as follows: **Step 1: Set Up the Arc Length Integral** The formula for the arc length \( L \) of a function \( y = f(x) \) from \( x = a \) to \( x = b \) is: \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] For \( y = x^{-1} = \frac{1}{x} \), the derivative is: \[ \frac{dy}{dx} = -x^{-2} \] So, the integrand becomes: \[ \sqrt{1 + \left( -x^{-2} \right)^2} = \sqrt{1 + x^{-4}} \] Therefore, the arc length is: \[ L = \int_{1}^{2} \sqrt{1 + x^{-4}} \, dx \] **Step 2: Divide the Interval into Subintervals** With \( n = 8 \) subdivisions: \[ \Delta x = \frac{b - a}{n} = \frac{2 - 1}{8} = 0.125 \] The \( x_i \) values are: \[ x_i = a + i\Delta x, \quad i = 0, 1, 2, \dots, 8 \] So: \[ x_0 = 1,\ x_1 = 1.125,\ x_2 = 1.25,\ x_3 = 1.375,\ x_4 = 1.5,\ x_5 = 1.625,\ x_6 = 1.75,\ x_7 = 1.875,\ x_8 = 2 \] **Step 3: Compute the Function Values at Each \( x_i \)** We compute \( f(x_i) = \sqrt{1 + x_i^{-4}} \): \[ \begin{align*} f(x_0) &= \sqrt{1 + (1)^{-4}} = \sqrt{1 + 1} = \sqrt{2} \approx 1.4142 \\ f(x_1) &= \sqrt{1 + (1.125)^{-4}} = \sqrt{1 + 0.6243} = \sqrt{1.6243} \approx 1.2748 \\ f(x_2) &= \sqrt{1 + (1.25)^{-4}} = \sqrt{1 + 0.4096} = \sqrt{1.4096} \approx 1.1874 \\ f(x_3) &= \sqrt{1 + (1.375)^{-4}} = \sqrt{1 + 0.2797} = \sqrt{1.2797} \approx 1.1310 \\ f(x_4) &= \sqrt{1 + (1.5)^{-4}} = \sqrt{1 + 0.1975} = \sqrt{1.1975} \approx 1.0943 \\ f(x_5) &= \sqrt{1 + (1.625)^{-4}} = \sqrt{1 + 0.1438} = \sqrt{1.1438} \approx 1.0693 \\ f(x_6) &= \sqrt{1 + (1.75)^{-4}} = \sqrt{1 + 0.1068} = \sqrt{1.1068} \approx 1.0528 \\ f(x_7) &= \sqrt{1 + (1.875)^{-4}} = \sqrt{1 + 0.0810} = \sqrt{1.0810} \approx 1.0396 \\ f(x_8) &= \sqrt{1 + (2)^{-4}} = \sqrt{1 + 0.0625} = \sqrt{1.0625} \approx 1.0308 \\ \end{align*} \] **Step 4: Apply Simpson's Rule** Simpson's Rule for \( n \) subdivisions is: \[ S_n = \frac{\Delta x}{3} \left[ f(x_0) + 4\sum_{\text{odd } i} f(x_i) + 2\sum_{\text{even } i} f(x_i) + f(x_n) \right] \] Compute the weighted sum: \[ \begin{align*} S &= f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + 4f(x_5) + 2f(x_6) + 4f(x_7) + f(x_8) \\ &\approx 1.4142 + 4(1.2748) + 2(1.1874) + 4(1.1310) + 2(1.0943) + 4(1.0693) + 2(1.0528) + 4(1.0396) + 1.0308 \\ &\approx 1.4142 + 5.0990 + 2.3748 + 4.5240 + 2.1886 + 4.2772 + 2.1056 + 4.1584 + 1.0308 \\ &\approx 27.1730 \end{align*} \] **Step 5: Multiply by \(\frac{\Delta x}{3}\) to Get the Approximate Integral** \[ L \approx \frac{\Delta x}{3} \times S = \frac{0.125}{3} \times 27.1730 \approx 0.0416667 \times 27.1730 \approx 1.1322 \] **Answer:** Approximately 1.1322 units—the arc length is ≈ 1.1322 using Simpson’s Rule with 8 intervals. |
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| GPT-4o | To find the arc length of the function \( y = x^{-1} \) over the interval \([1,2]\) using Simpson's Rule \( S_8 \), we can follow these steps: ### 1. Arc Length Formula The arc length \(L\) of a function \( y = f(x) \) from \( x = a \) to \( x = b \) is given by: \[ L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx. \] For \( y = x^{-1} \), we first need to find \(\frac{dy}{dx}\). ### 2. Derivative of \( y = x^{-1} \) \[ y = x^{-1} = \frac{1}{x}, \] \[ \frac{dy}{dx} = -\frac{1}{x^2}. \] ### 3. Expression Under the Square Root \[ \left( \frac{dy}{dx} \right)^2 = \left( -\frac{1}{x^2} \right)^2 = \frac{1}{x^4}. \] Thus, the arc length integral becomes: \[ L = \int_1^2 \sqrt{1 + \frac{1}{x^4}} \, dx. \] ### 4. Simpson's Rule \( S_8 \) Simpson's Rule approximates the integral by dividing the interval into an even number \( n \) of subintervals (here \(n = 8\)). The general form of Simpson's Rule is: \[ S_n = \frac{\Delta x}{3} \left[ f(x_0) + 4 \sum_{i=1,3,5,\ldots}^{n-1} f(x_i) + 2 \sum_{i=2,4,6,\ldots}^{n-2} f(x_i) + f(x_n) \right], \] where \(\Delta x = \frac{b-a}{n}\). For our problem: \[ a = 1, \quad b = 2, \quad n = 8, \] \[ \Delta x = \frac{2-1}{8} = \frac{1}{8}. \] ### 5. Evaluate the Function at Required Points We need to compute \( f(x) = \sqrt{1 + \frac{1}{x^4}} \) at the following points: \[ x_i = 1 + i \cdot \frac{1}{8}, \quad \text{for} \quad i = 0, 1, 2, \ldots, 8. \] So, the points \( x_i \) are: \[ x_0 = 1, \] \[ x_1 = 1 + \frac{1}{8} = 1.125, \] \[ x_2 = 1.25, \] \[ x_3 = 1.375, \] \[ x_4 = 1.5, \] \[ x_5 = 1.625, \] \[ x_6 = 1.75, \] \[ x_7 = 1.875, \] \[ x_8 = 2. \] ### 6. Compute \( f(x_i) \) Now, let's compute \( f(x_i) = \sqrt{1 + \frac{1}{x_i^4}} \) at each point. ### 7. Apply Simpson's Rule Let's denote \( f(x) = \sqrt{1 + \frac{1}{x^4}} \). Using Simpson's Rule \( S_8 \): \[ S_8 = \frac{\Delta x}{3} \left[ f(x_0) + 4 \sum_{i=1,3,5,7} f(x_i) + 2 \sum_{i=2,4,6} f(x_i) + f(x_8) \right]. \] Let's calculate each term: \[ f(x_0) = f(1) = \sqrt{1 + \frac{1}{1^4}} = \sqrt{2}, \] \[ f(x_1) = f(1.125) = \sqrt{1 + \frac{1}{1.125^4}}, \] \[ f(x_2) = f(1.25) = \sqrt{1 + \frac{1}{1.25^4}}, \] \[ f(x_3) = f(1.375) = \sqrt{1 + \frac{1}{1.375^4}}, \] \[ f(x_4) = f(1.5) = \sqrt{1 + \frac{1}{1.5^ |
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| llama3.1-70B-Instruct | To find the arc length of the curve y = x^{-1} over the interval [1,2] using Simpson's Rule S_8, we need to follow these steps: ## Step 1: Define the function and the interval The function is y = x^{-1} and the interval is [1,2]. ## Step 2: Calculate the derivative of the function The derivative of y = x^{-1} is y' = -x^{-2}. ## Step 3: Calculate the arc length integral The arc length integral is ∫[√(1 + (y')^2)] dx from 1 to 2. ## Step 4: Substitute the derivative into the arc length integral Substituting y' = -x^{-2} into the arc length integral, we get ∫[√(1 + (-x^{-2})^2)] dx from 1 to 2. ## Step 5: Simplify the integral Simplifying the integral, we get ∫[√(1 + x^{-4})] dx from 1 to 2. ## Step 6: Apply Simpson's Rule S_8 To apply Simpson's Rule S_8, we need to divide the interval [1,2] into 8 equal subintervals, each of width (2-1)/8 = 1/8. ## Step 7: Calculate the function values at the subinterval points We need to calculate the function values at the points x = 1, 1.125, 1.25, 1.375, 1.5, 1.625, 1.75, 1.875, and 2. ## Step 8: Calculate the arc length using Simpson's Rule S_8 Using Simpson's Rule S_8, we can approximate the arc length as: L ≈ (1/8) * [√(1 + 1^{-4}) + 4*√(1 + 1.125^{-4}) + 2*√(1 + 1.25^{-4}) + 4*√(1 + 1.375^{-4}) + 2*√(1 + 1.5^{-4}) + 4*√(1 + 1.625^{-4}) + 2*√(1 + 1.75^{-4}) + 4*√(1 + 1.875^{-4}) + √(1 + 2^{-4})]. ## Step 9: Calculate the numerical value of the arc length Using a calculator, we can evaluate the expression to get the numerical value of the arc length. The final answer is: $\boxed{1.0986}$ |
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| llama3.1-405B-Instruct | To find the arc length of y = x^{-1} over the interval [1,2] using Simpson's Rule S_8, we need to follow these steps: 1. Define the function and its derivative: f(x) = x^{-1} f'(x) = -x^{-2} 2. Define the interval and the number of subintervals: a = 1 b = 2 n = 8 3. Calculate the width of each subinterval: h = (b - a) / n = (2 - 1) / 8 = 1/8 4. Calculate the points of subdivision: x_0 = a = 1 x_1 = a + h = 1 + 1/8 = 9/8 x_2 = x_1 + h = 9/8 + 1/8 = 10/8 x_3 = x_2 + h = 10/8 + 1/8 = 11/8 x_4 = x_3 + h = 11/8 + 1/8 = 12/8 x_5 = x_4 + h = 12/8 + 1/8 = 13/8 x_6 = x_5 + h = 13/8 + 1/8 = 14/8 x_7 = x_6 + h = 14/8 + 1/8 = 15/8 x_8 = x_7 + h = 15/8 + 1/8 = 2 5. Calculate the function values at each point of subdivision: f(x_0) = f(1) = 1 f(x_1) = f(9/8) = 8/9 f(x_2) = f(10/8) = 8/10 f(x_3) = f(11/8) = 8/11 f(x_4) = f(12/8) = 8/12 f(x_5) = f(13/8) = 8/13 f(x_6) = f(14/8) = 8/14 f(x_7) = f(15/8) = 8/15 f(x_8) = f(2) = 1/2 6. Calculate the derivative values at each point of subdivision: f'(x_0) = f'(1) = -1 f'(x_1) = f'(9/8) = -64/81 f'(x_2) = f'(10/8) = -64/100 f'(x_3) = f'(11/8) = -64/121 f'(x_4) = f'(12/8) = -64/144 f'(x_5) = f'(13/8) = -64/169 f'(x_6) = f'(14/8) = -64/196 f'(x_7) = f'(15/8) = -64/225 f'(x_8) = f'(2) = -1/4 7. Apply Simpson's Rule S_8: S_8 = (h/3) \* [f(x_0) + 4\*f(x_1) + 2\*f(x_2) + 4\*f(x_3) + 2\*f(x_4) + 4\*f(x_5) + 2\*f(x_6) + 4\*f(x_7) + f(x_8)] S_8 = (1/24) \* [1 + 4\*(8/9) + 2\*(8/10) + 4\*(8/11) + 2\*(8/12) + 4\*(8/13) + 2\*(8/14) + 4\*(8/15) + 1/2] S_8 ≈ 0.693122 Therefore, the arc length of y = x^{-1} over the interval [1,2] using Simpson's Rule S_8 is approximately 0.693122. |
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| DeepSeek-Coder-V2 | To find the arc length of the function \( y = x^{-1} \) over the interval \([1, 2]\) using Simpson's Rule \( S_8 \), we need to follow these steps: 1. **Find the derivative of the function**: \[ y = x^{-1} \implies \frac{dy}{dx} = -x^{-2} \] 2. **Set up the arc length formula**: The arc length \( L \) of a function \( y = f(x) \) over an interval \([a, b]\) is given by: \[ L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] For our function, this becomes: \[ L = \int_1^2 \sqrt{1 + (-x^{-2})^2} \, dx = \int_1^2 \sqrt{1 + x^{-4}} \, dx \] 3. **Apply Simpson's Rule \( S_8 \)**: Simpson's Rule approximates the integral by dividing the interval into \( n \) subintervals (where \( n \) is even) and using quadratic interpolations. For \( S_8 \), \( n = 8 \). The formula for Simpson's Rule is: \[ S_n = \frac{\Delta x}{3} \left( f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \cdots + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_n) \right) \] where \( \Delta x = \frac{b - a}{n} \). 4. **Calculate \( \Delta x \)**: \[ \Delta x = \frac{2 - 1}{8} = \frac{1}{8} \] 5. **Determine the \( x_i \) values**: \[ x_i = 1 + i \cdot \Delta x \quad \text{for} \quad i = 0, 1, 2, \ldots, 8 \] Thus, the \( x_i \) values are: \[ x_0 = 1, \quad x_1 = 1 + \frac{1}{8}, \quad x_2 = 1 + \frac{2}{8}, \quad \ldots, \quad x_8 = 2 \] 6. **Evaluate the function \( f(x) = \sqrt{1 + x^{-4}} \) at these points**: \[ f(x_i) = \sqrt{1 + (x_i)^{-4}} \] 7. **Apply Simpson's Rule**: \[ S_8 = \frac{\Delta x}{3} \left( f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \cdots + 2f(x_{6}) + 4f(x_{7}) + f(x_{8}) \right) \] 8. **Compute the values**: - \( f(x_0) = \sqrt{1 + 1^{-4}} = \sqrt{2} \) - \( f(x_1) = \sqrt{1 + \left(1 + \frac{1}{8}\right)^{-4}} \) - \( f(x_2) = \sqrt{1 + \left(1 + \frac{2}{8}\right)^{-4}} \) - Continue this process for all \( x_i \) up to \( x_8 \). 9. **Sum the values and apply the Simpson's Rule formula**: \[ S_8 = \frac{1/8}{3} \left( \sqrt{2} + 4 \cdot f(x_1) + 2 \cdot f(x_2) + \cdots + 2 \cdot f(x_6) + 4 \cdot f(x_7) + \sqrt{1 + 2^{-4}} \right) \] 10. **Final result**: After computing all the values and summing them up, you will get the approximate arc length using Simpson's Rule \( S_8 \). This process involves numerical computation and should be performed using a calculator or a computer algebra system for accuracy. |
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| Qwen2-72B-Instruct | To find the arc length of a function over a given interval using Simpson's Rule, we first need to establish the formula for arc length, which is given by: \[L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} dx\] Given the function \(y = x^{-1}\), we first find its derivative: \[f'(x) = -x^{-2}\] Then, we square the derivative and add 1: \[1 + [f'(x)]^2 = 1 + (-x^{-2})^2 = 1 + x^{-4}\] Thus, the arc length formula becomes: \[L = \int_{1}^{2} \sqrt{1 + x^{-4}} dx\] To approximate this integral using Simpson's Rule with \(S_8\), we need to divide the interval [1, 2] into 8 equal subintervals. This means the width of each subinterval, \(h\), is: \[h = \frac{b - a}{n} = \frac{2 - 1}{8} = \frac{1}{8}\] The nodes at which we evaluate the function are: \[x_0 = 1, x_1 = 1.125, x_2 = 1.25, x_3 = 1.375, x_4 = 1.5,\] \[x_5 = 1.625, x_6 = 1.75, x_7 = 1.875, x_8 = 2\] Now, we evaluate the function \(\sqrt{1 + x^{-4}}\) at these nodes and apply Simpson's Rule formula: \[S_8 = \frac{h}{3} [f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + 4f(x_5) + 2f(x_6) + 4f(x_7) + f(x_8)]\] Let's calculate the values of \(f(x_i)\): \[f(1) = \sqrt{1 + 1^{-4}} = \sqrt{2}\] \[f(1.125) = \sqrt{1 + 1.125^{-4}} \approx 1.041\] \[f(1.25) = \sqrt{1 + 1.25^{-4}} \approx 1.075\] \[f(1.375) = \sqrt{1 + 1.375^{-4}} \approx 1.103\] \[f(1.5) = \sqrt{1 + 1.5^{-4}} \approx 1.125\] \[f(1.625) = \sqrt{1 + 1.625^{-4}} \approx 1.143\] \[f(1.75) = \sqrt{1 + 1.75^{-4}} \approx 1.158\] \[f(1.875) = \sqrt{1 + 1.875^{-4}} \approx 1.171\] \[f(2) = \sqrt{1 + 2^{-4}} = \sqrt{1.0625} \approx 1.031\] Now, plug these values into the Simpson's Rule formula: \[S_8 = \frac{1/8}{3} [\sqrt{2} + 4(1.041) + 2(1.075) + 4(1.103) + 2(1.125) + 4(1.143) + 2(1.158) + 4(1.171) + 1.031]\] \[S_8 = \frac{1}{24} [1.414 + 4.164 + 2.15 + 4.412 + 2.25 + 4.572 + 2.316 + 4.684 + 1.031]\] \[S_8 = \frac{1}{24} [26.183]\] \[S_8 \approx 1.091\] Hence, the approximate arc length of \(y = x^{-1}\) over the interval [1, 2] using Simpson's Rule \(S_8\) is approximately 1.091. |
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| baichuan4 | To find the arc length of the function \(y = x^{-1}\) over the interval \([1,2]\) using Simpson's Rule \(S_8\), we first need to determine the formula for the arc length and then apply Simpson's Rule to approximate the integral. The arc length \(L\) of a function \(y = f(x)\) over an interval \([a,b]\) is given by the integral: \[ L = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx \] For our function \(y = x^{-1}\), the derivative \(f'(x) = -x^{-2}\). Therefore, the arc length formula becomes: \[ L = \int_1^2 \sqrt{1 + (-x^{-2})^2} \, dx \] \[ L = \int_1^2 \sqrt{1 + x^{-4}} \, dx \] Now, we will apply Simpson's Rule with \(n=8\) subintervals of equal length to approximate this integral. Simpson's Rule with \(n\) subintervals is given by: \[ S_n = \frac{\Delta x}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \dots + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_n) \right] \] where \(\Delta x = \frac{b - a}{n}\) is the width of each subinterval, and \(x_i = a + i\Delta x\) for \(i = 0, 1, \dots, n\). For our problem, \(a = 1\), \(b = 2\), \(n = 8\), and \(\Delta x = \frac{2 - 1}{8} = \frac{1}{8}\). We need to calculate the function values at \(x_0, x_1, \dots, x_8\): - \(x_0 = 1\), \(x_8 = 2\) - \(x_1 = 1 + \frac{1}{8} = 1.125\), \(x_7 = 2 - \frac{1}{8} = 1.875\) - \(x_2 = 1 + 2\cdot\frac{1}{8} = 1.25\), \(x_6 = 2 - 2\cdot\frac{1}{8} = 1.75\) - \(x_3 = 1 + 3\cdot\frac{1}{8} = 1.375\), \(x_5 = 2 - 3\cdot\frac{1}{8} = 1.625\) - \(x_4 = 1 + 4\cdot\frac{1}{8} = 1.5\) Now, we calculate the corresponding function values \(f(x_i) = \sqrt{1 + x_i^{-4}}\) for each \(x_i\): - \(f(x_0) = \sqrt{1 + 1^{-4}} = \sqrt{1 + 0.0625} = \sqrt{1.0625}\) - \(f(x_8) = \sqrt{1 + 2^{-4}} = \sqrt{1 + 0.0625} = \sqrt{1.0625}\) - \(f(x_1) = f(1.125) = \sqrt{1 + (1.125)^{-4}}\) - \(f(x_7) = f(1.875) = \sqrt{1 + (1.875)^{-4}}\) - and so on for the other \(x_i\) values. After calculating all the necessary function values, we plug them into Simpson's Rule formula: \[ S_8 = \frac{1}{8} \cdot \frac{1}{3} \left[ \sqrt{1.0625} + 4\sqrt{1 + (1.125)^{-4}} + 2\sqrt{1 + (1.25)^{-4}} + \dots + 4\sqrt{1 + (1.875)^{-4}} + \sqrt{1.0625} \right] \] Calculating these values and summing them up according to Simpson's Rule will give us the approximate arc length of the function \(y = x^{-1}\) over the interval \([1,2]\). Let's calculate: - \(f(x_1) = f(1.125) = \sqrt{1 + (1.125)^{-4}} \approx \sqrt{1.0566} \) - \(f(x_2) = f(1.25) = \sqrt{1 + (1.25)^{-4}} \approx \sqrt{1.0506} \) - \(f(x_3) = f(1.375) = \sqrt{1 + (1.375)^{-4}} \approx \sqrt{1.0453} \) - \(f(x_4) = f(1.5) = \sqrt{1 + (1.5)^{-4}} \approx \sqrt{1.0408} \) - \(f(x_5) = f(1.625) = \sqrt{1 + (1.625)^{-4}} \approx \sqrt{1.0370} \) - \(f(x_6) = f(1.75) = \sqrt{1 + (1.75)^{-4}} \approx \sqrt{1.0339} \) - \(f(x_7) = f(1.875) = \sqrt{1 + (1.875)^{-4}} \approx \sqrt{1.0314} \) Now, plugging these values into the Simpson's Rule formula: \[ S_8 = \frac{1}{24} \left[ 2\sqrt{1.0625} + 4\sqrt{1.0566} + 2\sqrt{1.0506} + 4\sqrt{1.0453} + 2\sqrt{1.0408} + 4\sqrt{1.0370} + 2\sqrt{1.0339} + 4\sqrt{1.0314} \right] \] \[ S_8 = \frac{1}{24} \left[ 2\cdot1.0313 + 4\cdot1.0281 + 2\cdot1.0252 + 4\cdot1.0226 + 2\cdot1.0204 + 4\cdot1.0185 + 2\cdot1.0169 + 4\cdot1.0156 \right] \] \[ S_8 = \frac{1}{24} \left[ 2.0626 + 4.1124 + 2.0504 + 4.0904 + 2.0408 + 4.0740 + 2.0338 + 4.0624 \right] \] \[ S_8 = \frac{1}{24} \times 24.476 \] \[ S_8 \approx 1.020 \] Therefore, the arc length of \(y = x^{-1}\) over the interval \([1,2] |
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| doubao | First, we need to find the derivative of y = x^{-1} which is y' = -x^{-2} = -1/x^2. The arc length formula is given by: L = ∫[a,b] √(1 + (y')^2) dx. So, we have to evaluate the integral from 1 to 2 of √(1 + (-1/x^2)^2) dx. Using Simpson's Rule S_8, we divide the interval [1,2] into 8 subintervals. The width of each subinterval is (2-1)/8 = 1/8. The x-values for the subintervals are: 1, 1 + 1/8 = 9/8, 1 + 2(1/8) = 5/4, 1 + 3(1/8) = 13/8, 1 + 4(1/8) = 7/4, 1 + 5(1/8) = 17/8, 1 + 6(1/8) = 9/4, 1 + 7(1/8) = 25/8, 2. Now we evaluate the function √(1 + (-1/x^2)^2) at these x-values and use Simpson's Rule formula: S_8 = (1/3)[f(1) + 4f(9/8) + 2f(5/4) + 4f(13/8) + 2f(7/4) + 4f(17/8) + 2f(9/4) + 4f(25/8) + f(2)]. After doing the calculations, we get the approximate value of the arc length. |
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| minimax | To find the arc length of the curve defined by \( y = x^{-1} \) over the interval \([1,2]\) using Simpson's Rule, we first need to set up the integral that represents the arc length and then apply Simpson's Rule to approximate it. The arc length \( L \) of a curve defined by \( y = f(x) \) from \( x = a \) to \( x = b \) is given by the integral: \[ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] For the given function \( y = x^{-1} \), the derivative \( \frac{dy}{dx} \) is: \[ \frac{dy}{dx} = -x^{-2} \] Plugging this into the formula for arc length, we get: \[ L = \int_{1}^{2} \sqrt{1 + \left(-x^{-2}\right)^2} \, dx = \int_{1}^{2} \sqrt{1 + x^{-4}} \, dx \] Now, we need to approximate this integral using Simpson's Rule. Simpson's Rule for \( n \) subintervals (where \( n \) is even) is given by: \[ S_n = \frac{\Delta x}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \ldots + 4f(x_{n-1}) + f(x_n) \right] \] For \( S_8 \), we have \( n = 8 \) subintervals, so \( \Delta x = \frac{b-a}{n} = \frac{2-1}{8} = \frac{1}{8} \). The points \( x_i \) are then: \[ x_i = a + i\Delta x = 1 + \frac{i}{8} \quad \text{for} \quad i = 0, 1, \ldots, 8 \] Now we need to evaluate the function \( f(x) = \sqrt{1 + x^{-4}} \) at these points and apply Simpson's Rule: \[ S_8 = \frac{1/8}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + 4f(x_5) + 2f(x_6) + 4f(x_7) + f(x_8) \right] \] \[ S_8 = \frac{1}{24} \left[ f(1) + 4f\left(\frac{9}{8}\right) + 2f\left(\frac{5}{4}\right) + 4f\left(\frac{11}{8}\right) + 2f\left(\frac{3}{2}\right) + 4f\left(\frac{13}{8}\right) + 2f\left(\frac{7}{4}\right) + 4f\left(\frac{15}{8}\right) + f(2) \right] \] Now we need to calculate the values of \( f(x) \) at these points and plug them into the formula: \[ f(x) = \sqrt{1 + x^{-4}} \] \[ f(1) = \sqrt{1 + 1^4} = \sqrt{2} \] \[ f\left(\frac{9}{8}\right) = \sqrt{1 + \left(\frac{8}{9}\right)^4} \] \[ f\left(\frac{5}{4}\right) = \sqrt{1 + \left(\frac{4}{5}\right)^4} \] \[ f\left(\frac{11}{8}\right) = \sqrt{1 + \left(\frac{8}{11}\right)^4} \] \[ f\left(\frac{3}{2}\right) = \sqrt{1 + \left(\frac{2}{3}\right)^4} \] \[ f\left(\frac{13}{8}\right) = \sqrt{1 + \left(\frac{8}{13}\right)^4} \] \[ f\left(\frac{7}{4}\right) = \sqrt{1 + \left(\frac{4}{7}\right)^4} \] \[ f\left(\frac{15}{8}\right) = \sqrt{1 + \left(\frac{8}{15}\right)^4} \] \[ f(2) = \sqrt{1 + 2^{-4}} = \sqrt{1 + \frac{1}{16}} = \sqrt{\frac{17}{16}} = \frac{\sqrt{17}}{4} \] After calculating these values, we would plug them into the Simpson's Rule formula and perform the arithmetic to get the approximate arc length \( S_8 \). Since the calculations involve square roots and powers, they are best done with a calculator or computational tool. |
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| kimi | To find the arc length of the curve \( y = x^{-1} \) over the interval \([1, 2]\) using Simpson's Rule, we first need to recall the formula for arc length and the formula for Simpson's Rule. The arc length \( L \) of a curve \( y = f(x) \) from \( x = a \) to \( x = b \) is given by: \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] For the curve \( y = x^{-1} \), the derivative is: \[ \frac{dy}{dx} = -x^{-2} \] So, the integrand for the arc length becomes: \[ \sqrt{1 + \left( -x^{-2} \right)^2} = \sqrt{1 + x^{-4}} \] Now, let's apply Simpson's Rule. Simpson's Rule for a single interval \([a, b]\) is given by: \[ S_2 = \frac{b-a}{6} \left[ f(a) + 4f\left( \frac{a+b}{2} \right) + f(b) \right] \] For Simpson's Rule with \( n = 8 \) subintervals, we divide the interval \([1, 2]\) into \( 8 \) equal subintervals, each of width \( \Delta x = \frac{2 - 1}{8} = \frac{1}{8} \). The points are \( x_0 = 1, x_1 = 1 + \frac{1}{8}, x_2 = 1 + 2 \cdot \frac{1}{8}, \ldots, x_8 = 2 \). The Simpson's Rule for \( n = 8 \) subintervals is: \[ S_8 = \frac{\Delta x}{3} \left[ f(x_0) + 4 \sum_{i=1,5,7} f(x_i) + 2 \sum_{i=2,4,6} f(x_i) + 4 \sum_{i=3} f(x_i) + f(x_8) \right] \] However, since we are integrating the arc length, we need to use the integrand \( \sqrt{1 + x^{-4}} \) instead of \( f(x) \). So, let's calculate the values of \( \sqrt{1 + x^{-4}} \) at the necessary points. \[ f(x_0) = \sqrt{1 + 1^{-4}} = \sqrt{2} \] \[ f(x_1) = \sqrt{1 + (1 + \frac{1}{8})^{-4}} \] \[ f(x_2) = \sqrt{1 + (1 + 2 \cdot \frac{1}{8})^{-4}} \] \[ \ldots \] \[ f(x_8) = \sqrt{1 + 2^{-4}} = \sqrt{1 + \frac{1}{16}} \] Now, we can plug these values into the Simpson's Rule formula. However, calculating these values exactly would require a calculator or computer software. Let's do the calculation for the first few points to illustrate the process: \[ f(x_1) = \sqrt{1 + (1.125)^{-4}} \] \[ f(x_2) = \sqrt{1 + (1.25)^{-4}} \] \[ \ldots \] After calculating all the necessary values, we would sum them according to the Simpson's Rule formula and multiply by \( \frac{\Delta x}{3} \) to get the approximate arc length. For simplicity, let's assume we have calculated all the necessary values and denote them as \( f(x_i) \) for \( i = 0, 1, \ldots, 8 \). Then the Simpson's Rule approximation for the arc length would be: \[ S_8 = \frac{1/8}{3} \left[ \sqrt{2} + 4 \left( \sqrt{1 + (1.125)^{-4}} + \sqrt{1 + (1.5)^{-4}} \right) + 2 \left( \sqrt{1 + (1.25)^{-4}} + \sqrt{1 + (1.375)^{-4}} + \sqrt{1 + (1.75)^{-4}} \right) + 4 \left( \sqrt{1 + (1.625)^{-4}} \right) + \sqrt{1 + \frac{ |
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| wenxinyiyan | To find the arc length of y = x^{-1} over the interval [1,2] using Simpson's Rule S_8, we first need to determine the arc length formula. The arc length of a function y(x) over an interval [a,b] is given by: L = ∫[a,b] sqrt(1 + (y')^2) dx where y' is the derivative of y with respect to x. For y = x^{-1}, the derivative y' is: y' = -x^{-2} So, the arc length integral becomes: L = ∫[1,2] sqrt(1 + (-x^{-2})^2) dx = ∫[1,2] sqrt(1 + x^{-4}) dx Now, we will use Simpson's Rule S_8 to approximate this integral. Simpson's Rule S_8 approximates an integral over an interval [a,b] using 8 subintervals. It is given by: S_8 = (Δx/3) [f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + 4f(x_5) + 2f(x_6) + 4f(x_7) + f(x_8)] where Δx = (b - a)/8 and x_i = a + iΔx for i = 0, 1, ..., 8. In our case, a = 1, b = 2, and f(x) = sqrt(1 + x^{-4}). So, Δx = (2 - 1)/8 = 1/8. Now, we can calculate the values of f(x_i) for i = 0, 1, ..., 8: x_0 = 1 x_1 = 1 + 1/8 = 9/8 x_2 = 1 + 2/8 = 5/4 x_3 = 1 + 3/8 = 11/8 x_4 = 1 + 4/8 = 3/2 x_5 = 1 + 5/8 = 13/8 x_6 = 1 + 6/8 = 7/4 x_7 = 1 + 7/8 = 15/8 x_8 = 2 f(x_0) = sqrt(1 + 1^{-4}) = sqrt(2) f(x_1) = sqrt(1 + (9/8)^{-4}) f(x_2) = sqrt(1 + (5/4)^{-4}) f(x_3) = sqrt(1 + (11/8)^{-4}) f(x_4) = sqrt(1 + (3/2)^{-4}) f(x_5) = sqrt(1 + (13/8)^{-4}) f(x_6) = sqrt(1 + (7/4)^{-4}) f(x_7) = sqrt(1 + (15/8)^{-4}) f(x_8) = sqrt(1 + 2^{-4}) Now, we can plug these values into the Simpson's Rule S_8 formula: S_8 = (1/8/3) [sqrt(2) + 4sqrt(1 + (9/8)^{-4}) + 2sqrt(1 + (5/4)^{-4}) + 4sqrt(1 + (11/8)^{-4}) + 2sqrt(1 + (3/2)^{-4}) + 4sqrt(1 + (13/8)^{-4}) + 2sqrt(1 + (7/4)^{-4}) + 4sqrt(1 + (15/8)^{-4}) + sqrt(1 + 2^{-4})] |
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| zhipu | To find the arc length of the curve \( y = x^{-1} \) over the interval \([1, 2]\) using Simpson's Rule \( S_8 \), we need to follow these steps: 1. **Determine the Arc Length Formula:** The arc length \( L \) of a function \( y = f(x) \) from \( x = a \) to \( x = b \) is given by: \[ L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] 2. **Find the Derivative \( \frac{dy}{dx} \):** For \( y = x^{-1} \), we have: \[ \frac{dy}{dx} = -x^{-2} = -\frac{1}{x^2} \] 3. **Substitute into the Arc Length Formula:** \[ L = \int_1^2 \sqrt{1 + \left( -\frac{1}{x^2} \right)^2} \, dx = \int_1^2 \sqrt{1 + \frac{1}{x^4}} \, dx \] 4. **Set Up Simpson's Rule:** Simpson's Rule for approximating integrals is given by: \[ S_n = \frac{\Delta x}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \ldots + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_n) \right] \] where \( \Delta x = \frac{b - a}{n} \). For \( S_8 \), \( n = 8 \), \( a = 1 \), and \( b = 2 \): \[ \Delta x = \frac{2 - 1}{8} = \frac{1}{8} \] 5. **Calculate the Function Values:** We need to evaluate \( f(x) = \sqrt{1 + \frac{1}{x^4}} \) at the points \( x_0, x_1, \ldots, x_8 \) where \( x_i = 1 + i \cdot \Delta x \). \[ \begin{align*} x_0 &= 1, \\ x_1 &= 1 + \frac{1}{8} = 1.125, \\ x_2 &= 1 + 2 \cdot \frac{1}{8} = 1.25, \\ x_3 &= 1 + 3 \cdot \frac{1}{8} = 1.375, \\ x_4 &= 1 + 4 \cdot \frac{1}{8} = 1.5, \\ x_5 &= 1 + 5 \cdot \frac{1}{8} = 1.625, \\ x_6 &= 1 + 6 \cdot \frac{1}{8} = 1.75, \\ x_7 &= 1 + 7 \cdot \frac{1}{8} = 1.875, \\ x_8 &= 2. \end{align*} \] Now, calculate \( f(x_i) \) for each \( x_i \): \[ \begin{align*} f(x_0) &= \sqrt{1 + \frac{1}{1^4}} = \sqrt{2}, \\ f(x_1) &= \sqrt{1 + \frac{1}{1.125^4}}, \\ f(x_2) &= \sqrt{1 + \frac{1}{1.25^4}}, \\ f(x_3) &= \sqrt{1 + \frac{1}{1.375^4}}, \\ f(x_4) &= \sqrt{1 + \frac{1}{1.5^4}}, \\ f(x_5) &= \sqrt{1 + \frac{1}{1.625^4}}, \\ f(x_6) &= \sqrt{1 + \frac{1}{1.75^4}}, \\ f(x_7) &= \sqrt{1 + \frac{1}{1.875^4}}, \\ f(x_8) &= \sqrt{1 + \frac{1}{2^4}} = \sqrt{1 + \frac{1}{16}} = \sqrt{\frac{17}{16}} = \frac{\sqrt{17}}{4}. \end{align*} \] 6. **Apply Simpson's Rule:** \[ S_8 = \frac{\Delta x}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + 4f(x_5) + 2f(x_6) + 4f(x_7) + f(x_8) \right] \] Substitute \( \Delta x = \frac{1}{8} \) and the calculated function values: \[ S_8 = \frac{1}{24} \left[ \sqrt{2} + 4\sqrt{1 + \frac{1}{1.125^4}} + 2\sqrt{1 + \frac{1}{1.25^4}} + 4\sqrt{1 + \frac{1}{1.375^4}} + 2\sqrt{1 + \frac{1}{1.5^4}} + 4\sqrt{1 + \frac{1}{1.625^4}} + 2\sqrt{1 + \frac{1}{1.75^4}} + 4\sqrt{1 + \frac{1}{1.875^4}} + \frac{\sqrt{17}}{4} \right] \] 7. **Calculate the Numerical Value:** Use a calculator to compute the values of the square roots and the final sum. \[ S_8 \approx \frac{1}{24} \left[ \sqrt{2} + 4 \cdot 1.0037 + 2 \cdot 1.0062 + 4 \cdot 1.0096 + 2 \cdot 1.0139 + 4 \cdot 1.0182 + 2 \cdot 1.0235 + 4 \cdot 1.0290 + \frac{\sqrt{17}}{4} \right] \] \[ S_8 \approx \frac{1}{24} \left[ 1.4142 + 4.0148 + 2.0124 + 4.0384 + 2.0278 + 4.0728 + 2.0470 + 4.1600 + 0.4125 \right] \] \[ S_8 \approx \frac{1}{24} \left[ 27.8099 \right] \] \[ S_8 \approx 1.1571 \] Therefore, the approximate arc length of \( y = x^{-1} \) over the interval \([1, 2]\) using Simpson's Rule \( S_8 \) is approximately \( 1.1571 \). |
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| gpto1 | To find the arc length of the function \( y = x^{-1} \) over the interval \([1, 2]\) using Simpson's Rule with \( n = 8 \) subdivisions, we proceed as follows: **Step 1: Set Up the Arc Length Integral** The formula for the arc length \( L \) of a function \( y = f(x) \) from \( x = a \) to \( x = b \) is: \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] For \( y = x^{-1} = \frac{1}{x} \), the derivative is: \[ \frac{dy}{dx} = -x^{-2} \] So, the integrand becomes: \[ \sqrt{1 + \left( -x^{-2} \right)^2} = \sqrt{1 + x^{-4}} \] Therefore, the arc length is: \[ L = \int_{1}^{2} \sqrt{1 + x^{-4}} \, dx \] **Step 2: Divide the Interval into Subintervals** With \( n = 8 \) subdivisions: \[ \Delta x = \frac{b - a}{n} = \frac{2 - 1}{8} = 0.125 \] The \( x_i \) values are: \[ x_i = a + i\Delta x, \quad i = 0, 1, 2, \dots, 8 \] So: \[ x_0 = 1,\ x_1 = 1.125,\ x_2 = 1.25,\ x_3 = 1.375,\ x_4 = 1.5,\ x_5 = 1.625,\ x_6 = 1.75,\ x_7 = 1.875,\ x_8 = 2 \] **Step 3: Compute the Function Values at Each \( x_i \)** We compute \( f(x_i) = \sqrt{1 + x_i^{-4}} \): \[ \begin{align*} f(x_0) &= \sqrt{1 + (1)^{-4}} = \sqrt{1 + 1} = \sqrt{2} \approx 1.4142 \\ f(x_1) &= \sqrt{1 + (1.125)^{-4}} = \sqrt{1 + 0.6243} = \sqrt{1.6243} \approx 1.2748 \\ f(x_2) &= \sqrt{1 + (1.25)^{-4}} = \sqrt{1 + 0.4096} = \sqrt{1.4096} \approx 1.1874 \\ f(x_3) &= \sqrt{1 + (1.375)^{-4}} = \sqrt{1 + 0.2797} = \sqrt{1.2797} \approx 1.1310 \\ f(x_4) &= \sqrt{1 + (1.5)^{-4}} = \sqrt{1 + 0.1975} = \sqrt{1.1975} \approx 1.0943 \\ f(x_5) &= \sqrt{1 + (1.625)^{-4}} = \sqrt{1 + 0.1438} = \sqrt{1.1438} \approx 1.0693 \\ f(x_6) &= \sqrt{1 + (1.75)^{-4}} = \sqrt{1 + 0.1068} = \sqrt{1.1068} \approx 1.0528 \\ f(x_7) &= \sqrt{1 + (1.875)^{-4}} = \sqrt{1 + 0.0810} = \sqrt{1.0810} \approx 1.0396 \\ f(x_8) &= \sqrt{1 + (2)^{-4}} = \sqrt{1 + 0.0625} = \sqrt{1.0625} \approx 1.0308 \\ \end{align*} \] **Step 4: Apply Simpson's Rule** Simpson's Rule for \( n \) subdivisions is: \[ S_n = \frac{\Delta x}{3} \left[ f(x_0) + 4\sum_{\text{odd } i} f(x_i) + 2\sum_{\text{even } i} f(x_i) + f(x_n) \right] \] Compute the weighted sum: \[ \begin{align*} S &= f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + 4f(x_5) + 2f(x_6) + 4f(x_7) + f(x_8) \\ &\approx 1.4142 + 4(1.2748) + 2(1.1874) + 4(1.1310) + 2(1.0943) + 4(1.0693) + 2(1.0528) + 4(1.0396) + 1.0308 \\ &\approx 1.4142 + 5.0990 + 2.3748 + 4.5240 + 2.1886 + 4.2772 + 2.1056 + 4.1584 + 1.0308 \\ &\approx 27.1730 \end{align*} \] **Step 5: Multiply by \(\frac{\Delta x}{3}\) to Get the Approximate Integral** \[ L \approx \frac{\Delta x}{3} \times S = \frac{0.125}{3} \times 27.1730 \approx 0.0416667 \times 27.1730 \approx 1.1322 \] **Answer:** Approximately 1.1322 units—the arc length is ≈ 1.1322 using Simpson’s Rule with 8 intervals. |
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| deepseekv3 | To find the arc length of the curve \( y = x^{-1} \) over the interval \([1, 2]\) using Simpson's Rule \( S_8 \), follow these steps: ### Step 1: Arc Length Formula The arc length \( L \) of a curve \( y = f(x) \) over the interval \([a, b]\) is given by: \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] For \( y = x^{-1} \), we first compute the derivative: \[ \frac{dy}{dx} = -\frac{1}{x^2} \] Thus, the integrand becomes: \[ \sqrt{1 + \left( -\frac{1}{x^2} \right)^2} = \sqrt{1 + \frac{1}{x^4}} \] ### Step 2: Simpson's Rule \( S_8 \) Simpson's Rule approximates the integral of a function \( f(x) \) over an interval \([a, b]\) using a weighted sum of function values at equally spaced points. For \( S_8 \), we divide the interval \([1, 2]\) into 8 subintervals. Let \( h = \frac{b - a}{n} = \frac{2 - 1}{8} = \frac{1}{8} \). The points \( x_i \) are: \[ x_i = 1 + i \cdot h \quad \text{for} \quad i = 0, 1, 2, \ldots, 8 \] The Simpson's Rule \( S_8 \) is given by: \[ S_8 = \frac{h}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + 4f(x_5) + 2f(x_6) + 4f(x_7) + f(x_8) \right] \] ### Step 3: Compute \( f(x_i) \) We need to evaluate the integrand \( f(x) = \sqrt{1 + \frac{1}{x^4}} \) at each \( x_i \): \[ \begin{align*} x_0 &= 1.0000, & f(x_0) &= \sqrt{1 + \frac{1}{1^4}} = \sqrt{2} \approx 1.4142 \\ x_1 &= 1.1250, & f(x_1) &= \sqrt{1 + \frac{1}{1.1250^4}} \approx 1.2448 \\ x_2 &= 1.2500, & f(x_2) &= \sqrt{1 + \frac{1}{1.2500^4}} \approx 1.1471 \\ x_3 &= 1.3750, & f(x_3) &= \sqrt{1 + \frac{1}{1.3750^4}} \approx 1.0855 \\ x_4 &= 1.5000, & f(x_4) &= \sqrt{1 + \frac{1}{1.5000^4}} \approx 1.0426 \\ x_5 &= 1.6250, & f(x_5) &= \sqrt{1 + \frac{1}{1.6250^4}} \approx 1.0113 \\ x_6 &= 1.7500, & f(x_6) &= \sqrt{1 + \frac{1}{1.7500^4}} \approx 0.9887 \\ x_7 &= 1.8750, & f(x_7) &= \sqrt{1 + \frac{1}{1.8750^4}} \approx 0.9718 \\ x_8 &= 2.0000, & f(x_8) &= \sqrt{1 + \frac{1}{2.0000^4}} \approx 0.9682 \\ \end{align*} \] ### Step 4: Apply Simpson's Rule Substitute the values into Simpson's Rule: \[ S_8 = \frac{1/8}{3} \left[ 1.4142 + 4(1.2448) + 2(1.1471) + 4(1.0855) + 2(1.0426) + 4(1.0113) + 2(0.9887) + 4(0.9718) + 0.9682 \right] \] Calculate the weighted sum: \[ S_8 \approx \frac{1}{24} \left[ 1.4142 + 4.9792 + 2.2942 + 4.3420 + 2.0852 + 4.0452 + 1.9774 + 3.8872 + 0.9682 \right] \] \[ S_8 \approx \frac{1}{24} \times 26.9928 \approx 1.1247 \] ### Final Answer The arc length of \( y = x^{-1} \) over the interval \([1, 2]\) using Simpson's Rule \( S_8 \) is approximately: \[ L \approx 1.1247 \] |
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