| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To approximate the area under the curve y=f(x) between x=0 and x=8 using the Trapezoidal rule with n=4 subintervals, we will follow these steps: 1. Calculate the width of each subinterval: Since the interval [0, 8] is divided into 4 subintervals, the width of each subinterval (h) is given by: h = (b - a) / n h = (8 - 0) / 4 h = 2 2. Calculate the values of f(x) at the endpoints of each subinterval: We are given the values of f(x) at x=0, x=2, x=4, x=6, and x=8. 3. Apply the Trapezoidal rule formula: The formula for the Trapezoidal rule with n subintervals is: A ≈ (h/2) * [f(x0) + 2f(x1) + 2f(x2) + ... + 2f(xn-1) + f(xn)] where A is the approximate area under the curve, h is the width of each subinterval, and f(xi) represents the values of the function at the endpoints of each subinterval. 4. Substitute the values into the formula: A ≈ (2/2) * [f(0) + 2f(2) + 2f(4) + 2f(6) + f(8)] A ≈ (1) * [3 + 2(7) + 2(11) + 2(9) + 3] A ≈ 1 * [3 + 14 + 22 + 18 + 3] A ≈ 1 * 50 A ≈ 50 Therefore, the approximate area under the curve y=f(x) between x=0 and x=8 using the Trapezoidal rule with n=4 subintervals is 50 square units. |
60.0 |
| deepseekr1 | To approximate the area under the curve \( y = f(x) \) between \( x = 0 \) and \( x = 8 \) using the Trapezoidal rule with \( n = 4 \) subintervals, we follow these steps: 1. **Determine the width of each subinterval (\( \Delta x \))**: \[ \Delta x = \frac{8 - 0}{4} = 2 \] 2. **List the function values at the given points**: \[ f(0) = 3, \quad f(2) = 7, \quad f(4) = 11, \quad f(6) = 9, \quad f(8) = 3 \] 3. **Apply the Trapezoidal rule formula**: \[ \text{Area} = \frac{\Delta x}{2} \left[ f(x_0) + 2 \sum_{i=1}^{n-1} f(x_i) + f(x_n) \right] \] Substituting the values: \[ \text{Area} = \frac{2}{2} \left[ 3 + 2(7 + 11 + 9) + 3 \right] \] Simplify the expression inside the brackets: \[ 3 + 2(7) + 2(11) + 2(9) + 3 = 3 + 14 + 22 + 18 + 3 = 60 \] Therefore: \[ \text{Area} = 1 \times 60 = 60 \] 4. **Verification by calculating individual trapezoid areas**: - First trapezoid (0 to 2): \(\frac{3 + 7}{2} \times 2 = 10\) - Second trapezoid (2 to 4): \(\frac{7 + 11}{2} \times 2 = 18\) - Third trapezoid (4 to 6): \(\frac{11 + 9}{2} \times 2 = 20\) - Fourth trapezoid (6 to 8): \(\frac{9 + 3}{2} \times 2 = 12\) Summing these areas: \[ 10 + 18 + 20 + 12 = 60 \] Thus, the approximate area under the curve using the Trapezoidal rule with \( n = 4 \) subintervals is \(\boxed{60}\). |
60.0 |
| gpto1 | To approximate the area under the curve \( y = f(x) \) from \( x = 0 \) to \( x = 8 \) using the Trapezoidal Rule with \( n = 4 \) subintervals, we'll follow these steps: ### **Step 1: Determine the Width of Each Subinterval (\( h \))** The width \( h \) of each subinterval is calculated using the formula: \[ h = \frac{b - a}{n} \] Where: - \( a = 0 \) (starting point) - \( b = 8 \) (ending point) - \( n = 4 \) (number of subintervals) Calculating \( h \): \[ h = \frac{8 - 0}{4} = \frac{8}{4} = 2 \] ### **Step 2: Identify the \( x \)-values and Corresponding \( f(x) \)-values** Since \( h = 2 \), the \( x \)-values at the endpoints of the subintervals are: \[ \begin{align*} x_0 &= 0 \\ x_1 &= x_0 + h = 0 + 2 = 2 \\ x_2 &= x_1 + h = 2 + 2 = 4 \\ x_3 &= x_2 + h = 4 + 2 = 6 \\ x_4 &= x_3 + h = 6 + 2 = 8 \\ \end{align*} \] Corresponding \( f(x) \)-values: \[ \begin{align*} f(x_0) &= f(0) = 3 \\ f(x_1) &= f(2) = 7 \\ f(x_2) &= f(4) = 11 \\ f(x_3) &= f(6) = 9 \\ f(x_4) &= f(8) = 3 \\ \end{align*} \] ### **Step 3: Apply the Trapezoidal Rule** The Trapezoidal Rule formula for approximating the area is: \[ \text{Area} \approx \frac{h}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4) \right] \] Plugging in the values: \[ \text{Area} \approx \frac{2}{2} \left[ f(0) + 2f(2) + 2f(4) + 2f(6) + f(8) \right] \] Simplify the fraction: \[ \text{Area} \approx 1 \times \left[ 3 + 2(7) + 2(11) + 2(9) + 3 \right] \] Compute the products: \[ \text{Area} \approx 1 \times \left[ 3 + 14 + 22 + 18 + 3 \right] \] Sum the values inside the brackets: \[ \text{Area} \approx 1 \times 60 \] \[ \text{Area} \approx 60 \] ### **Final Answer** The approximate area under the curve between \( x = 0 \) and \( x = 8 \) is **60**. |
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| GPT-4o | To approximate the area under the curve \( y = f(x) \) between \( x = 0 \) and \( x = 8 \) using the Trapezoidal rule with \( n = 4 \) subintervals, we follow these steps: 1. **Determine the width of each subinterval:** \[ \Delta x = \frac{b - a}{n} = \frac{8 - 0}{4} = 2 \] So, each subinterval has a width of 2 units. 2. **Identify the function values at the endpoints of each subinterval:** \[ x_0 = 0, \quad x_1 = 2, \quad x_2 = 4, \quad x_3 = 6, \quad x_4 = 8 \] \[ f(x_0) = f(0) = 3, \quad f(x_1) = f(2) = 7, \quad f(x_2) = f(4) = 11, \quad f(x_3) = f(6) = 9, \quad f(x_4) = f(8) = 3 \] 3. **Apply the Trapezoidal rule formula:** The Trapezoidal rule for \( n \) subintervals is given by: \[ \text{Area} \approx \frac{\Delta x}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4) \right] \] Substituting the values we have: \[ \text{Area} \approx \frac{2}{2} \left[ f(0) + 2f(2) + 2f(4) + 2f(6) + f(8) \right] \] \[ \text{Area} \approx 1 \left[ 3 + 2(7) + 2(11) + 2(9) + 3 \right] \] \[ \text{Area} \approx 1 \left[ 3 + 14 + 22 + 18 + 3 \right] \] \[ \text{Area} \approx 1 \left[ 60 \right] \] \[ \text{Area} \approx 60 \] So, the approximate area under the curve \( y = f(x) \) between \( x = 0 \) and \( x = 8 \) using the Trapezoidal rule with \( n = 4 \) subintervals is \( \boxed{60} \). |
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| llama3.1-70B-Instruct | To approximate the area under the curve y=f(x) between x=0 and x=8 using the Trapezoidal rule with n=4 subintervals, we can follow these steps: 1. Divide the interval [0,8] into 4 equal subintervals, each of width 2. The subintervals are [0,2], [2,4], [4,6], and [6,8]. 2. Calculate the function values at the endpoints of each subinterval: f(0) = 3 f(2) = 7 f(4) = 11 f(6) = 9 f(8) = 3 3. Apply the Trapezoidal rule formula: Area ≈ (h/2) \* (f(x0) + 2\*f(x1) + 2\*f(x2) + 2\*f(x3) + f(x4)) where h is the width of each subinterval (2 in this case), and x0, x1, x2, x3, and x4 are the endpoints of the subintervals. Plugging in the values, we get: Area ≈ (2/2) \* (3 + 2\*7 + 2\*11 + 2\*9 + 3) = 1 \* (3 + 14 + 22 + 18 + 3) = 1 \* 60 = 60 Therefore, the approximate area under the curve y=f(x) between x=0 and x=8 using the Trapezoidal rule with n=4 subintervals is 60. |
60.0 |
| llama3.1-405B-Instruct | To approximate the area under the curve y=f(x) between x=0 and x=8 using the Trapezoidal rule with n=4 subintervals, we can use the following formula: Area ≈ (h/2) \* (f(x0) + 2\*f(x1) + 2\*f(x2) + 2\*f(x3) + f(x4)) where h is the width of each subinterval, and x0, x1, x2, x3, and x4 are the points that divide the interval [0,8] into 4 equal subintervals. First, we need to find the width of each subinterval: h = (8-0)/4 = 2 Next, we can find the points that divide the interval [0,8] into 4 equal subintervals: x0 = 0 x1 = 2 x2 = 4 x3 = 6 x4 = 8 Now, we can plug in the values of f(x) at these points into the formula: Area ≈ (2/2) \* (3 + 2\*7 + 2\*11 + 2\*9 + 3) Area ≈ 1 \* (3 + 14 + 22 + 18 + 3) Area ≈ 1 \* 60 Area ≈ 60 Therefore, the approximate area under the curve y=f(x) between x=0 and x=8 using the Trapezoidal rule with n=4 subintervals is 60. |
60.0 |
| DeepSeek-Coder-V2 | To approximate the area under the curve \( y = f(x) \) between \( x = 0 \) and \( x = 8 \) using the Trapezoidal rule with \( n = 4 \) subintervals, we follow these steps: 1. **Determine the width of each subinterval (h):** \[ h = \frac{b - a}{n} = \frac{8 - 0}{4} = 2 \] 2. **Identify the function values at the endpoints and midpoints of the subintervals:** - \( f(0) = 3 \) - \( f(2) = 7 \) - \( f(4) = 11 \) - \( f(6) = 9 \) - \( f(8) = 3 \) 3. **Apply the Trapezoidal rule formula:** \[ \text{Area} \approx \frac{h}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4) \right] \] where \( x_0 = 0 \), \( x_1 = 2 \), \( x_2 = 4 \), \( x_3 = 6 \), and \( x_4 = 8 \). 4. **Substitute the function values into the formula:** \[ \text{Area} \approx \frac{2}{2} \left[ f(0) + 2f(2) + 2f(4) + 2f(6) + f(8) \right] \] \[ \text{Area} \approx 1 \left[ 3 + 2 \cdot 7 + 2 \cdot 11 + 2 \cdot 9 + 3 \right] \] \[ \text{Area} \approx 1 \left[ 3 + 14 + 22 + 18 + 3 \right] \] \[ \text{Area} \approx 1 \left[ 60 \right] \] \[ \text{Area} \approx 60 \] Therefore, the approximate area under the curve \( y = f(x) \) between \( x = 0 \) and \( x = 8 \) using the Trapezoidal rule with \( n = 4 \) subintervals is \( \boxed{60} \). |
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| Qwen2-72B-Instruct | To approximate the area under the curve y=f(x) between x=0 and x=8 using the Trapezoidal rule with n=4 subintervals, we will follow these steps: 1. Calculate the width of each subinterval: Since the interval [0, 8] is divided into 4 subintervals, the width of each subinterval (h) is given by: h = (b - a) / n h = (8 - 0) / 4 h = 2 2. Calculate the values of f(x) at the endpoints of each subinterval: We are given the values of f(x) at x=0, x=2, x=4, x=6, and x=8. 3. Apply the Trapezoidal rule formula: The formula for the Trapezoidal rule with n subintervals is: A ≈ (h/2) * [f(x0) + 2f(x1) + 2f(x2) + ... + 2f(xn-1) + f(xn)] where A is the approximate area under the curve, h is the width of each subinterval, and f(xi) represents the values of the function at the endpoints of each subinterval. 4. Substitute the values into the formula: A ≈ (2/2) * [f(0) + 2f(2) + 2f(4) + 2f(6) + f(8)] A ≈ (1) * [3 + 2(7) + 2(11) + 2(9) + 3] A ≈ 1 * [3 + 14 + 22 + 18 + 3] A ≈ 1 * 50 A ≈ 50 Therefore, the approximate area under the curve y=f(x) between x=0 and x=8 using the Trapezoidal rule with n=4 subintervals is 50 square units. |
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| baichuan4 | To approximate the area under the curve y=f(x) between x=0 and x=8 using the Trapezoidal rule with n=4 subintervals, we first need to determine the width of each subinterval. Since the total width is 8-0=8, each subinterval has a width of 8/4=2. Now, we can calculate the area of each trapezoid formed by the function values at the endpoints of each subinterval. The formula for the area of a trapezoid is (1/2)h(a+b), where h is the height (width of the subinterval) and a and b are the lengths of the parallel sides. The subintervals and corresponding function values are as follows: 1. [0, 2]: f(0)=3, f(2)=7 2. [2, 4]: f(2)=7, f(4)=11 3. [4, 6]: f(4)=11, f(6)=9 4. [6, 8]: f(6)=9, f(8)=3 Now, we calculate the area for each trapezoid: 1. Area of trapezoid 1: (1/2)(2)(3+7) = (1/2)(2)(10) = 10 2. Area of trapezoid 2: (1/2)(2)(7+11) = (1/2)(2)(18) = 18 3. Area of trapezoid 3: (1/2)(2)(11+9) = (1/2)(2)(20) = 20 4. Area of trapezoid 4: (1/2)(2)(9+3) = (1/2)(2)(12) = 12 Finally, we add up the areas of all the trapezoids to get the total area under the curve: Total area = 10 + 18 + 20 + 12 = 60 Therefore, the approximate area under the curve y=f(x) between x=0 and x=8 using the Trapezoidal rule with n=4 subintervals is 60 square units. |
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| doubao | The width of each subinterval is 8/4 = 2. The intervals are [0, 2], [2, 4], [4, 6], [6, 8]. The area using the Trapezoidal rule is: [(f(0) + f(2))/2 * 2 + (f(2) + f(4))/2 * 2 + (f(4) + f(6))/2 * 2 + (f(6) + f(8))/2 * 2] = [(3 + 7)/2 * 2 + (7 + 11)/2 * 2 + (11 + 9)/2 * 2 + (9 + 3)/2 * 2] = [10 + 18 + 20 + 12] = 60. |
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| minimax | The Trapezoidal Rule is a numerical method for approximating the definite integral of a function. It works by dividing the area under the curve into trapezoids and summing their areas. Given the function values at specific points and the requirement to use n=4 subintervals, we can calculate the approximate area under the curve from x=0 to x=8. First, we need to determine the width of each subinterval. Since we have 4 subintervals over the range from x=0 to x=8, the width (h) of each subinterval is: h = (8 - 0) / 4 = 2 Now, we can apply the Trapezoidal Rule formula: Area ≈ (h/2) * [f(x0) + 2f(x1) + 2f(x2) + ... + 2f(xn-1) + f(xn)] Here, x0, x1, x2, ..., xn are the points at which we know the function values, and n is the number of subintervals. Given the function values: f(0) = 3 f(2) = 7 f(4) = 11 f(6) = 9 f(8) = 3 We can plug these values into the formula: Area ≈ (2/2) * [f(0) + 2f(2) + 2f(4) + 2f(6) + f(8)] Area ≈ 1 * [3 + 2*7 + 2*11 + 2*9 + 3] Area ≈ 1 * [3 + 14 + 22 + 18 + 3] Area ≈ 1 * [60] Area ≈ 60 Therefore, the approximate area under the curve y=f(x) between x=0 and x=8 using the Trapezoidal Rule with n=4 subintervals is 60 square units. |
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| kimi | To approximate the area under the curve \( y = f(x) \) between \( x = 0 \) and \( x = 8 \) using the Trapezoidal rule with \( n = 4 \) subintervals, we first need to determine the width of each subinterval, \( \Delta x \), and then calculate the area of each trapezoid formed by the function values at the endpoints of each subinterval. Given: - \( f(0) = 3 \) - \( f(2) = 7 \) - \( f(4) = 11 \) - \( f(6) = 9 \) - \( f(8) = 3 \) Since we are using \( n = 4 \) subintervals, the width of each subinterval is: \[ \Delta x = \frac{b - a}{n} = \frac{8 - 0}{4} = 2 \] Now, we calculate the area of each trapezoid: 1. For the first subinterval from \( x = 0 \) to \( x = 2 \): - \( A_1 = \frac{f(0) + f(2)}{2} \cdot \Delta x = \frac{3 + 7}{2} \cdot 2 = 10 \) 2. For the second subinterval from \( x = 2 \) to \( x = 4 \): - \( A_2 = \frac{f(2) + f(4)}{2} \cdot \Delta x = \frac{7 + 11}{2} \cdot 2 = 18 \) 3. For the third subinterval from \( x = 4 \) to \( x = 6 \): - \( A_3 = \frac{f(4) + f(6)}{2} \cdot \Delta x = \frac{11 + 9}{2} \cdot 2 = 20 \) 4. For the fourth subinterval from \( x = 6 \) to \( x = 8 \): - \( A_4 = \frac{f(6) + f(8)}{2} \cdot \Delta x = \frac{9 + 3}{2} \cdot 2 = 12 \) The total area under the curve is the sum of the areas of all trapezoids: \[ \text{Total Area} = A_1 + A_2 + A_3 + A_4 = 10 + 18 + 20 + 12 = 60 \] So, the approximate area under the curve \( y = f(x) \) between \( x = 0 \) and \( x = 8 \) using the Trapezoidal rule with \( n = 4 \) subintervals is \( 60 \) square units. |
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| wenxinyiyan | The Trapezoidal Rule is a method of numerical integration, used to approximate the area under a curve. When using n subintervals, the area under the curve y=f(x) between x=a and x=b is approximated by: Area ≈ (h/2) * [f(a) + 2*f(a+h) + 2*f(a+2h) + ... + 2*f(b-h) + f(b)] where h = (b-a)/n is the width of each subinterval. For the given problem, a=0, b=8, and n=4. Thus, h = (8-0)/4 = 2. The given values of f(x) are: f(0) = 3 f(2) = 7 f(4) = 11 f(6) = 9 f(8) = 3 Using the Trapezoidal Rule formula with these values: Area ≈ (2/2) * [f(0) + 2*f(2) + 2*f(4) + 2*f(6) + f(8)] = 0.5 * [3 + 2*7 + 2*11 + 2*9 + 3] = 0.5 * [3 + 14 + 22 + 18 + 3] = 0.5 * 60 = 30 So, the approximate area under the curve y=f(x) between x=0 and x=8 using the Trapezoidal Rule with n=4 subintervals is 30 square units. |
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| zhipu | The Trapezoidal Rule is a numerical method to approximate the definite integral, which gives the area under a curve. The formula for the Trapezoidal Rule is: \[ \int_{a}^{b} f(x) \, dx \approx \frac{\Delta x}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + \ldots + 2f(x_{n-1}) + f(x_n) \right] \] where: - \( \Delta x = \frac{b-a}{n} \) is the width of each subinterval, - \( a \) and \( b \) are the lower and upper limits of integration, respectively, - \( n \) is the number of subintervals, - \( x_i \) are the endpoints of the subintervals, - \( f(x_i) \) are the values of the function at the endpoints. Given the problem, we have: - \( a = 0 \) - \( b = 8 \) - \( n = 4 \) First, we calculate \( \Delta x \): \[ \Delta x = \frac{b - a}{n} = \frac{8 - 0}{4} = 2 \] The subintervals will be: \[ [0, 2], [2, 4], [4, 6], [6, 8] \] The corresponding function values at these points are already given as: - \( f(0) = 3 \) - \( f(2) = 7 \) - \( f(4) = 11 \) - \( f(6) = 9 \) - \( f(8) = 3 \) Now, we apply the Trapezoidal Rule formula: \[ \int_{0}^{8} f(x) \, dx \approx \frac{\Delta x}{2} \left[ f(0) + 2f(2) + 2f(4) + 2f(6) + f(8) \right] \] Substitute the values: \[ \int_{0}^{8} f(x) \, dx \approx \frac{2}{2} \left[ 3 + 2(7) + 2(11) + 2(9) + 3 \right] \] Simplify the expression: \[ \int_{0}^{8} f(x) \, dx \approx 1 \left[ 3 + 14 + 22 + 18 + 3 \right] \] \[ \int_{0}^{8} f(x) \, dx \approx 1 \left[ 60 \right] \] \[ \int_{0}^{8} f(x) \, dx \approx 60 \] Therefore, the approximate area under the curve \( y = f(x) \) between \( x = 0 \) and \( x = 8 \) using the Trapezoidal Rule with \( n = 4 \) subintervals is 60 square units. |
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| gpto1 | To approximate the area under the curve \( y = f(x) \) from \( x = 0 \) to \( x = 8 \) using the Trapezoidal Rule with \( n = 4 \) subintervals, we'll follow these steps: ### **Step 1: Determine the Width of Each Subinterval (\( h \))** The width \( h \) of each subinterval is calculated using the formula: \[ h = \frac{b - a}{n} \] Where: - \( a = 0 \) (starting point) - \( b = 8 \) (ending point) - \( n = 4 \) (number of subintervals) Calculating \( h \): \[ h = \frac{8 - 0}{4} = \frac{8}{4} = 2 \] ### **Step 2: Identify the \( x \)-values and Corresponding \( f(x) \)-values** Since \( h = 2 \), the \( x \)-values at the endpoints of the subintervals are: \[ \begin{align*} x_0 &= 0 \\ x_1 &= x_0 + h = 0 + 2 = 2 \\ x_2 &= x_1 + h = 2 + 2 = 4 \\ x_3 &= x_2 + h = 4 + 2 = 6 \\ x_4 &= x_3 + h = 6 + 2 = 8 \\ \end{align*} \] Corresponding \( f(x) \)-values: \[ \begin{align*} f(x_0) &= f(0) = 3 \\ f(x_1) &= f(2) = 7 \\ f(x_2) &= f(4) = 11 \\ f(x_3) &= f(6) = 9 \\ f(x_4) &= f(8) = 3 \\ \end{align*} \] ### **Step 3: Apply the Trapezoidal Rule** The Trapezoidal Rule formula for approximating the area is: \[ \text{Area} \approx \frac{h}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4) \right] \] Plugging in the values: \[ \text{Area} \approx \frac{2}{2} \left[ f(0) + 2f(2) + 2f(4) + 2f(6) + f(8) \right] \] Simplify the fraction: \[ \text{Area} \approx 1 \times \left[ 3 + 2(7) + 2(11) + 2(9) + 3 \right] \] Compute the products: \[ \text{Area} \approx 1 \times \left[ 3 + 14 + 22 + 18 + 3 \right] \] Sum the values inside the brackets: \[ \text{Area} \approx 1 \times 60 \] \[ \text{Area} \approx 60 \] ### **Final Answer** The approximate area under the curve between \( x = 0 \) and \( x = 8 \) is **60**. |
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| deepseekv3 | To approximate the area under the curve \( y = f(x) \) between \( x = 0 \) and \( x = 8 \) using the Trapezoidal Rule with \( n = 4 \) subintervals, follow these steps: ### Given Data: - \( f(0) = 3 \) - \( f(2) = 7 \) - \( f(4) = 11 \) - \( f(6) = 9 \) - \( f(8) = 3 \) ### Step 1: Determine the width of each subinterval (\( h \)) The interval \([0, 8]\) is divided into \( n = 4 \) subintervals. The width \( h \) of each subinterval is: \[ h = \frac{b - a}{n} = \frac{8 - 0}{4} = 2 \] ### Step 2: Apply the Trapezoidal Rule formula The Trapezoidal Rule formula is: \[ \text{Area} \approx \frac{h}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4) \right] \] Here, \( x_0 = 0, x_1 = 2, x_2 = 4, x_3 = 6, \) and \( x_4 = 8 \). Substitute the values: \[ \text{Area} \approx \frac{2}{2} \left[ f(0) + 2f(2) + 2f(4) + 2f(6) + f(8) \right] \] \[ \text{Area} \approx 1 \left[ 3 + 2(7) + 2(11) + 2(9) + 3 \right] \] ### Step 3: Simplify the expression \[ \text{Area} \approx 1 \left[ 3 + 14 + 22 + 18 + 3 \right] \] \[ \text{Area} \approx 1 \left[ 60 \right] \] \[ \text{Area} \approx 60 \] ### Final Answer: The approximate area under the curve \( y = f(x) \) between \( x = 0 \) and \( x = 8 \) using the Trapezoidal Rule with \( n = 4 \) subintervals is: \[ \boxed{60} \] |
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