| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To approximate the area under the curve \( y = 2^x \) between \( x = -1 \) and \( x = 3 \) using the Trapezoidal rule with \( n = 4 \) subintervals, we need to divide the interval \([-1, 3]\) into 4 equal subintervals. The width of each subinterval (\(\Delta x\)) is given by: \[ \Delta x = \frac{b - a}{n} \] \[ \Delta x = \frac{3 - (-1)}{4} \] \[ \Delta x = \frac{4}{4} \] \[ \Delta x = 1 \] Now, we can apply the Trapezoidal rule formula: \[ \text{Area} \approx \frac{\Delta x}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4) \right] \] Let's calculate the function values at the endpoints of each subinterval: \[ f(x_0) = f(-1) = 2^{-1} = 0.5 \] \[ f(x_1) = f(0) = 2^0 = 1 \] \[ f(x_2) = f(1) = 2^1 = 2 \] \[ f(x_3) = f(2) = 2^2 = 4 \] \[ f(x_4) = f(3) = 2^3 = 8 \] Plugging in these values into the formula: \[ \text{Area} \approx \frac{1}{2} \left[ 0.5 + 2(1) + 2(2) + 2(4) + 8 \right] \] \[ \text{Area} \approx \frac{1}{2} \left[ 0.5 + 2 + 4 + 8 + 8 \right] \] \[ \text{Area} \approx \frac{1}{2} \left[ 22.5 \right] \] \[ \text{Area} \approx 11.25 \] Therefore, the approximate area under the curve \( y = 2^x \) between \( x = -1 \) and \( x = 3 \) using the Trapezoidal rule with 4 subintervals is \( 11.25 \) square units. |
11.25 |
| deepseekr1 | To approximate the area under the curve \( y = 2^x \) between \( x = -1 \) and \( x = 3 \) using the Trapezoidal rule with \( n = 4 \) subintervals, follow these steps: 1. **Determine the width of each subinterval (\( \Delta x \))**: \[ \Delta x = \frac{b - a}{n} = \frac{3 - (-1)}{4} = \frac{4}{4} = 1 \] 2. **Identify the x-values**: \[ x_0 = -1,\ x_1 = 0,\ x_2 = 1,\ x_3 = 2,\ x_4 = 3 \] 3. **Calculate the function values at these points**: \[ \begin{align*} f(x_0) &= 2^{-1} = 0.5 \\ f(x_1) &= 2^{0} = 1 \\ f(x_2) &= 2^{1} = 2 \\ f(x_3) &= 2^{2} = 4 \\ f(x_4) &= 2^{3} = 8 \\ \end{align*} \] 4. **Apply the Trapezoidal rule formula**: \[ \text{Area} = \frac{\Delta x}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4) \right] \] Plugging in the values: \[ \text{Area} = \frac{1}{2} \left[ 0.5 + 2(1) + 2(2) + 2(4) + 8 \right] \] Simplify the expression inside the brackets: \[ 0.5 + 2 + 4 + 8 + 8 = 22.5 \] Multiply by \( \frac{1}{2} \): \[ \text{Area} = \frac{1}{2} \times 22.5 = 11.25 \] **Final Answer**: \[ \boxed{\frac{45}{4}} \quad \text{or} \quad \boxed{11.25} \] |
11.25 |
| gpto1 | To approximate the area under the curve \( y = 2^x \) between \( x = -1 \) and \( x = 3 \) 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 given by: \[ h = \frac{b - a}{n} \] Where \( a = -1 \) and \( b = 3 \). \[ h = \frac{3 - (-1)}{4} = \frac{4}{4} = 1 \] **Step 2: Identify the x-values at the endpoints of the subintervals** Starting from \( x_0 = a = -1 \), we find each subsequent \( x_i \) by adding \( h \): \[ \begin{align*} x_0 &= -1 \\ x_1 &= x_0 + h = -1 + 1 = 0 \\ x_2 &= x_0 + 2h = -1 + 2(1) = 1 \\ x_3 &= x_0 + 3h = -1 + 3(1) = 2 \\ x_4 &= x_0 + 4h = -1 + 4(1) = 3 \\ \end{align*} \] So the x-values are \( -1, 0, 1, 2, 3 \). **Step 3: Calculate the corresponding y-values using \( y = 2^x \)** \[ \begin{align*} f(x_0) &= 2^{-1} = \frac{1}{2} = 0.5 \\ f(x_1) &= 2^0 = 1 \\ f(x_2) &= 2^1 = 2 \\ f(x_3) &= 2^2 = 4 \\ f(x_4) &= 2^3 = 8 \\ \end{align*} \] **Step 4: Apply the Trapezoidal Rule formula** The Trapezoidal Rule approximation is: \[ \int_{a}^{b} f(x) \, dx \approx \frac{h}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4) \right] \] Plugging in the calculated values: \[ \text{Approximate area} = \frac{1}{2} \left[ 0.5 + 2(1) + 2(2) + 2(4) + 8 \right] \] **Step 5: Simplify the expression** Compute the sum inside the brackets: \[ \begin{align*} \text{Sum} &= 0.5 + 2(1) + 2(2) + 2(4) + 8 \\ &= 0.5 + 2 + 4 + 8 + 8 \\ &= (0.5 + 2) + (4 + 8 + 8) \\ &= 2.5 + 20 \\ &= 22.5 \\ \end{align*} \] **Step 6: Calculate the approximate area** \[ \text{Approximate area} = \frac{1}{2} \times 22.5 = 11.25 \] **Final Answer:** The approximate area under the curve \( y = 2^x \) from \( x = -1 \) to \( x = 3 \) using the Trapezoidal Rule with \( n = 4 \) subintervals is **11.25**. --- **Therefore, the estimated area is 11.25 square units.** |
11.25 |
| GPT-4o | To approximate the area under the curve \( y = 2^x \) between \( x = -1 \) and \( x = 3 \) using the Trapezoidal rule with \( n = 4 \) subintervals, follow these steps: ### Step-by-Step Solution: 1. **Determine the width of each subinterval (\( h \)):** \[ h = \frac{b - a}{n} = \frac{3 - (-1)}{4} = \frac{4}{4} = 1 \] 2. **Identify the x-values at the endpoints of each subinterval:** \[ x_0 = -1, \quad x_1 = 0, \quad x_2 = 1, \quad x_3 = 2, \quad x_4 = 3 \] 3. **Evaluate the function \( y = 2^x \) at each of these x-values:** \[ f(x_0) = 2^{-1} = \frac{1}{2} = 0.5 \] \[ f(x_1) = 2^0 = 1 \] \[ f(x_2) = 2^1 = 2 \] \[ f(x_3) = 2^2 = 4 \] \[ f(x_4) = 2^3 = 8 \] 4. **Apply the Trapezoidal rule formula:** \[ \text{Area} \approx \frac{h}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4) \right] \] Substituting the values we obtained: \[ \text{Area} \approx \frac{1}{2} \left[ 0.5 + 2(1) + 2(2) + 2(4) + 8 \right] \] 5. **Simplify the expression inside the brackets:** \[ 0.5 + 2(1) + 2(2) + 2(4) + 8 = 0.5 + 2 + 4 + 8 + 8 = 22.5 \] 6. **Multiply by \(\frac{h}{2}\):** \[ \text{Area} \approx \frac{1}{2} \times 22.5 = 11.25 \] ### Final Answer: The approximate area under the curve \( y = 2^x \) between \( x = -1 \) and \( x = 3 \) using the Trapezoidal rule with \( n = 4 \) subintervals is \( 11.25 \). |
11.25 |
| llama3.1-70B-Instruct | To approximate the area under the curve y=2^{x} between x=-1 and x=3 using the Trapezoidal rule with n=4 subintervals, we can follow these steps: ## Step 1: Define the function and the limits of integration The function is y=2^{x} and the limits of integration are x=-1 and x=3. ## Step 2: Determine the width of each subinterval The width of each subinterval is given by (b-a)/n, where b is the upper limit of integration (3), a is the lower limit of integration (-1), and n is the number of subintervals (4). Therefore, the width of each subinterval is (3-(-1))/4 = 4/4 = 1. ## Step 3: Calculate the x-values for each subinterval The x-values for each subinterval are given by x_{i} = a + i\*h, where i is the index of the subinterval (0, 1, 2, 3), a is the lower limit of integration (-1), and h is the width of each subinterval (1). Therefore, the x-values for each subinterval are: x_{0} = -1 + 0\*1 = -1 x_{1} = -1 + 1\*1 = 0 x_{2} = -1 + 2\*1 = 1 x_{3} = -1 + 3\*1 = 2 x_{4} = -1 + 4\*1 = 3 ## Step 4: Calculate the corresponding y-values for each subinterval The y-values for each subinterval are given by y_{i} = 2^{x_{i}}. Therefore, the y-values for each subinterval are: y_{0} = 2^{-1} = 0.5 y_{1} = 2^{0} = 1 y_{2} = 2^{1} = 2 y_{3} = 2^{2} = 4 y_{4} = 2^{3} = 8 ## Step 5: Apply the Trapezoidal rule The Trapezoidal rule states that the area under the curve is approximately equal to the sum of the areas of the trapezoids formed by the subintervals. The area of each trapezoid is given by (h/2)\*(y_{i} + y_{i+1}), where h is the width of each subinterval (1). Therefore, the area under the curve is approximately equal to: A ≈ (1/2)\*(0.5 + 1) + (1/2)\*(1 + 2) + (1/2)\*(2 + 4) + (1/2)\*(4 + 8) A ≈ 0.75 + 1.5 + 3 + 6 A ≈ 11.25 The final answer is: $\boxed{11.25}$ |
11.25 |
| llama3.1-405B-Instruct | To approximate the area under the curve y=2^x between x=-1 and x=3 using the Trapezoidal rule with n=4 subintervals, we can follow these steps: 1. Divide the interval [-1, 3] into 4 equal subintervals: x_0 = -1, x_1 = 0, x_2 = 1, x_3 = 2, x_4 = 3 2. Calculate the width of each subinterval: Δx = (x_4 - x_0) / n = (3 - (-1)) / 4 = 1 3. Calculate the function values at each subinterval point: y_0 = 2^(-1) = 0.5 y_1 = 2^0 = 1 y_2 = 2^1 = 2 y_3 = 2^2 = 4 y_4 = 2^3 = 8 4. Apply the Trapezoidal rule formula: Area ≈ (Δx/2) \* (y_0 + 2\*y_1 + 2\*y_2 + 2\*y_3 + y_4) = (1/2) \* (0.5 + 2\*1 + 2\*2 + 2\*4 + 8) = (1/2) \* (0.5 + 2 + 4 + 8 + 8) = (1/2) \* 22.5 = 11.25 Therefore, the approximate area under the curve y=2^x between x=-1 and x=3 using the Trapezoidal rule with n=4 subintervals is 11.25. |
11.25 |
| DeepSeek-Coder-V2 | To approximate the area under the curve \( y = 2^x \) between \( x = -1 \) and \( x = 3 \) using the Trapezoidal rule with \( n = 4 \) subintervals, follow these steps: 1. **Determine the width of each subinterval (h):** \[ h = \frac{b - a}{n} = \frac{3 - (-1)}{4} = \frac{4}{4} = 1 \] 2. **Identify the x-values at the endpoints and the midpoints of the subintervals:** \[ x_0 = -1, \quad x_1 = 0, \quad x_2 = 1, \quad x_3 = 2, \quad x_4 = 3 \] 3. **Calculate the corresponding y-values (i.e., \( 2^x \) at each x-value):** \[ y_0 = 2^{-1} = \frac{1}{2}, \quad y_1 = 2^0 = 1, \quad y_2 = 2^1 = 2, \quad y_3 = 2^2 = 4, \quad y_4 = 2^3 = 8 \] 4. **Apply the Trapezoidal rule formula:** \[ \text{Area} \approx \frac{h}{2} \left( y_0 + 2(y_1 + y_2 + y_3) + y_4 \right) \] Substituting the values: \[ \text{Area} \approx \frac{1}{2} \left( \frac{1}{2} + 2(1 + 2 + 4) + 8 \right) \] Simplify inside the parentheses: \[ \text{Area} \approx \frac{1}{2} \left( \frac{1}{2} + 2 \cdot 7 + 8 \right) = \frac{1}{2} \left( \frac{1}{2} + 14 + 8 \right) = \frac{1}{2} \left( \frac{1}{2} + 22 \right) = \frac{1}{2} \left( 22.5 \right) = 11.25 \] Therefore, the approximate area under the curve \( y = 2^x \) between \( x = -1 \) and \( x = 3 \) using the Trapezoidal rule with \( n = 4 \) subintervals is: \[ \boxed{11.25} \] |
11.25 |
| Qwen2-72B-Instruct | To approximate the area under the curve y = 2^x between x = -1 and x = 3 using the Trapezoidal rule with n = 4 subintervals, we need to divide the interval [-1, 3] into 4 equal subintervals. The width of each subinterval (Δx) is given by: Δx = (b - a) / n Δx = (3 - (-1)) / 4 Δx = 4 / 4 Δx = 1 Now, we can apply the Trapezoidal rule formula: Area ≈ (Δx / 2) * [f(x0) + 2f(x1) + 2f(x2) + 2f(x3) + f(x4)] Let's calculate the function values at the endpoints of each subinterval: f(x0) = f(-1) = 2^(-1) = 0.5 f(x1) = f(0) = 2^(0) = 1 f(x2) = f(1) = 2^(1) = 2 f(x3) = f(2) = 2^(2) = 4 f(x4) = f(3) = 2^(3) = 8 Plugging in these values into the formula: Area ≈ (1 / 2) * [0.5 + 2(1) + 2(2) + 2(4) + 8] Area ≈ (1 / 2) * [0.5 + 2 + 4 + 8 + 8] Area ≈ (1 / 2) * [22.5] Area ≈ 11.25 Therefore, the approximate area under the curve y = 2^x between x = -1 and x = 3 using the Trapezoidal rule with 4 subintervals is 11.25 square units. |
11.25 |
| baichuan4 | To approximate the area under the curve y=2^{x} between x=-1 and x=3 using the Trapezoidal rule with n=4 subintervals, we first need to determine the width of each subinterval, which is given by: Width of each subinterval (Δx) = (b - a) / n Where 'b' is the upper limit of x (3 in this case), 'a' is the lower limit of x (-1 in this case), and 'n' is the number of subintervals (4 in this case). Δx = (3 - (-1)) / 4 Δx = 4 / 4 Δx = 1 Now, we need to calculate the function values at the endpoints of each subinterval. The subintervals are: [-1, 0], [0, 1], [1, 2], [2, 3] We calculate the function values at the endpoints of each subinterval: f(-1) = 2^{-1} = 1/2 f(0) = 2^{0} = 1 f(1) = 2^{1} = 2 f(2) = 2^{2} = 4 f(3) = 2^{3} = 8 Next, we use the Trapezoidal rule formula to approximate the area under the curve: Area ≈ (Δx / 2) * [f(a) + 2f(a+Δx) + 2f(a+2Δx) + ... + 2f(b-Δx) + f(b)] Substituting the values we calculated earlier: Area ≈ (1 / 2) * [1/2 + 2*1 + 2*2 + 2*4 + 8] Area ≈ (1 / 2) * [1/2 + 2 + 4 + 8 + 8] Area ≈ (1 / 2) * [23.5] Area ≈ 11.75 Therefore, the approximate area under the curve y=2^{x} between x=-1 and x=3 using the Trapezoidal rule with n=4 subintervals is 11.75 square units. |
11.25 |
| doubao | The Trapezoidal rule formula is: ∫[a,b] f(x) dx ≈ (b - a) * [f(a) + 2(f(x1) + f(x2) +... + f(xn-1)) + f(b)] / (2n) Here, a = -1, b = 3, and n = 4. The width of each subinterval is (3 - (-1)) / 4 = 1. The x-values of the subintervals are -1, 0, 1, 2, 3. Evaluating the function at these points: f(-1) = 2^{-1} = 1/2; f(0) = 2^0 = 1; f(1) = 2^1 = 2; f(2) = 2^2 = 4; f(3) = 2^3 = 8. Applying the Trapezoidal rule: ∫[-1,3] 2^x dx ≈ (3 - (-1)) * [(1/2) + 2(1 + 2 + 4) + 8] / (2 * 4) = 4 * [(1/2) + 2(7) + 8] / 8 = 4 * [(1/2) + 14 + 8] / 8 = 4 * [(1/2) + 22] / 8 = 4 * (45/2) / 8 = 45 / 4. |
11.25 |
| minimax | To approximate the area under the curve \( y = 2^x \) between \( x = -1 \) and \( x = 3 \) using the Trapezoidal rule with \( n = 4 \) subintervals, we first need to determine the width of each subinterval and then apply the Trapezoidal rule formula. The width of each subinterval, \( \Delta x \), is given by the total interval length divided by the number of subintervals: \[ \Delta x = \frac{b - a}{n} \] where \( a = -1 \), \( b = 3 \), and \( n = 4 \). So, \[ \Delta x = \frac{3 - (-1)}{4} = \frac{4}{4} = 1 \] Now, we need to evaluate the function \( y = 2^x \) at the endpoints of each subinterval. The subintervals are: \[ [-1, 0], [0, 1], [1, 2], [2, 3] \] We evaluate \( y = 2^x \) at \( x = -1, 0, 1, 2, 3 \): \[ y(-1) = 2^{-1} = \frac{1}{2} \] \[ y(0) = 2^{0} = 1 \] \[ y(1) = 2^{1} = 2 \] \[ y(2) = 2^{2} = 4 \] \[ y(3) = 2^{3} = 8 \] The Trapezoidal rule formula for approximating the area under the curve is: \[ A \approx \frac{\Delta x}{2} [y_0 + 2(y_1 + y_2 + \ldots + y_{n-1}) + y_n] \] where \( y_0, y_1, \ldots, y_n \) are the function values at the endpoints of the subintervals. Plugging in the values we have: \[ A \approx \frac{1}{2} \left[ \frac{1}{2} + 2(1 + 2 + 4) + 8 \right] \] \[ A \approx \frac{1}{2} \left[ \frac{1}{2} + 2(7) + 8 \right] \] \[ A \approx \frac{1}{2} \left[ \frac{1}{2} + 14 + 8 \right] \] \[ A \approx \frac{1}{2} \left[ \frac{1}{2} + 22 \right] \] \[ A \approx \frac{1}{2} \left[ 22.5 \right] \] \[ A \approx 11.25 \] Therefore, the approximate area under the curve \( y = 2^x \) between \( x = -1 \) and \( x = 3 \) using the Trapezoidal rule with \( n = 4 \) subintervals is 11.25 square units. |
11.25 |
| kimi | The Trapezoidal rule is a numerical method for approximating the definite integral of a function. It works by dividing the interval [a, b] into n equal subintervals and then approximating the area under the curve by the sum of the areas of trapezoids formed by the function values at the endpoints of the subintervals. Given the function y = 2^x, the interval [a, b] = [-1, 3], and n = 4 subintervals, we can find the approximate area under the curve using the following steps: 1. Determine the width of each subinterval (Δx): Δx = (b - a) / n = (3 - (-1)) / 4 = 4 / 4 = 1 2. Calculate the x-values of the endpoints of each subinterval: x0 = -1, x1 = 0, x2 = 1, x3 = 2, x4 = 3 3. Evaluate the function y = 2^x at each of these x-values: y0 = 2^(-1) = 0.5 y1 = 2^0 = 1 y2 = 2^1 = 2 y3 = 2^2 = 4 y4 = 2^3 = 8 4. Apply the Trapezoidal rule formula: Area ≈ (Δx / 2) * (y0 + 2*(y1 + y2 + y3) + y4) 5. Substitute the values into the formula: Area ≈ (1 / 2) * (0.5 + 2*(1 + 2 + 4) + 8) Area ≈ (1 / 2) * (0.5 + 2*7 + 8) Area ≈ (1 / 2) * (0.5 + 14 + 8) Area ≈ (1 / 2) * (22.5) Area ≈ 11.25 So, the approximate area under the curve y = 2^x between x = -1 and x = 3 using the Trapezoidal rule with n = 4 subintervals is 11.25. |
11.25 |
| wenxinyiyan | To approximate the area under the curve \(y = 2^x\) between \(x = -1\) and \(x = 3\) using the Trapezoidal rule with \(n = 4\) subintervals, we follow these steps: Step 1: Determine the subinterval width. The width of each subinterval, \(h\), is given by: \[ h = \frac{b - a}{n} \] where \(a = -1\) is the lower limit, \(b = 3\) is the upper limit, and \(n = 4\) is the number of subintervals. \[ h = \frac{3 - (-1)}{4} = \frac{4}{4} = 1 \] Step 2: Calculate the function values at the endpoints of each subinterval. The subintervals are \([-1, 0]\), \([0, 1]\), \([1, 2]\), and \([2, 3]\). We evaluate \(y = 2^x\) at the endpoints of each subinterval: * \(x = -1\): \(y = 2^{-1} = \frac{1}{2}\) * \(x = 0\): \(y = 2^0 = 1\) * \(x = 1\): \(y = 2^1 = 2\) * \(x = 2\): \(y = 2^2 = 4\) * \(x = 3\): \(y = 2^3 = 8\) Step 3: Apply the Trapezoidal rule. The Trapezoidal rule for approximating the area under a curve is given by: \[ \text{Area} \approx \frac{h}{2} \left( y_0 + 2y_1 + 2y_2 + \ldots + 2y_{n-1} + y_n \right) \] where \(y_0, y_1, \ldots, y_n\) are the function values at the endpoints of each subinterval. Using the function values calculated in Step 2, we have: \[ \text{Area} \approx \frac{1}{2} \left( \frac{1}{2} + 2 \times 1 + 2 \times 2 + 2 \times 4 + 8 \right) \] \[ \text{Area} \approx \frac{1}{2} \left( \frac{1}{2} + 2 + 4 + 8 + 8 \right) \] \[ \text{Area} \approx \frac{1}{2} \times 22.5 \] \[ \text{Area} \approx 11.25 \] Therefore, using the Trapezoidal rule with \(n = 4\) subintervals, the approximate area under the curve \(y = 2^x\) between \(x = -1\) and \(x = 3\) is 11.25. |
11.25 |
| zhipu | To approximate the area under the curve \( y = 2^x \) between \( x = -1 \) and \( x = 3 \) using the Trapezoidal Rule with \( n = 4 \) subintervals, we follow these steps: ### Step 1: Determine the width of each subinterval The total interval is from \( x = -1 \) to \( x = 3 \), which gives a range of \( 3 - (-1) = 4 \). With \( n = 4 \) subintervals, the width \( \Delta x \) of each subinterval is: \[ \Delta x = \frac{4}{4} = 1 \] ### Step 2: Find the x-values for the subintervals The x-values at the endpoints of the subintervals are: \[ x_0 = -1, \quad x_1 = 0, \quad x_2 = 1, \quad x_3 = 2, \quad x_4 = 3 \] ### Step 3: Calculate the y-values at these x-values Using the function \( y = 2^x \), we find: \[ y_0 = 2^{-1} = \frac{1}{2} \] \[ y_1 = 2^0 = 1 \] \[ y_2 = 2^1 = 2 \] \[ y_3 = 2^2 = 4 \] \[ y_4 = 2^3 = 8 \] ### Step 4: Apply the Trapezoidal Rule formula The Trapezoidal Rule formula for approximating the area under the curve is: \[ \text{Area} \approx \frac{\Delta x}{2} \left( y_0 + 2(y_1 + y_2 + y_3) + y_4 \right) \] Plugging in the values we have: \[ \text{Area} \approx \frac{1}{2} \left( \frac{1}{2} + 2(1 + 2 + 4) + 8 \right) \] \[ \text{Area} \approx \frac{1}{2} \left( \frac{1}{2} + 2 \times 7 + 8 \right) \] \[ \text{Area} \approx \frac{1}{2} \left( \frac{1}{2} + 14 + 8 \right) \] \[ \text{Area} \approx \frac{1}{2} \left( \frac{1}{2} + 22 \right) \] \[ \text{Area} \approx \frac{1}{2} \left( 22.5 \right) \] \[ \text{Area} \approx 11.25 \] ### Conclusion The approximate area under the curve \( y = 2^x \) between \( x = -1 \) and \( x = 3 \) using the Trapezoidal Rule with \( n = 4 \) subintervals is \( 11.25 \). |
11.25 |
| gpto1 | To approximate the area under the curve \( y = 2^x \) between \( x = -1 \) and \( x = 3 \) 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 given by: \[ h = \frac{b - a}{n} \] Where \( a = -1 \) and \( b = 3 \). \[ h = \frac{3 - (-1)}{4} = \frac{4}{4} = 1 \] **Step 2: Identify the x-values at the endpoints of the subintervals** Starting from \( x_0 = a = -1 \), we find each subsequent \( x_i \) by adding \( h \): \[ \begin{align*} x_0 &= -1 \\ x_1 &= x_0 + h = -1 + 1 = 0 \\ x_2 &= x_0 + 2h = -1 + 2(1) = 1 \\ x_3 &= x_0 + 3h = -1 + 3(1) = 2 \\ x_4 &= x_0 + 4h = -1 + 4(1) = 3 \\ \end{align*} \] So the x-values are \( -1, 0, 1, 2, 3 \). **Step 3: Calculate the corresponding y-values using \( y = 2^x \)** \[ \begin{align*} f(x_0) &= 2^{-1} = \frac{1}{2} = 0.5 \\ f(x_1) &= 2^0 = 1 \\ f(x_2) &= 2^1 = 2 \\ f(x_3) &= 2^2 = 4 \\ f(x_4) &= 2^3 = 8 \\ \end{align*} \] **Step 4: Apply the Trapezoidal Rule formula** The Trapezoidal Rule approximation is: \[ \int_{a}^{b} f(x) \, dx \approx \frac{h}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4) \right] \] Plugging in the calculated values: \[ \text{Approximate area} = \frac{1}{2} \left[ 0.5 + 2(1) + 2(2) + 2(4) + 8 \right] \] **Step 5: Simplify the expression** Compute the sum inside the brackets: \[ \begin{align*} \text{Sum} &= 0.5 + 2(1) + 2(2) + 2(4) + 8 \\ &= 0.5 + 2 + 4 + 8 + 8 \\ &= (0.5 + 2) + (4 + 8 + 8) \\ &= 2.5 + 20 \\ &= 22.5 \\ \end{align*} \] **Step 6: Calculate the approximate area** \[ \text{Approximate area} = \frac{1}{2} \times 22.5 = 11.25 \] **Final Answer:** The approximate area under the curve \( y = 2^x \) from \( x = -1 \) to \( x = 3 \) using the Trapezoidal Rule with \( n = 4 \) subintervals is **11.25**. --- **Therefore, the estimated area is 11.25 square units.** |
11.25 |
| deepseekv3 | To approximate the area under the curve \( y = 2^x \) between \( x = -1 \) and \( x = 3 \) using the **Trapezoidal Rule** with \( n = 4 \) subintervals, follow these steps: --- ### **Step 1: Determine the width of each subinterval (\( \Delta x \))** The interval from \( x = -1 \) to \( x = 3 \) has a total width of: \[ \Delta x = \frac{b - a}{n} = \frac{3 - (-1)}{4} = \frac{4}{4} = 1 \] --- ### **Step 2: Identify the endpoints of the subintervals** The subintervals are: \[ x_0 = -1, \quad x_1 = 0, \quad x_2 = 1, \quad x_3 = 2, \quad x_4 = 3 \] --- ### **Step 3: Evaluate the function at each endpoint** Compute \( y = 2^x \) at each \( x \): \[ \begin{align*} y_0 &= 2^{-1} = \frac{1}{2} \\ y_1 &= 2^{0} = 1 \\ y_2 &= 2^{1} = 2 \\ y_3 &= 2^{2} = 4 \\ y_4 &= 2^{3} = 8 \\ \end{align*} \] --- ### **Step 4: Apply the Trapezoidal Rule formula** The Trapezoidal Rule formula is: \[ \text{Area} \approx \frac{\Delta x}{2} \left( y_0 + 2y_1 + 2y_2 + 2y_3 + y_4 \right) \] Substitute the values: \[ \text{Area} \approx \frac{1}{2} \left( \frac{1}{2} + 2(1) + 2(2) + 2(4) + 8 \right) \] Simplify: \[ \text{Area} \approx \frac{1}{2} \left( \frac{1}{2} + 2 + 4 + 8 + 8 \right) = \frac{1}{2} \left( 22.5 \right) = 11.25 \] --- ### **Final Answer** The approximate area under the curve \( y = 2^x \) between \( x = -1 \) and \( x = 3 \) using the Trapezoidal Rule with \( n = 4 \) subintervals is: \[ \boxed{11.25} \] |
11.25 |