| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the area between the graphs of the functions \( f(x) = x^2 - 4x + 10 \) and \( g(x) = 4x - x^2 \) on the interval \( 1 \leq x \leq 3 \), we need to find the absolute difference between the two functions and then integrate that difference over the given interval. First, let's find the difference between the two functions: \[ h(x) = |f(x) - g(x)| = |(x^2 - 4x + 10) - (4x - x^2)| \] \[ h(x) = |x^2 - 4x + 10 - 4x + x^2| = |2x^2 - 8x + 10| \] Since we're dealing with absolute values, we need to consider where the expression inside the absolute value is positive or negative. However, for our purposes of finding the area between two curves, we can simply integrate the expression inside the absolute value without considering the sign, because the area is always positive. Let's integrate \( h(x) = 2x^2 - 8x + 10 \) over the interval \( 1 \leq x \leq 3 \): \[ \int_{1}^{3} (2x^2 - 8x + 10) dx \] \[ = \left[ \frac{2}{3}x^3 - 4x^2 + 10x \right]_{1}^{3} \] \[ = \left( \frac{2}{3}(3)^3 - 4(3)^2 + 10(3) \right) - \left( \frac{2}{3}(1)^3 - 4(1)^2 + 10(1) \right) \] \[ = \left( \frac{2}{3}(27) - 4(9) + 30 \right) - \left( \frac{2}{3} - 4 + 10 \right) \] \[ = \left( 18 - 36 + 30 \right) - \left( \frac{2}{3} + 6 \right) \] \[ = (12) - \left( 6 + \frac{2}{3} \right) \] \[ = 12 - 6 - \frac{2}{3} \] \[ = 6 - \frac{2}{3} \] \[ = \frac{18}{3} - \frac{2}{3} \] \[ = \frac{16}{3} \] Therefore, the area between the graphs of the functions \( f(x) = x^2 - 4x + 10 \) and \( g(x) = 4x - x^2 \) on the interval \( 1 \leq x \leq 3 \) is \( \frac{16}{3} \) square units. |
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| deepseekr1 | To find the area of the region between the graphs of \( f(x) = x^2 - 4x + 10 \) and \( g(x) = 4x - x^2 \) from \( x = 1 \) to \( x = 3 \), we first determine which function is on top within this interval. We set \( f(x) = g(x) \) to check for intersections: \[ x^2 - 4x + 10 = 4x - x^2 \] Simplifying, we get: \[ 2x^2 - 8x + 10 = 0 \] The discriminant of this quadratic equation is negative (\( D = -4 \)), indicating no real solutions. Thus, the functions do not intersect, and one is always above the other in the interval. Testing values in the interval (e.g., \( x = 2 \), \( x = 1 \), \( x = 3 \)) shows \( f(x) \) is always above \( g(x) \). The area is calculated by integrating the difference \( f(x) - g(x) \) from 1 to 3: \[ f(x) - g(x) = (x^2 - 4x + 10) - (4x - x^2) = 2x^2 - 8x + 10 \] The integral is: \[ \int_{1}^{3} (2x^2 - 8x + 10) \, dx \] Computing the antiderivative: \[ \left[ \frac{2}{3}x^3 - 4x^2 + 10x \right]_{1}^{3} \] Evaluating at the bounds: - At \( x = 3 \): \[ \frac{2}{3}(27) - 4(9) + 10(3) = 18 - 36 + 30 = 12 \] - At \( x = 1 \): \[ \frac{2}{3}(1) - 4(1) + 10(1) = \frac{2}{3} - 4 + 10 = \frac{20}{3} \] Subtracting the lower limit from the upper limit: \[ 12 - \frac{20}{3} = \frac{36}{3} - \frac{20}{3} = \frac{16}{3} \] The area of the region between the graphs is \boxed{\dfrac{16}{3}}. |
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| gpto1 | To find the area of the region between the graphs of the functions \( f(x) = x^2 - 4x + 10 \) and \( g(x) = 4x - x^2 \) over the interval \( 1 \leq x \leq 3 \), we need to compute the definite integral of the difference between the functions over that interval. **Step 1: Determine which function is on top.** First, evaluate both functions at several points in the interval to see which one is greater. For \( x = 1 \): \[ \begin{align*} f(1) &= (1)^2 - 4(1) + 10 = 1 - 4 + 10 = 7 \\ g(1) &= 4(1) - (1)^2 = 4 - 1 = 3 \end{align*} \] For \( x = 2 \): \[ \begin{align*} f(2) &= (2)^2 - 4(2) + 10 = 4 - 8 + 10 = 6 \\ g(2) &= 4(2) - (2)^2 = 8 - 4 = 4 \end{align*} \] For \( x = 3 \): \[ \begin{align*} f(3) &= (3)^2 - 4(3) + 10 = 9 - 12 + 10 = 7 \\ g(3) &= 4(3) - (3)^2 = 12 - 9 = 3 \end{align*} \] In each case, \( f(x) > g(x) \) on the interval \( [1, 3] \). Therefore, \( f(x) \) is the upper function and \( g(x) \) is the lower function over this interval. **Step 2: Compute the difference between the functions.** Find \( f(x) - g(x) \): \[ \begin{align*} f(x) - g(x) &= \left( x^2 - 4x + 10 \right) - \left( 4x - x^2 \right) \\ &= x^2 - 4x + 10 - 4x + x^2 \\ &= 2x^2 - 8x + 10 \end{align*} \] **Step 3: Set up the definite integral.** The area \( A \) between the two curves from \( x = 1 \) to \( x = 3 \) is: \[ A = \int_{1}^{3} [f(x) - g(x)]\, dx = \int_{1}^{3} (2x^2 - 8x + 10)\, dx \] **Step 4: Integrate the function.** Compute the integral: \[ \begin{align*} \int (2x^2 - 8x + 10)\, dx &= 2 \int x^2\, dx - 8 \int x\, dx + 10 \int 1\, dx \\ &= 2 \left( \frac{x^3}{3} \right) - 8 \left( \frac{x^2}{2} \right) + 10x + C \\ &= \frac{2x^3}{3} - 4x^2 + 10x + C \end{align*} \] **Step 5: Evaluate the definite integral.** Compute \( F(3) - F(1) \), where \( F(x) \) is the antiderivative: \[ \begin{align*} F(3) &= \frac{2(3)^3}{3} - 4(3)^2 + 10(3) \\ &= \frac{2(27)}{3} - 4(9) + 30 \\ &= \frac{54}{3} - 36 + 30 \\ &= 18 - 36 + 30 \\ &= 12 \\ \\ F(1) &= \frac{2(1)^3}{3} - 4(1)^2 + 10(1) \\ &= \frac{2(1)}{3} - 4(1) + 10 \\ &= \frac{2}{3} - 4 + 10 \\ &= \left( \frac{2}{3} + 6 \right) \\ &= \frac{2}{3} + \frac{18}{3} \\ &= \frac{20}{3} \end{align*} \] Subtract: \[ A = F(3) - F(1) = 12 - \frac{20}{3} = \frac{36}{3} - \frac{20}{3} = \frac{16}{3} \] **Answer:** The area between the two curves from \( x = 1 \) to \( x = 3 \) is \( \frac{16}{3} \) square units. \(\boxed{\dfrac{16}{3}}\) |
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| GPT-4o | To find the area of the region between the graphs of the functions \( f(x) = x^2 - 4x + 10 \) and \( g(x) = 4x - x^2 \) over the interval \( 1 \leq x \leq 3 \), we need to integrate the difference of the functions over this interval. First, let's determine which function is on top and which is on the bottom in the interval \( 1 \leq x \leq 3 \). We do this by comparing the values of \( f(x) \) and \( g(x) \) at several points within the interval. 1. Evaluate \( f(x) \) and \( g(x) \) at \( x = 1 \): \[ f(1) = 1^2 - 4(1) + 10 = 1 - 4 + 10 = 7 \] \[ g(1) = 4(1) - 1^2 = 4 - 1 = 3 \] 2. Evaluate \( f(x) \) and \( g(x) \) at \( x = 3 \): \[ f(3) = 3^2 - 4(3) + 10 = 9 - 12 + 10 = 7 \] \[ g(3) = 4(3) - 3^2 = 12 - 9 = 3 \] From these values, we see that \( f(x) \geq g(x) \) at both endpoints of the interval. To confirm \( f(x) \geq g(x) \) throughout the interval, we can analyze the difference: \[ f(x) - g(x) = (x^2 - 4x + 10) - (4x - x^2) = x^2 - 4x + 10 - 4x + x^2 = 2x^2 - 8x + 10 \] Since \( 2x^2 - 8x + 10 \) is a quadratic with a positive leading coefficient, it is a parabola opening upwards. The minimum value of this quadratic can be found by completing the square or using the vertex formula \( x = -\frac{b}{2a} \) where \( a = 2 \) and \( b = -8 \): \[ x = \frac{8}{4} = 2 \] \[ f(2) - g(2) = 2(2)^2 - 8(2) + 10 = 8 - 16 + 10 = 2 \] Since \( f(x) - g(x) \geq 2 \) in the interval \( 1 \leq x \leq 3 \), \( f(x) \) is always above \( g(x) \) in this interval. The area between the curves is given by the integral of the difference \( f(x) - g(x) \) from 1 to 3: \[ \text{Area} = \int_1^3 \left( f(x) - g(x) \right) \, dx = \int_1^3 \left( (x^2 - 4x + 10) - (4x - x^2) \right) \, dx \] \[ = \int_1^3 (x^2 - 4x + 10 - 4x + x^2) \, dx = \int_1^3 (2x^2 - 8x + 10) \, dx \] Now, we integrate: \[ \int_1^3 (2x^2 - 8x + 10) \, dx = \left[ \frac{2x^3}{3} - 4x^2 + 10x \right]_1^3 \] Evaluating at the endpoints: \[ \left( \frac{2(3)^3}{3} - 4(3)^2 + 10(3) \right) - \left( \frac{2(1)^3}{3} - 4(1)^2 + 10(1) \right) \] Calculate each term: \[ \left( \frac{2 \cdot 27}{3} - 4 \cdot 9 + 10 \cdot 3 \right) - \left( \frac{2 \cdot 1}{3} - 4 \cdot 1 + 10 \cdot 1 \right) \] \[ = \left( 18 - 36 + 30 \right) - \left( \frac{2 |
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| llama3.1-70B-Instruct | To find the area of the region between the graphs of the functions f(x) = x^2 - 4x + 10 and g(x) = 4x - x^2, we need to integrate the difference between the two functions over the interval [1, 3]. First, let's find the difference between the two functions: f(x) - g(x) = (x^2 - 4x + 10) - (4x - x^2) = x^2 - 4x + 10 - 4x + x^2 = 2x^2 - 8x + 10 Now, we can integrate this difference over the interval [1, 3]: ∫[1,3] (2x^2 - 8x + 10) dx = [2/3 x^3 - 4x^2 + 10x] from 1 to 3 = [2/3 (3)^3 - 4(3)^2 + 10(3)] - [2/3 (1)^3 - 4(1)^2 + 10(1)] = [2/3 (27) - 4(9) + 30] - [2/3 (1) - 4(1) + 10] = [18 - 36 + 30] - [2/3 - 4 + 10] = 12 - 2/3 + 4 - 10 = 12 - 2/3 + 4 - 10 = 12 - 2/3 - 6 = 6 - 2/3 = 16/3 Therefore, the area of the region between the graphs of the functions f(x) = x^2 - 4x + 10 and g(x) = 4x - x^2 over the interval [1, 3] is 16/3. |
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| llama3.1-405B-Instruct | To find the area of the region between the graphs of the functions f(x) = x^2 - 4x + 10 and g(x) = 4x - x^2, we need to integrate the difference between the two functions over the interval [1, 3]. First, let's find the difference between the two functions: f(x) - g(x) = (x^2 - 4x + 10) - (4x - x^2) = x^2 - 4x + 10 - 4x + x^2 = 2x^2 - 8x + 10 Now, we need to integrate this difference over the interval [1, 3]: Area = ∫[1,3] (2x^2 - 8x + 10) dx To evaluate this integral, we can use the power rule of integration and the constant multiple rule of integration: Area = ∫[1,3] 2x^2 dx - ∫[1,3] 8x dx + ∫[1,3] 10 dx = 2/3 x^3 | [1,3] - 4x^2 | [1,3] + 10x | [1,3] = 2/3 (3^3 - 1^3) - 4(3^2 - 1^2) + 10(3 - 1) = 2/3 (27 - 1) - 4(9 - 1) + 10(2) = 2/3 (26) - 4(8) + 20 = 52/3 - 32 + 20 = 52/3 - 12 = 40/3 Therefore, the area of the region between the graphs of the functions f(x) = x^2 - 4x + 10 and g(x) = 4x - x^2 over the interval [1, 3] is 40/3. |
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| DeepSeek-Coder-V2 | To find the area of the region between the graphs of the functions \( f(x) = x^2 - 4x + 10 \) and \( g(x) = 4x - x^2 \) over the interval \( 1 \leq x \leq 3 \), we need to follow these steps: 1. **Determine the points of intersection**: We need to find where \( f(x) = g(x) \). \[ x^2 - 4x + 10 = 4x - x^2 \] Combine like terms: \[ 2x^2 - 8x + 10 = 0 \] Divide the entire equation by 2: \[ x^2 - 4x + 5 = 0 \] Solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -4 \), and \( c = 5 \): \[ x = \frac{4 \pm \sqrt{16 - 20}}{2} = \frac{4 \pm \sqrt{-4}}{2} = \frac{4 \pm 2i}{2} = 2 \pm i \] Since the solutions are complex, there are no real points of intersection within the given interval. 2. **Set up the integral for the area**: Since \( f(x) \) and \( g(x) \) do not intersect within the interval, we need to determine which function is greater over the interval \( [1, 3] \). Evaluate \( f(x) \) and \( g(x) \) at the endpoints: \[ f(1) = 1^2 - 4 \cdot 1 + 10 = 7 \] \[ g(1) = 4 \cdot 1 - 1^2 = 3 \] \[ f(3) = 3^2 - 4 \cdot 3 + 10 = 7 \] \[ g(3) = 4 \cdot 3 - 3^2 = 3 \] Since \( f(x) \geq g(x) \) over the interval \( [1, 3] \), the area between the curves is given by: \[ A = \int_{1}^{3} [f(x) - g(x)] \, dx \] 3. **Evaluate the integral**: \[ f(x) - g(x) = (x^2 - 4x + 10) - (4x - x^2) = 2x^2 - 8x + 10 \] \[ A = \int_{1}^{3} (2x^2 - 8x + 10) \, dx \] Find the antiderivative of \( 2x^2 - 8x + 10 \): \[ \int (2x^2 - 8x + 10) \, dx = \frac{2}{3}x^3 - 4x^2 + 10x \] Evaluate the antiderivative from 1 to 3: \[ \left[ \frac{2}{3}x^3 - 4x^2 + 10x \right]_{1}^{3} = \left( \frac{2}{3}(3)^3 - 4(3)^2 + 10(3) \right) - \left( \frac{2}{3}(1)^3 - 4(1)^2 + 10(1) \right) \] \[ = \left( \frac{2}{3}(27) - 4(9) + 30 \right) - \left( \frac{2}{3}(1) - 4(1) + 10 \right) \] \[ = \left( 18 - 36 + 30 \right) - \left( \frac{2}{3} - 4 + 10 \right) \] \[ = 12 - \left( \frac{2}{3} + 6 \right) \] \[ = 12 - \frac{20}{3} \] \[ = \frac{36}{3} - \frac{20}{3} = \frac{16}{3} \] Thus, the area of the region between the graphs of the functions \( f(x) \) and \( g(x) \) over the interval \( 1 \leq x \leq 3 \) is \( \frac{16}{3} \). |
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| Qwen2-72B-Instruct | To find the area between the graphs of the functions \( f(x) = x^2 - 4x + 10 \) and \( g(x) = 4x - x^2 \) on the interval \( 1 \leq x \leq 3 \), we need to find the absolute difference between the two functions and then integrate that difference over the given interval. First, let's find the difference between the two functions: \[ h(x) = |f(x) - g(x)| = |(x^2 - 4x + 10) - (4x - x^2)| \] \[ h(x) = |x^2 - 4x + 10 - 4x + x^2| = |2x^2 - 8x + 10| \] Since we're dealing with absolute values, we need to consider where the expression inside the absolute value is positive or negative. However, for our purposes of finding the area between two curves, we can simply integrate the expression inside the absolute value without considering the sign, because the area is always positive. Let's integrate \( h(x) = 2x^2 - 8x + 10 \) over the interval \( 1 \leq x \leq 3 \): \[ \int_{1}^{3} (2x^2 - 8x + 10) dx \] \[ = \left[ \frac{2}{3}x^3 - 4x^2 + 10x \right]_{1}^{3} \] \[ = \left( \frac{2}{3}(3)^3 - 4(3)^2 + 10(3) \right) - \left( \frac{2}{3}(1)^3 - 4(1)^2 + 10(1) \right) \] \[ = \left( \frac{2}{3}(27) - 4(9) + 30 \right) - \left( \frac{2}{3} - 4 + 10 \right) \] \[ = \left( 18 - 36 + 30 \right) - \left( \frac{2}{3} + 6 \right) \] \[ = (12) - \left( 6 + \frac{2}{3} \right) \] \[ = 12 - 6 - \frac{2}{3} \] \[ = 6 - \frac{2}{3} \] \[ = \frac{18}{3} - \frac{2}{3} \] \[ = \frac{16}{3} \] Therefore, the area between the graphs of the functions \( f(x) = x^2 - 4x + 10 \) and \( g(x) = 4x - x^2 \) on the interval \( 1 \leq x \leq 3 \) is \( \frac{16}{3} \) square units. |
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| baichuan4 | To find the area between the graphs of the functions \(f(x) = x^2 - 4x + 10\) and \(g(x) = 4x - x^2\), we need to integrate the difference of these functions over the specified interval \(1 \leq x \leq 3\). First, let's simplify the difference of the functions: \[f(x) - g(x) = (x^2 - 4x + 10) - (4x - x^2)\] \[= x^2 - 4x + 10 - 4x + x^2\] \[= 2x^2 - 8x + 10\] Now, we integrate this expression with respect to \(x\) over the interval \([1, 3]\): \[\int_{1}^{3} (2x^2 - 8x + 10) dx\] Let's compute the integral: \[= \int_{1}^{3} 2x^2 dx - \int_{1}^{3} 8x dx + \int_{1}^{3} 10 dx\] \[= 2 \cdot \frac{1}{3}x^3 \Big|_{1}^{3} - 4x^2 \Big|_{1}^{3} + 10x \Big|_{1}^{3}\] \[= \frac{2}{3}(27 - 1) - 4(9 - 1) + 10(3 - 1)\] \[= \frac{2}{3} \cdot 26 - 4 \cdot 8 + 10 \cdot 2\] \[= \frac{52}{3} - 32 + 20\] \[= \frac{52}{3} - 12\] \[= \frac{52 - 36}{3}\] \[= \frac{16}{3}\] So, the area between the graphs of the functions \(f(x) = x^2 - 4x + 10\) and \(g(x) = 4x - x^2\) over the interval \(1 \leq x \leq 3\) is \(\frac{16}{3}\) square units. |
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| doubao | First find the points of intersection of the two functions: x^2 - 4x + 10 = 4x - x^2 2x^2 - 8x + 10 = 0 x^2 - 4x + 5 = 0 Using the quadratic formula: x = [4 ± √(16 - 4*1*5)] / (2*1) = [4 ± √(-4)] / 2, which has no real solutions. So the functions do not intersect within the given interval. The area between the curves is the integral of the absolute difference of the functions from 1 to 3. Area = ∫[1 to 3] |(x^2 - 4x + 10) - (4x - x^2)| dx = ∫[1 to 3] |2x^2 - 8x + 10| dx = ∫[1 to 3] (2x^2 - 8x + 10) dx = [2*(x^3/3) - 4*(x^2) + 10x] from 1 to 3 = [2*(3^3/3) - 4*(3^2) + 10*3] - [2*(1^3/3) - 4*(1^2) + 10*1] = [18 - 36 + 30] - [2/3 - 4 + 10] = 12 - [2/3 + 6] = 12 - 20/3 = 16/3. |
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| minimax | To find the area of the region between the graphs of the functions f(x) = x^2 - 4x + 10 and g(x) = 4x - x^2 over the interval [1, 3], we need to: 1. Find the points of intersection between the two functions within the given interval. 2. Integrate the difference between the two functions over the interval from the leftmost point of intersection to the rightmost point of intersection. First, let's find the points of intersection by setting f(x) equal to g(x): x^2 - 4x + 10 = 4x - x^2 Combine like terms: 2x^2 - 8x + 10 = 0 This is a quadratic equation, which we can solve using the quadratic formula: x = [-b ± sqrt(b^2 - 4ac)] / (2a) Here, a = 2, b = -8, and c = 10. Plugging these values into the quadratic formula gives us: x = [8 ± sqrt((-8)^2 - 4(2)(10))] / (2(2)) x = [8 ± sqrt(64 - 80)] / 4 x = [8 ± sqrt(-16)] / 4 Since the discriminant (the part under the square root) is negative, there are no real solutions to this equation. This means that the two functions do not intersect within the interval [1, 3]. Therefore, the area between the two curves over this interval is simply the integral of the difference between the functions from x = 1 to x = 3. To find the area, we integrate the absolute value of the difference between the functions: Area = ∫ from 1 to 3 of |g(x) - f(x)| dx Since g(x) is above f(x) in the given interval, we can just integrate the difference without the absolute value: Area = ∫ from 1 to 3 of (g(x) - f(x)) dx Area = ∫ from 1 to 3 of ((4x - x^2) - (x^2 - 4x + 10)) dx Area = ∫ from 1 to 3 of (8x - 2x^2 - 10) dx Now we integrate term by term: Area = [4x^2 - (2/3)x^3 - 10x] from 1 to 3 Evaluate the integral at the upper and lower limits and subtract: Area = [(4(3)^2 - (2/3)(3)^3 - 10(3)) - (4(1)^2 - (2/3)(1)^3 - 10(1))] Area = [(4(9) - (2/3)(27) - 30) - (4 - (2/3) - 10)] Area = [(36 - 18 - 30) - (4 - 2/3 - 10)] Area = [(-6) - (-6 - 2/3)] Area = -6 + 6 + 2/3 Area = 2/3 So the area of the region between the graphs of the functions f(x) = x^2 - 4x + 10 and g(x) = 4x - x^2 from x = 1 to x = 3 is 2/3 square units. |
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| kimi | To find the area between the two curves, we first need to determine where the two functions intersect within the given interval [1, 3]. We do this by setting f(x) equal to g(x): \[ f(x) = g(x) \] \[ x^2 - 4x + 10 = 4x - x^2 \] Solving for x, we combine like terms: \[ 2x^2 - 8x + 10 = 0 \] \[ x^2 - 4x + 5 = 0 \] This quadratic equation can be factored: \[ (x - 2)(x - 3) = 0 \] So the solutions are x = 2 and x = 3. Since we are only interested in the interval [1, 3], we will consider the intersection at x = 2. Now, to find the area between the curves, we need to integrate the difference between the two functions over the interval [1, 3]. The area A is given by: \[ A = \int_{1}^{3} (g(x) - f(x)) \, dx \] Substitute the functions into the integral: \[ A = \int_{1}^{3} ((4x - x^2) - (x^2 - 4x + 10)) \, dx \] \[ A = \int_{1}^{3} (4x - x^2 - x^2 + 4x - 10) \, dx \] \[ A = \int_{1}^{3} (8x - 2x^2 - 10) \, dx \] Now, integrate each term: \[ A = \left[ 4x^2 - \frac{2}{3}x^3 - 10x \right]_{1}^{3} \] Evaluate the antiderivative at the upper and lower limits of integration: \[ A = \left( 4(3)^2 - \frac{2}{3}(3)^3 - 10(3) \right) - \left( 4(1)^2 - \frac{2}{3}(1)^3 - 10(1) \right) \] \[ A = \left( 36 - 18 - 30 \right) - \left( 4 - \frac{2}{3} - 10 \right) \] \[ A = (-12) - (-8\frac{2}{3}) \] \[ A = -12 + 8\frac{2}{3} \] \[ A = -3\frac{1}{3} \] Since the area cannot be negative, we take the absolute value: \[ A = 3\frac{1}{3} \] So the area between the two curves from x = 1 to x = 3 is \(\boxed{3\frac{1}{3}}\) square units. |
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| wenxinyiyan | 为了找到函数 $f(x) = x^2 - 4x + 10$ 和 $g(x) = 4x - x^2$ 在区间 $1 \leq x \leq 3$ 之间围成的面积，我们需要计算这两个函数在这个区间上的定积分，并找出它们之间的差值。 首先，我们确定哪个函数在这个区间上位于上方。通过观察或者计算特定点的函数值，我们可以发现 $f(x)$ 在这个区间上总是大于 $g(x)$。 面积 $A$ 可以通过以下定积分来计算： $$A = \int_{1}^{3} [f(x) - g(x)] dx$$将 $f(x)$ 和 $g(x)$ 的表达式代入上式，得到： $$A = \int_{1}^{3} [(x^2 - 4x + 10) - (4x - x^2)] dx$$化简得： $$A = \int_{1}^{3} (2x^2 - 8x + 10) dx$$接下来，我们计算这个定积分。对于多项式 $2x^2 - 8x + 10$，其原函数 $F(x)$ 是： $$F(x) = \frac{2}{3}x^3 - 4x^2 + 10x$$因此，面积 $A$ 为： $$A = F(3) - F(1) = \left( \frac{2}{3} \cdot 3^3 - 4 \cdot 3^2 + 10 \cdot 3 \right) - \left( \frac{2}{3} \cdot 1^3 - 4 \cdot 1^2 + 10 \cdot 1 \right)$$计算后得到： $$A = \left( 18 - 36 + 30 \right) - \left( \frac{2}{3} - 4 + 10 \right) = \frac{28}{3}$$所以，函数 $f(x)$ 和 $g(x)$ 在区间 $1 \leq x \leq 3$ 之间围成的面积是 $\frac{28}{3}$ 平方单位。 |
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| zhipu | To find the area of the region between the graphs of the functions \( f(x) = x^2 - 4x + 10 \) and \( g(x) = 4x - x^2 \) over the interval \( 1 \leq x \leq 3 \), we need to determine which function is on top (has the greater value) over this interval and then integrate the difference between the two functions. First, let's find the points of intersection of the two functions by setting them equal to each other: \[ f(x) = g(x) \] \[ x^2 - 4x + 10 = 4x - x^2 \] Combining like terms, we get: \[ 2x^2 - 8x + 10 = 0 \] \[ x^2 - 4x + 5 = 0 \] This is a quadratic equation, but it doesn't factor nicely, so we use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For the equation \( x^2 - 4x + 5 = 0 \), \( a = 1 \), \( b = -4 \), and \( c = 5 \). Plugging these values into the quadratic formula gives: \[ x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(5)}}{2(1)} \] \[ x = \frac{4 \pm \sqrt{16 - 20}}{2} \] \[ x = \frac{4 \pm \sqrt{-4}}{2} \] Since the discriminant (\( b^2 - 4ac \)) is negative, there are no real roots, which means the functions do not intersect over the real numbers. Therefore, one function is entirely above the other over the interval \( 1 \leq x \leq 3 \). Next, we need to determine which function is on top over the interval \( 1 \leq x \leq 3 \). We can do this by evaluating each function at a point within the interval, say \( x = 2 \): \[ f(2) = 2^2 - 4(2) + 10 = 4 - 8 + 10 = 6 \] \[ g(2) = 4(2) - 2^2 = 8 - 4 = 4 \] Since \( f(2) > g(2) \), we can conclude that \( f(x) \) is the top function over the interval \( 1 \leq x \leq 3 \). Now, we integrate the difference between the two functions from \( x = 1 \) to \( x = 3 \): \[ \text{Area} = \int_{1}^{3} (f(x) - g(x)) \, dx \] \[ \text{Area} = \int_{1}^{3} ((x^2 - 4x + 10) - (4x - x^2)) \, dx \] \[ \text{Area} = \int_{1}^{3} (2x^2 - 8x + 10) \, dx \] Now we integrate term by term: \[ \int 2x^2 \, dx = \frac{2}{3}x^3 \] \[ \int -8x \, dx = -4x^2 \] \[ \int 10 \, dx = 10x \] Evaluating these from \( x = 1 \) to \( x = 3 \), we get: \[ \text{Area} = \left[ \frac{2}{3}x^3 - 4x^2 + 10x \right]_{1}^{3} \] \[ \text{Area} = \left( \frac{2}{3}(3)^3 - 4(3)^2 + 10(3) \right) - \left( \frac{2}{3}(1)^3 - 4(1)^2 + 10(1) \right) \] \[ \text{Area} = \left( \frac{2}{3}(27) - 4(9) + 30 \right) - \left( \frac{2}{3}(1) - 4(1) + 10 \right) \] \[ \text{Area} = \left( 18 - 36 + 30 \right) - \left( \frac{2}{3} - 4 + 10 \right) \] \[ \text{Area} = (12) - \left( \frac{2}{3} + 6 \right) \] \[ \text{Area} = 12 - \frac{20}{3} \] \[ \text{Area} = \frac{36}{3} - \frac{20}{3} \] \[ \text{Area} = \frac{16}{3} \] So, the area of the region between the graphs of the functions \( f(x) \) and \( g(x) \) over the interval \( 1 \leq x \leq 3 \) is \( \frac{16}{3} \) square units. |
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| gpto1 | To find the area of the region between the graphs of the functions \( f(x) = x^2 - 4x + 10 \) and \( g(x) = 4x - x^2 \) over the interval \( 1 \leq x \leq 3 \), we need to compute the definite integral of the difference between the functions over that interval. **Step 1: Determine which function is on top.** First, evaluate both functions at several points in the interval to see which one is greater. For \( x = 1 \): \[ \begin{align*} f(1) &= (1)^2 - 4(1) + 10 = 1 - 4 + 10 = 7 \\ g(1) &= 4(1) - (1)^2 = 4 - 1 = 3 \end{align*} \] For \( x = 2 \): \[ \begin{align*} f(2) &= (2)^2 - 4(2) + 10 = 4 - 8 + 10 = 6 \\ g(2) &= 4(2) - (2)^2 = 8 - 4 = 4 \end{align*} \] For \( x = 3 \): \[ \begin{align*} f(3) &= (3)^2 - 4(3) + 10 = 9 - 12 + 10 = 7 \\ g(3) &= 4(3) - (3)^2 = 12 - 9 = 3 \end{align*} \] In each case, \( f(x) > g(x) \) on the interval \( [1, 3] \). Therefore, \( f(x) \) is the upper function and \( g(x) \) is the lower function over this interval. **Step 2: Compute the difference between the functions.** Find \( f(x) - g(x) \): \[ \begin{align*} f(x) - g(x) &= \left( x^2 - 4x + 10 \right) - \left( 4x - x^2 \right) \\ &= x^2 - 4x + 10 - 4x + x^2 \\ &= 2x^2 - 8x + 10 \end{align*} \] **Step 3: Set up the definite integral.** The area \( A \) between the two curves from \( x = 1 \) to \( x = 3 \) is: \[ A = \int_{1}^{3} [f(x) - g(x)]\, dx = \int_{1}^{3} (2x^2 - 8x + 10)\, dx \] **Step 4: Integrate the function.** Compute the integral: \[ \begin{align*} \int (2x^2 - 8x + 10)\, dx &= 2 \int x^2\, dx - 8 \int x\, dx + 10 \int 1\, dx \\ &= 2 \left( \frac{x^3}{3} \right) - 8 \left( \frac{x^2}{2} \right) + 10x + C \\ &= \frac{2x^3}{3} - 4x^2 + 10x + C \end{align*} \] **Step 5: Evaluate the definite integral.** Compute \( F(3) - F(1) \), where \( F(x) \) is the antiderivative: \[ \begin{align*} F(3) &= \frac{2(3)^3}{3} - 4(3)^2 + 10(3) \\ &= \frac{2(27)}{3} - 4(9) + 30 \\ &= \frac{54}{3} - 36 + 30 \\ &= 18 - 36 + 30 \\ &= 12 \\ \\ F(1) &= \frac{2(1)^3}{3} - 4(1)^2 + 10(1) \\ &= \frac{2(1)}{3} - 4(1) + 10 \\ &= \frac{2}{3} - 4 + 10 \\ &= \left( \frac{2}{3} + 6 \right) \\ &= \frac{2}{3} + \frac{18}{3} \\ &= \frac{20}{3} \end{align*} \] Subtract: \[ A = F(3) - F(1) = 12 - \frac{20}{3} = \frac{36}{3} - \frac{20}{3} = \frac{16}{3} \] **Answer:** The area between the two curves from \( x = 1 \) to \( x = 3 \) is \( \frac{16}{3} \) square units. \(\boxed{\dfrac{16}{3}}\) |
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| deepseekv3 | server error: Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-7388056c-ddff-93f1-859a-04b517d5fe48', 'request_id': '7388056c-ddff-93f1-859a-04b517d5fe48'} | 5.333 |