| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the given double integral over the rectangle $D$, we need to integrate the function $f(x, y) = xy^2$ over the region defined by $0 \leq x \leq 2$ and $0 \leq y \leq 1$. We can set up the integral as follows: \[ \iint_D xy^2 \,dA = \int_0^1 \int_0^2 xy^2 \,dx\,dy \] First, we integrate with respect to $x$: \[ \int_0^2 xy^2 \,dx = \left. \frac{1}{2}x^2y^2 \right|_0^2 = \frac{1}{2} \cdot 2^2 \cdot y^2 = 2y^2 \] Now, we integrate the result with respect to $y$: \[ \int_0^1 2y^2 \,dy = \left. \frac{2}{3}y^3 \right|_0^1 = \frac{2}{3} \cdot 1^3 = \frac{2}{3} \] Therefore, the solution to the integral $\iint_D xy^2 \,dA$ over the rectangle $D$ is $\boxed{\frac{2}{3}}$. |
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| deepseekr1 | To compute the integral \(\iint_D xy^2 \, dA\) where \(D\) is the rectangle defined by \(0 \leq x \leq 2\) and \(0 \leq y \leq 1\), we can use Fubini's theorem to express the double integral as an iterated integral. First, we integrate with respect to \(x\): \[ \int_0^2 x \, dx = \left[ \frac{1}{2} x^2 \right]_0^2 = \frac{1}{2} (2^2 - 0^2) = \frac{1}{2} \cdot 4 = 2 \] Next, we integrate with respect to \(y\): \[ \int_0^1 y^2 \, dy = \left[ \frac{1}{3} y^3 \right]_0^1 = \frac{1}{3} (1^3 - 0^3) = \frac{1}{3} \] Multiplying the results of these two integrals gives the value of the double integral: \[ 2 \cdot \frac{1}{3} = \frac{2}{3} \] Thus, the value of the integral is \(\boxed{\dfrac{2}{3}}\). |
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| gpto1 | To compute the double integral \(\iint_D xy^2 \, dA\) over the rectangle \(D\) defined by \(0 \leq x \leq 2\) and \(0 \leq y \leq 1\), we'll integrate with respect to \(y\) first and then \(x\): **Step 1: Set up the integral** The integral becomes: \[ \iint_D xy^2 \, dA = \int_{x=0}^2 \int_{y=0}^1 xy^2 \, dy \, dx \] **Step 2: Integrate with respect to \(y\)** Compute the inner integral: \[ \int_{y=0}^1 xy^2 \, dy = x \int_{y=0}^1 y^2 \, dy = x \left[ \frac{y^3}{3} \right]_0^1 = x \left( \frac{1^3}{3} - \frac{0^3}{3} \right) = x \left( \frac{1}{3} \right) = \frac{x}{3} \] **Step 3: Integrate with respect to \(x\)** Now, compute the outer integral: \[ \int_{x=0}^2 \frac{x}{3} \, dx = \frac{1}{3} \int_{x=0}^2 x \, dx = \frac{1}{3} \left[ \frac{x^2}{2} \right]_0^2 = \frac{1}{3} \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = \frac{1}{3} \left( \frac{4}{2} \right) = \frac{1}{3} (2) = \frac{2}{3} \] **Conclusion:** The value of the integral is \(\dfrac{2}{3}\). **Answer:** \(\displaystyle \frac{2}{3}\) |
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| GPT-4o | To compute the integral \(\iint_D xy^2 \, dA\), where \(D\) is the rectangle defined by \(0 \leq x \leq 2\) and \(0 \leq y \leq 1\), we can set up the double integral in the following way: \[ \iint_D xy^2 \, dA = \int_{0}^{2} \int_{0}^{1} xy^2 \, dy \, dx \] Let's first integrate with respect to \(y\): \[ \int_{0}^{1} xy^2 \, dy \] Since \(x\) is treated as a constant in this inner integral, we can factor it out: \[ x \int_{0}^{1} y^2 \, dy \] Now, compute the integral of \(y^2\) with respect to \(y\): \[ \int_{0}^{1} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \] So the inner integral becomes: \[ x \cdot \frac{1}{3} = \frac{x}{3} \] Now, we need to integrate this result with respect to \(x\): \[ \int_{0}^{2} \frac{x}{3} \, dx \] We can factor out the constant \(\frac{1}{3}\): \[ \frac{1}{3} \int_{0}^{2} x \, dx \] Next, compute the integral of \(x\) with respect to \(x\): \[ \int_{0}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{2} = \frac{2^2}{2} - \frac{0^2}{2} = \frac{4}{2} - 0 = 2 \] So the outer integral becomes: \[ \frac{1}{3} \cdot 2 = \frac{2}{3} \] Therefore, the value of the integral \(\iint_D xy^2 \, dA\) is: \[ \boxed{\frac{2}{3}} \] |
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| llama3.1-70B-Instruct | To compute the integral $\iint_D xy^2 dA$, where $D$ is the rectangle defined by $0 \leq x \leq 2$ and $0 \leq y \leq 1$, we can use the following steps: ## Step 1: Define the integral The integral is given by $\iint_D xy^2 dA$, where $D$ is the rectangle defined by $0 \leq x \leq 2$ and $0 \leq y \leq 1$. ## Step 2: Evaluate the inner integral First, we evaluate the inner integral with respect to $y$. We have $\int_0^1 xy^2 dy = \frac{1}{3}xy^3 \Big|_0^1 = \frac{1}{3}x$. ## Step 3: Evaluate the outer integral Next, we evaluate the outer integral with respect to $x$. We have $\int_0^2 \frac{1}{3}x dx = \frac{1}{6}x^2 \Big|_0^2 = \frac{1}{6}(2^2) = \frac{2}{3}$. The final answer is: $\boxed{\frac{2}{3}}$ |
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| llama3.1-405B-Instruct | To compute the integral $\iint_D xy^2 dA$, where $D$ is the rectangle defined by $0 \leq x \leq 2$ and $0 \leq y \leq 1$, we can use the following steps: 1. Evaluate the inner integral with respect to $y$: $$\int_0^1 xy^2 dy = \left[\frac{xy^3}{3}\right]_0^1 = \frac{x}{3}$$ 2. Evaluate the outer integral with respect to $x$: $$\int_0^2 \frac{x}{3} dx = \left[\frac{x^2}{6}\right]_0^2 = \frac{4}{6} = \frac{2}{3}$$ Therefore, the value of the integral is $\frac{2}{3}$. |
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| DeepSeek-Coder-V2 | To compute the integral \(\iint_D xy^2 dA\), where \(D\) is the rectangle defined by \(0 \leq x \leq 2\) and \(0 \leq y \leq 1\), we can follow these steps: 1. **Set up the integral:** The region \(D\) is defined by the bounds \(0 \leq x \leq 2\) and \(0 \leq y \leq 1\). Therefore, the integral can be written as: \[ \iint_D xy^2 dA = \int_{0}^{2} \int_{0}^{1} xy^2 \, dy \, dx \] 2. **Evaluate the inner integral with respect to \(y\):** We first integrate \(xy^2\) with respect to \(y\): \[ \int_{0}^{1} xy^2 \, dy \] Since \(x\) is treated as a constant with respect to \(y\), we can factor \(x\) out of the integral: \[ x \int_{0}^{1} y^2 \, dy \] Now, integrate \(y^2\) with respect to \(y\): \[ \int_{0}^{1} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \] Therefore, the inner integral becomes: \[ x \cdot \frac{1}{3} = \frac{x}{3} \] 3. **Evaluate the outer integral with respect to \(x\):** Now, we need to integrate \(\frac{x}{3}\) with respect to \(x\) from 0 to 2: \[ \int_{0}^{2} \frac{x}{3} \, dx \] Factor out the constant \(\frac{1}{3}\): \[ \frac{1}{3} \int_{0}^{2} x \, dx \] Integrate \(x\) with respect to \(x\): \[ \int_{0}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{2} = \frac{2^2}{2} - \frac{0^2}{2} = \frac{4}{2} = 2 \] Therefore, the outer integral becomes: \[ \frac{1}{3} \cdot 2 = \frac{2}{3} \] 4. **Final result:** The value of the integral \(\iint_D xy^2 dA\) is: \[ \boxed{\frac{2}{3}} \] |
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| Qwen2-72B-Instruct | To solve the given double integral over the rectangle $D$, we need to integrate the function $f(x, y) = xy^2$ over the region defined by $0 \leq x \leq 2$ and $0 \leq y \leq 1$. We can set up the integral as follows: \[ \iint_D xy^2 \,dA = \int_0^1 \int_0^2 xy^2 \,dx\,dy \] First, we integrate with respect to $x$: \[ \int_0^2 xy^2 \,dx = \left. \frac{1}{2}x^2y^2 \right|_0^2 = \frac{1}{2} \cdot 2^2 \cdot y^2 = 2y^2 \] Now, we integrate the result with respect to $y$: \[ \int_0^1 2y^2 \,dy = \left. \frac{2}{3}y^3 \right|_0^1 = \frac{2}{3} \cdot 1^3 = \frac{2}{3} \] Therefore, the solution to the integral $\iint_D xy^2 \,dA$ over the rectangle $D$ is $\boxed{\frac{2}{3}}$. |
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| baichuan4 | To compute the integral $\iint_D xy^2 dA$, we need to perform a double integral over the region $D$, which is the rectangle defined by $0 \leq x \leq 2$ and $0 \leq y \leq 1$. Since the region $D$ is a rectangle, we can choose to integrate either with respect to $x$ first or with respect to $y$ first. Let's choose to integrate with respect to $y$ first. The inner integral (with respect to $y$) is: $$\int_{0}^{1} xy^2 dy$$ Integrating this with respect to $y$, we get: $$\left[\frac{xy^3}{3}\right]_{0}^{1} = \frac{x}{3}$$ Now, we integrate the result with respect to $x$ over the interval $[0, 2]$: $$\int_{0}^{2} \frac{x}{3} dx$$ Integrating this with respect to $x$, we get: $$\left[\frac{x^2}{6}\right]_{0}^{2} = \frac{4}{6} = \frac{2}{3}$$ Therefore, the value of the double integral $\iint_D xy^2 dA$ is $\frac{2}{3}$. |
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| doubao | The integral is evaluated as follows: $\iint_D xy^2 dA = \int_0^2 \int_0^1 xy^2 dy dx$. First integrate with respect to $y$: $\int_0^1 xy^2 dy = x[\frac{y^3}{3}]_0^1 = x(\frac{1}{3} - 0) = \frac{x}{3}$. Then integrate the result with respect to $x$: $\int_0^2 \frac{x}{3} dx = \frac{1}{3}[\frac{x^2}{2}]_0^2 = \frac{1}{3}(\frac{4}{2} - 0) = \frac{2}{3}$. |
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| minimax | To compute the integral $\iint_D xy^2 dA$ over the rectangle $D$ defined by $0 \leq x \leq 2$ and $0 \leq y \leq 1$, we can use the fact that the limits of integration are constants and the integrand is a product of functions of $x$ and $y$. Since the limits for $x$ and $y$ are constants, we can integrate each variable separately. This is possible because the integral over a rectangle is equivalent to iterated integrals, and the order of integration does not matter due to Fubini's theorem (assuming the integrand is continuous over the region $D$). The integral can be computed as follows: $$ \iint_D xy^2 dA = \int_{0}^{2} \int_{0}^{1} xy^2 dy dx $$ First, we integrate with respect to $y$: $$ \int_{0}^{1} xy^2 dy = x \int_{0}^{1} y^2 dy = x \left[ \frac{y^3}{3} \right]_{0}^{1} = x \left( \frac{1^3}{3} - \frac{0^3}{3} \right) = \frac{x}{3} $$ Now, we integrate the result with respect to $x$: $$ \int_{0}^{2} \frac{x}{3} dx = \frac{1}{3} \int_{0}^{2} x dx = \frac{1}{3} \left[ \frac{x^2}{2} \right]_{0}^{2} = \frac{1}{3} \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = \frac{1}{3} \cdot 2 = \frac{2}{3} $$ Therefore, the value of the integral is: $$ \iint_D xy^2 dA = \frac{2}{3} $$ |
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| kimi | To solve the integral $\iint_D xy^2 \, dA$ over the rectangle $D$ defined by $0 \leq x \leq 2$ and $0 \leq y \leq 1$, we can set up the double integral as follows: $$\iint_D xy^2 \, dA = \int_{0}^{2} \int_{0}^{1} xy^2 \, dy \, dx.$$ We'll first integrate with respect to $y$ and then with respect to $x$. 1. Integrate with respect to $y$: $$\int_{0}^{1} xy^2 \, dy = x \int_{0}^{1} y^2 \, dy = x \left[ \frac{y^3}{3} \right]_{0}^{1} = x \left( \frac{1^3}{3} - \frac{0^3}{3} \right) = \frac{x}{3}.$$ 2. Now integrate the result with respect to $x$: $$\int_{0}^{2} \frac{x}{3} \, dx = \frac{1}{3} \int_{0}^{2} x \, dx = \frac{1}{3} \left[ \frac{x^2}{2} \right]_{0}^{2} = \frac{1}{3} \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = \frac{1}{3} \left( 2 \right) = \frac{2}{3}.$$ So the solution to the integral $\iint_D xy^2 \, dA$ is $\boxed{\frac{2}{3}}$. |
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| wenxinyiyan | 首先，我们确定积分的区域D，它是0 <= x <= 2和0 <= y <= 1围成的矩形。 然后，我们将给定的函数f(x, y) = xy^2在区域D上进行积分。 使用二重积分的计算公式，我们可以将f(x, y)在D上的积分表示为： ∬D xy^2 dA = ∫[0,2] ∫[0,1] xy^2 dy dx 首先，我们对y进行积分，将x视为常数： ∫[0,1] xy^2 dy = x * ∫[0,1] y^2 dy = x * [y^3/3] [0,1] = x * (1^3/3 - 0^3/3) = x/3 然后，我们将上述结果对x进行积分： ∫[0,2] x/3 dx = 1/3 * ∫[0,2] x dx = 1/3 * [x^2/2] [0,2] = 1/3 * (2^2/2 - 0^2/2) = 2/3 所以，积分的结果是2/3。 |
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| zhipu | To compute the double integral \(\iint_D xy^2 \, dA\) over the rectangle \(D\) defined by \(0 \leq x \leq 2\) and \(0 \leq y \leq 1\), we can set up the integral with the given bounds and integrate with respect to \(y\) first and then \(x\). Here's the step-by-step solution: 1. **Set up the double integral:** \[ \iint_D xy^2 \, dA = \int_{x=0}^{2} \int_{y=0}^{1} xy^2 \, dy \, dx \] 2. **Integrate with respect to \(y\):** \[ \int_{y=0}^{1} xy^2 \, dy \] Since \(x\) is a constant with respect to \(y\), we can pull it out of the integral: \[ x \int_{y=0}^{1} y^2 \, dy \] Now, compute the integral of \(y^2\): \[ \int_{y=0}^{1} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \] So, we have: \[ x \cdot \frac{1}{3} = \frac{x}{3} \] 3. **Integrate with respect to \(x\):** \[ \int_{x=0}^{2} \frac{x}{3} \, dx \] Factor out the constant \(\frac{1}{3}\): \[ \frac{1}{3} \int_{x=0}^{2} x \, dx \] Now, compute the integral of \(x\): \[ \int_{x=0}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{2} = \frac{2^2}{2} - \frac{0^2}{2} = \frac{4}{2} = 2 \] So, we have: \[ \frac{1}{3} \cdot 2 = \frac{2}{3} \] Therefore, the value of the double integral \(\iint_D xy^2 \, dA\) over the rectangle \(D\) is: \[ \boxed{\frac{2}{3}} \] |
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| gpto1 | To compute the double integral \(\iint_D xy^2 \, dA\) over the rectangle \(D\) defined by \(0 \leq x \leq 2\) and \(0 \leq y \leq 1\), we'll integrate with respect to \(y\) first and then \(x\): **Step 1: Set up the integral** The integral becomes: \[ \iint_D xy^2 \, dA = \int_{x=0}^2 \int_{y=0}^1 xy^2 \, dy \, dx \] **Step 2: Integrate with respect to \(y\)** Compute the inner integral: \[ \int_{y=0}^1 xy^2 \, dy = x \int_{y=0}^1 y^2 \, dy = x \left[ \frac{y^3}{3} \right]_0^1 = x \left( \frac{1^3}{3} - \frac{0^3}{3} \right) = x \left( \frac{1}{3} \right) = \frac{x}{3} \] **Step 3: Integrate with respect to \(x\)** Now, compute the outer integral: \[ \int_{x=0}^2 \frac{x}{3} \, dx = \frac{1}{3} \int_{x=0}^2 x \, dx = \frac{1}{3} \left[ \frac{x^2}{2} \right]_0^2 = \frac{1}{3} \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = \frac{1}{3} \left( \frac{4}{2} \right) = \frac{1}{3} (2) = \frac{2}{3} \] **Conclusion:** The value of the integral is \(\dfrac{2}{3}\). **Answer:** \(\displaystyle \frac{2}{3}\) |
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| deepseekv3 | To compute the double integral \(\iint_D xy^2 \, dA\) over the rectangle \(D\) defined by \(0 \leq x \leq 2\) and \(0 \leq y \leq 1\), follow these steps: ### Step 1: Set Up the Integral The rectangle \(D\) allows us to express the double integral as an iterated integral. We can integrate with respect to \(y\) first and then \(x\), or vice versa. Here, we'll integrate with respect to \(y\) first and then \(x\): \[ \iint_D xy^2 \, dA = \int_{x=0}^{2} \left( \int_{y=0}^{1} xy^2 \, dy \right) dx \] ### Step 2: Integrate with Respect to \(y\) First, integrate \(xy^2\) with respect to \(y\) from \(0\) to \(1\): \[ \int_{y=0}^{1} xy^2 \, dy = x \int_{y=0}^{1} y^2 \, dy \] Compute the integral of \(y^2\): \[ \int_{y=0}^{1} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \] Thus, the inner integral becomes: \[ x \cdot \frac{1}{3} = \frac{x}{3} \] ### Step 3: Integrate with Respect to \(x\) Now, integrate \(\frac{x}{3}\) with respect to \(x\) from \(0\) to \(2\): \[ \int_{x=0}^{2} \frac{x}{3} \, dx = \frac{1}{3} \int_{x=0}^{2} x \, dx \] Compute the integral of \(x\): \[ \int_{x=0}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{2} = \frac{2^2}{2} - \frac{0^2}{2} = \frac{4}{2} = 2 \] Thus, the outer integral becomes: \[ \frac{1}{3} \cdot 2 = \frac{2}{3} \] ### Final Answer \[ \iint_D xy^2 \, dA = \boxed{\dfrac{2}{3}} \] |
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