| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the given double integral over the indicated rectangle \( R = [-5,4] \times [0, 3] \), we have to integrate the function \( f(x,y) = 2x - 4y^3 \) with respect to \( x \) and \( y \) over the given ranges. The double integral can be written as: \[ \iint\limits_{R}{{2x - 4{y^3}\,dA}} = \int_{-5}^{4}\int_{0}^{3}(2x - 4y^3)\,dy\,dx \] First, we integrate with respect to \( y \): \[ \int_{0}^{3}(2x - 4y^3)\,dy = \left[2xy - y^4\right]_{0}^{3} = 2x(3) - 3^4 = 6x - 81 \] Then, we integrate the result with respect to \( x \): \[ \int_{-5}^{4}(6x - 81)\,dx = \left[3x^2 - 81x\right]_{-5}^{4} = (3(4)^2 - 81(4)) - (3(-5)^2 - 81(-5)) \] \[ = (48 - 324) - (75 + 405) = -276 - 480 = -756 \] Therefore, the solution to the given double integral is \(-756\). |
-756 |
| deepseekr1 | To compute the double integral \(\iint\limits_{R}{{2x - 4{y^3}\,dA}}\) over the rectangle \(R = [-5, 4] \times [0, 3]\), we can convert the double integral into iterated integrals and compute each part separately. First, we split the integral into two parts: \[ \iint\limits_{R} (2x - 4y^3) \, dA = \iint\limits_{R} 2x \, dA - \iint\limits_{R} 4y^3 \, dA \] **First Integral:** \[ \iint\limits_{R} 2x \, dA = \int_{0}^{3} \int_{-5}^{4} 2x \, dx \, dy \] Compute the inner integral with respect to \(x\): \[ \int_{-5}^{4} 2x \, dx = \left[ x^2 \right]_{-5}^{4} = 16 - 25 = -9 \] Then integrate with respect to \(y\): \[ \int_{0}^{3} -9 \, dy = -9 \int_{0}^{3} dy = -9 \cdot 3 = -27 \] **Second Integral:** \[ \iint\limits_{R} 4y^3 \, dA = \int_{0}^{3} \int_{-5}^{4} 4y^3 \, dx \, dy \] Compute the inner integral with respect to \(x\): \[ \int_{-5}^{4} 4y^3 \, dx = 4y^3 \cdot (4 - (-5)) = 4y^3 \cdot 9 = 36y^3 \] Then integrate with respect to \(y\): \[ \int_{0}^{3} 36y^3 \, dy = 36 \left[ \frac{y^4}{4} \right]_{0}^{3} = 36 \left( \frac{81}{4} - 0 \right) = 36 \cdot \frac{81}{4} = 9 \cdot 81 = 729 \] **Combining the Results:** \[ \iint\limits_{R} (2x - 4y^3) \, dA = -27 - 729 = -756 \] Thus, the value of the double integral is \(\boxed{-756}\). |
-756 |
| gpto1 | To compute the double integral \[ \iint_R (2x - 4y^3) \, dA, \] where \( R = [-5,4] \times [0,3] \), we'll proceed by evaluating the integral step by step. **Step 1: Write the Double Integral as Iterated Integrals** Since \( R \) is a rectangle, we can write the double integral as an iterated integral: \[ \iint_R (2x - 4y^3) \, dA = \int_{y=0}^{3} \int_{x=-5}^{4} (2x - 4y^3) \, dx \, dy. \] **Step 2: Separate the Integral** Notice that the integrand \( 2x - 4y^3 \) can be split into two parts: \[ \iint_R (2x - 4y^3) \, dA = \iint_R 2x \, dA - \iint_R 4y^3 \, dA. \] We will compute these two integrals separately. **Part I: Compute \( I_1 = \iint_R 2x \, dA \)** **a. Evaluate the Inner Integral with respect to \( x \):** \[ \int_{x=-5}^{4} 2x \, dx = 2 \int_{x=-5}^{4} x \, dx = 2 \left[ \frac{1}{2} x^2 \right]_{x=-5}^{x=4} = 2 \left[ \frac{1}{2} (4^2) - \frac{1}{2} (-5)^2 \right] = 2 \left( 8 - 12.5 \right) = 2(-4.5) = -9. \] **b. Evaluate the Outer Integral with respect to \( y \):** \[ I_1 = \int_{y=0}^{3} (-9) \, dy = -9 \int_{y=0}^{3} 1 \, dy = -9 [y]_{0}^{3} = -9 (3 - 0) = -27. \] **Part II: Compute \( I_2 = \iint_R 4y^3 \, dA \)** **a. Evaluate the Inner Integral with respect to \( x \):** Since \( 4y^3 \) does not depend on \( x \), the inner integral simplifies to: \[ \int_{x=-5}^{4} 4y^3 \, dx = 4y^3 (4 - (-5)) = 4y^3 (9) = 36 y^3. \] **b. Evaluate the Outer Integral with respect to \( y \):** \[ I_2 = \int_{y=0}^{3} 36 y^3 \, dy = 36 \int_{y=0}^{3} y^3 \, dy = 36 \left[ \frac{1}{4} y^4 \right]_{0}^{3} = 36 \left( \frac{1}{4} (3^4) - 0 \right) = 36 \left( \frac{81}{4} \right) = 9 \times 81 = 729. \] **Step 3: Combine the Results** Subtract \( I_2 \) from \( I_1 \): \[ \iint_R (2x - 4y^3) \, dA = I_1 - I_2 = (-27) - 729 = -756. \] **Final Answer:** \[ \iint_R (2x - 4y^3) \, dA = -756. \] **Answer:** \(-756\) |
-756 |
| GPT-4o | To solve the double integral \(\iint\limits_{R} (2x - 4y^3) \, dA\) over the rectangle \(R = [-5, 4] \times [0, 3]\), we can proceed as follows: The region \(R\) is defined by the Cartesian product of the intervals: \[ R = \{ (x, y) \mid -5 \leq x \leq 4, \; 0 \leq y \leq 3 \} \] We can set up the double integral as: \[ \iint\limits_{R} (2x - 4y^3) \, dA = \int_{0}^{3} \int_{-5}^{4} (2x - 4y^3) \, dx \, dy \] We first integrate with respect to \(x\): \[ \int_{-5}^{4} (2x - 4y^3) \, dx \] Let's split this into two separate integrals: \[ \int_{-5}^{4} 2x \, dx - \int_{-5}^{4} 4y^3 \, dx \] First, compute \(\int_{-5}^{4} 2x \, dx\): \[ \int_{-5}^{4} 2x \, dx = 2 \left[ \frac{x^2}{2} \right]_{-5}^{4} = \left[ x^2 \right]_{-5}^{4} = 4^2 - (-5)^2 = 16 - 25 = -9 \] Next, compute \(\int_{-5}^{4} 4y^3 \, dx\). Since \(4y^3\) is a constant with respect to \(x\): \[ \int_{-5}^{4} 4y^3 \, dx = 4y^3 \int_{-5}^{4} 1 \, dx = 4y^3 (4 - (-5)) = 4y^3 \cdot 9 = 36y^3 \] Now, combining these results: \[ \int_{-5}^{4} (2x - 4y^3) \, dx = -9 - 36y^3 \] Next, we integrate this result with respect to \(y\): \[ \int_{0}^{3} (-9 - 36y^3) \, dy \] This can again be split into two integrals: \[ \int_{0}^{3} -9 \, dy - \int_{0}^{3} 36y^3 \, dy \] First, compute \(\int_{0}^{3} -9 \, dy\): \[ \int_{0}^{3} -9 \, dy = -9 \left[ y \right]_{0}^{3} = -9 (3 - 0) = -27 \] Next, compute \(\int_{0}^{3} 36y^3 \, dy\): \[ \int_{0}^{3} 36y^3 \, dy = 36 \left[ \frac{y^4}{4} \right]_{0}^{3} = 36 \left[ \frac{3^4}{4} - \frac{0^4}{4} \right] = 36 \left[ \frac{81}{4} \right] = 36 \cdot \frac{81}{4} = 9 \cdot 81 = 729 \] Combining these results: \[ \int_{0}^{3} (-9 - 36y^3) \, dy = -27 - 729 = -756 \] Therefore, the value of the double integral \(\iint\limits_{R} (2x - 4y^3) \, dA\) over the rectangle \(R = [-5, 4] \times [0, 3]\) is \(\boxed{-756}\). |
-756 |
| llama3.1-70B-Instruct | To compute the double integral \(\iint\limits_{R}{{2x - 4{y^3}\,dA}}\) over the rectangle \(R = [-5,4] \times [0, 3]\), we follow these steps: ## Step 1: Understand the Double Integral The double integral \(\iint\limits_{R}{{2x - 4{y^3}\,dA}}\) represents the volume under the surface defined by \(z = 2x - 4{y^3}\) over the rectangle \(R\). ## Step 2: Set Up the Integral To integrate over the rectangle \(R = [-5,4] \times [0, 3]\), we set up the integral as follows: \[ \int_{-5}^{4} \int_{0}^{3} (2x - 4{y^3}) \,dy\,dx \] ## Step 3: Evaluate the Inner Integral First, we integrate with respect to \(y\), treating \(x\) as a constant: \[ \int_{0}^{3} (2x - 4{y^3}) \,dy = \left[2xy - y^4\right]_{0}^{3} = (2x \cdot 3 - 3^4) - (2x \cdot 0 - 0^4) = 6x - 81 \] ## Step 4: Evaluate the Outer Integral Next, we integrate the result with respect to \(x\): \[ \int_{-5}^{4} (6x - 81) \,dx = \left[3x^2 - 81x\right]_{-5}^{4} = (3 \cdot 4^2 - 81 \cdot 4) - (3 \cdot (-5)^2 - 81 \cdot (-5)) \] \[ = (48 - 324) - (75 + 405) = -276 - 480 = -756 \] The final answer is: $\boxed{-756}$ |
-756 |
| llama3.1-405B-Instruct | To compute the double integral over the indicated rectangle, we can use the following steps: 1. Evaluate the inner integral with respect to y: $$\int_{0}^{3} (2x - 4y^3) dy$$ Using the power rule of integration, we get: $$\int_{0}^{3} (2x - 4y^3) dy = \left[2xy - y^4\right]_{0}^{3}$$ Evaluating the limits, we get: $$\int_{0}^{3} (2x - 4y^3) dy = (6x - 81) - (0 - 0) = 6x - 81$$ 2. Evaluate the outer integral with respect to x: $$\int_{-5}^{4} (6x - 81) dx$$ Using the power rule of integration, we get: $$\int_{-5}^{4} (6x - 81) dx = \left[3x^2 - 81x\right]_{-5}^{4}$$ Evaluating the limits, we get: $$\int_{-5}^{4} (6x - 81) dx = (48 - 324) - (75 + 405) = -756$$ Therefore, the value of the double integral is: $$\iint\limits_{R}{{2x - 4{y^3}\,dA}} = -756$$ |
-756 |
| DeepSeek-Coder-V2 | To solve the double integral \(\iint\limits_{R}{{2x - 4{y^3}\,dA}}\) over the rectangle \(R = [-5, 4] \times [0, 3]\), we will use Fubini's theorem, which allows us to evaluate the integral as an iterated integral. The integral can be written as: \[ \iint\limits_{R}{{2x - 4{y^3}\,dA}} = \int_{-5}^{4} \int_{0}^{3} (2x - 4y^3) \, dy \, dx \] First, we evaluate the inner integral with respect to \(y\): \[ \int_{0}^{3} (2x - 4y^3) \, dy \] We can split this into two separate integrals: \[ \int_{0}^{3} 2x \, dy - \int_{0}^{3} 4y^3 \, dy \] Evaluating each integral separately: 1. \(\int_{0}^{3} 2x \, dy\): Since \(2x\) is a constant with respect to \(y\), we can factor it out: \[ 2x \int_{0}^{3} dy = 2x [y]_{0}^{3} = 2x (3 - 0) = 6x \] 2. \(\int_{0}^{3} 4y^3 \, dy\): \[ 4 \int_{0}^{3} y^3 \, dy = 4 \left[ \frac{y^4}{4} \right]_{0}^{3} = 4 \left( \frac{3^4}{4} - \frac{0^4}{4} \right) = 4 \left( \frac{81}{4} \right) = 81 \] Combining these results, we get: \[ \int_{0}^{3} (2x - 4y^3) \, dy = 6x - 81 \] Now, we evaluate the outer integral with respect to \(x\): \[ \int_{-5}^{4} (6x - 81) \, dx \] We can split this into two separate integrals: \[ \int_{-5}^{4} 6x \, dx - \int_{-5}^{4} 81 \, dx \] Evaluating each integral separately: 1. \(\int_{-5}^{4} 6x \, dx\): \[ 6 \int_{-5}^{4} x \, dx = 6 \left[ \frac{x^2}{2} \right]_{-5}^{4} = 6 \left( \frac{4^2}{2} - \frac{(-5)^2}{2} \right) = 6 \left( \frac{16}{2} - \frac{25}{2} \right) = 6 \left( 8 - 12.5 \right) = 6 \left( -4.5 \right) = -27 \] 2. \(\int_{-5}^{4} 81 \, dx\): Since \(81\) is a constant with respect to \(x\), we can factor it out: \[ 81 \int_{-5}^{4} dx = 81 [x]_{-5}^{4} = 81 (4 - (-5)) = 81 (4 + 5) = 81 \times 9 = 729 \] Combining these results, we get: \[ \int_{-5}^{4} (6x - 81) \, dx = -27 - 729 = -756 \] Thus, the value of the double integral is: \[ \iint\limits_{R}{{2x - 4{y^3}\,dA}} = -756 \] |
-756 |
| Qwen2-72B-Instruct | To solve the given double integral over the indicated rectangle $R = [-5,4] \times [0, 3]$, we have to integrate the function $f(x,y) = 2x - 4y^3$ with respect to $x$ and $y$ over the given ranges. The double integral can be written as: \[ \iint\limits_{R}{{2x - 4{y^3}\,dA}} = \int_{-5}^{4}\int_{0}^{3}(2x - 4y^3)\,dy\,dx \] First, we integrate with respect to $y$: \[ \int_{0}^{3}(2x - 4y^3)\,dy = \left[2xy - y^4\right]_{0}^{3} = 2x(3) - 3^4 = 6x - 81 \] Then, we integrate the result with respect to $x$: \[ \int_{-5}^{4}(6x - 81)\,dx = \left[3x^2 - 81x\right]_{-5}^{4} = (3(4)^2 - 81(4)) - (3(-5)^2 - 81(-5)) \] \[ = (48 - 324) - (75 + 405) = -276 - 480 = -756 \] Therefore, the solution to the given double integral is \(-756\). |
-756 |
| baichuan4 | To compute the double integral over the rectangle $R = [-5,4] \times [0, 3]$, we need to integrate the function $f(x,y) = 2x - 4y^3$ with respect to $y$ first and then with respect to $x$. The integral can be written as: $$\iint\limits_{R} (2x - 4y^3) \, dA = \int_{-5}^{4} \int_{0}^{3} (2x - 4y^3) \, dy \, dx$$ First, let's integrate with respect to $y$: $$\int_{0}^{3} (2x - 4y^3) \, dy = 2xy - y^4 \Big|_{0}^{3} = 2x(3) - 3^4 - (2x(0) - 0^4) = 6x - 81$$ Now, integrate the result with respect to $x$: $$\int_{-5}^{4} (6x - 81) \, dx = 3x^2 - 81x \Big|_{-5}^{4} = (3(4)^2 - 81(4)) - (3(-5)^2 - 81(-5))$$ $$= 48 - 324 - (75 + 405) = 48 - 324 - 480 + 75 = -781$$ So, the value of the double integral is $-781$. |
-756 |
| doubao | The double integral is evaluated as follows: \[ \begin{align*} &\int_{-5}^{4}\int_{0}^{3}(2x-4y^3)dydx\\ =&\int_{-5}^{4}[2xy-y^4]_{0}^{3}dx\\ =&\int_{-5}^{4}(2x\times3-3^4)-(0)dx\\ =&\int_{-5}^{4}(6x-81)dx\\ =&[3x^2-81x]_{-5}^{4}\\ =&(3\times4^2-81\times4)-(3\times(-5)^2-81\times(-5))\\ =&(48-324)-(75+405)\\ =&-276-480\\ =&-756 \end{align*} \] |
-756 |
| minimax | To compute the double integral $\iint\limits_{R}{{2x - 4{y^3}\,dA}}$ over the rectangle $R = [-5,4] \times [0, 3]$, we can break it into two separate integrals, one with respect to $x$ and the other with respect to $y$. Since the limits for $y$ are constants and the limits for $x$ are constants, we can integrate each variable separately. First, we integrate with respect to $x$ from $-5$ to $4$, and then we integrate with respect to $y$ from $0$ to $3$. $$ \begin{align*} \iint\limits_{R}{{2x - 4{y^3}\,dA}} &= \int_{0}^{3} \int_{-5}^{4} (2x - 4y^3) \,dx\,dy \\ &= \int_{0}^{3} \left[ x^2 - 4y^3x \right]_{-5}^{4} dy \\ &= \int_{0}^{3} \left[ (4^2 - 4y^3 \cdot 4) - ((-5)^2 - 4y^3 \cdot (-5)) \right] dy \\ &= \int_{0}^{3} \left[ (16 - 16y^3) - (25 + 20y^3) \right] dy \\ &= \int_{0}^{3} \left[ 16 - 16y^3 - 25 - 20y^3 \right] dy \\ &= \int_{0}^{3} \left[ -9 - 36y^3 \right] dy \\ &= \left[ -9y - 9y^4 \right]_{0}^{3} \\ &= \left[ -9(3) - 9(3)^4 \right] - \left[ -9(0) - 9(0)^4 \right] \\ &= \left[ -27 - 9(81) \right] - [0] \\ &= -27 - 729 \\ &= -756. \end{align*} $$ Therefore, the value of the double integral over the rectangle $R$ is $-756$. |
-756 |
| kimi | To compute the double integral over the rectangle \( R = [-5,4] \times [0, 3] \), we first set up the integral in terms of \( x \) and \( y \): \[ \iint\limits_{R}{{2x - 4{y^3}\,dA}} = \int_{-5}^{4} \int_{0}^{3} (2x - 4y^3) \, dy \, dx \] We will integrate with respect to \( y \) first and then with respect to \( x \). 1. Integrate with respect to \( y \): \[ \int_{0}^{3} (2x - 4y^3) \, dy = \left[ 2xy - y^4 \right]_{0}^{3} = 2x(3) - 3^4 - (2x(0) - 0^4) = 6x - 81 \] 2. Now, integrate the result with respect to \( x \): \[ \int_{-5}^{4} (6x - 81) \, dx = \left[ 3x^2 - 81x \right]_{-5}^{4} = (3(4)^2 - 81(4)) - (3(-5)^2 - 81(-5)) \] 3. Calculate the values: \[ = (3(16) - 324) - (3(25) + 405) = (48 - 324) - (75 + 405) = -276 - 480 = -756 \] So, the value of the double integral is: \[ \boxed{-756} \] |
-756 |
| wenxinyiyan | 为了解决这个问题，我们首先需要理解题目中的双积分。双积分可以看作是先对一个变量积分，然后对另一个变量积分的过程。在这个问题中，给定的函数是$f(x, y) = 2x - 4y^3$，需要在矩形区域$R = [-5,4] \times [0, 3]$上进行积分。 步骤1: 首先对$x$进行积分，将$y$视为常数。积分区间为$[-5, 4]$。 $\int_{x=-5}^{4}{(2x - 4y^3)}\,dx = \left[ x^2 - 4y^3x \right]_{-5}^{4} = 16 - 16y^3 - (25 + 20y^3) = -9 - 36y^3$步骤2: 接下来，我们对上一步得到的结果关于$y$在$[0, 3]$区间上进行积分。 $\int_{y=0}^{3}{(-9 - 36y^3)}\,dy = \left[ -9y - 9y^4 \right]_{0}^{3} = -9 \times 3 - 9 \times 3^4 = -27 - 9 \times 81 = -27 - 729 = -756$所以，$\iint\limits_{R}{{2x - 4{y^3}\,dA}} = -756$。 |
-756 |
| zhipu | To compute the double integral \(\iint\limits_{R}{{2x - 4{y^3}\,dA}}\) over the rectangle \(R = [-5,4] \times [0, 3]\), we will set up the integral with the appropriate bounds and then evaluate it step-by-step. The region \(R\) is defined by \(x\) ranging from \(-5\) to \(4\) and \(y\) ranging from \(0\) to \(3\). Therefore, the double integral can be written as: \[ \iint\limits_{R} (2x - 4y^3) \, dA = \int_{-5}^{4} \int_{0}^{3} (2x - 4y^3) \, dy \, dx \] We start by integrating with respect to \(y\): \[ \int_{-5}^{4} \left( \int_{0}^{3} (2x - 4y^3) \, dy \right) dx \] First, evaluate the inner integral \(\int_{0}^{3} (2x - 4y^3) \, dy\): \[ \int_{0}^{3} (2x - 4y^3) \, dy = \int_{0}^{3} 2x \, dy - \int_{0}^{3} 4y^3 \, dy \] Since \(2x\) is a constant with respect to \(y\), we can pull it out of the integral: \[ \int_{0}^{3} 2x \, dy = 2x \int_{0}^{3} \, dy = 2x [y]_{0}^{3} = 2x (3 - 0) = 6x \] Next, evaluate \(\int_{0}^{3} 4y^3 \, dy\): \[ \int_{0}^{3} 4y^3 \, dy = 4 \int_{0}^{3} y^3 \, dy = 4 \left[ \frac{y^4}{4} \right]_{0}^{3} = \left[ y^4 \right]_{0}^{3} = 3^4 - 0^4 = 81 \] Putting these results together: \[ \int_{0}^{3} (2x - 4y^3) \, dy = 6x - 81 \] Now, we need to integrate this result with respect to \(x\): \[ \int_{-5}^{4} (6x - 81) \, dx \] Split this into two separate integrals: \[ \int_{-5}^{4} 6x \, dx - \int_{-5}^{4} 81 \, dx \] Evaluate each integral separately: \[ \int_{-5}^{4} 6x \, dx = 6 \int_{-5}^{4} x \, dx = 6 \left[ \frac{x^2}{2} \right]_{-5}^{4} = 6 \left( \frac{4^2}{2} - \frac{(-5)^2}{2} \right) = 6 \left( \frac{16}{2} - \frac{25}{2} \right) = 6 \left( 8 - 12.5 \right) = 6 \left( -4.5 \right) = -27 \] \[ \int_{-5}^{4} 81 \, dx = 81 \int_{-5}^{4} \, dx = 81 [x]_{-5}^{4} = 81 (4 - (-5)) = 81 (4 + 5) = 81 \times 9 = 729 \] Combine the results: \[ \int_{-5}^{4} (6x - 81) \, dx = -27 - 729 = -756 \] Therefore, the value of the double integral is: \[ \boxed{-756} \] |
-756 |
| gpto1 | To compute the double integral \[ \iint_R (2x - 4y^3) \, dA, \] where \( R = [-5,4] \times [0,3] \), we'll proceed by evaluating the integral step by step. **Step 1: Write the Double Integral as Iterated Integrals** Since \( R \) is a rectangle, we can write the double integral as an iterated integral: \[ \iint_R (2x - 4y^3) \, dA = \int_{y=0}^{3} \int_{x=-5}^{4} (2x - 4y^3) \, dx \, dy. \] **Step 2: Separate the Integral** Notice that the integrand \( 2x - 4y^3 \) can be split into two parts: \[ \iint_R (2x - 4y^3) \, dA = \iint_R 2x \, dA - \iint_R 4y^3 \, dA. \] We will compute these two integrals separately. **Part I: Compute \( I_1 = \iint_R 2x \, dA \)** **a. Evaluate the Inner Integral with respect to \( x \):** \[ \int_{x=-5}^{4} 2x \, dx = 2 \int_{x=-5}^{4} x \, dx = 2 \left[ \frac{1}{2} x^2 \right]_{x=-5}^{x=4} = 2 \left[ \frac{1}{2} (4^2) - \frac{1}{2} (-5)^2 \right] = 2 \left( 8 - 12.5 \right) = 2(-4.5) = -9. \] **b. Evaluate the Outer Integral with respect to \( y \):** \[ I_1 = \int_{y=0}^{3} (-9) \, dy = -9 \int_{y=0}^{3} 1 \, dy = -9 [y]_{0}^{3} = -9 (3 - 0) = -27. \] **Part II: Compute \( I_2 = \iint_R 4y^3 \, dA \)** **a. Evaluate the Inner Integral with respect to \( x \):** Since \( 4y^3 \) does not depend on \( x \), the inner integral simplifies to: \[ \int_{x=-5}^{4} 4y^3 \, dx = 4y^3 (4 - (-5)) = 4y^3 (9) = 36 y^3. \] **b. Evaluate the Outer Integral with respect to \( y \):** \[ I_2 = \int_{y=0}^{3} 36 y^3 \, dy = 36 \int_{y=0}^{3} y^3 \, dy = 36 \left[ \frac{1}{4} y^4 \right]_{0}^{3} = 36 \left( \frac{1}{4} (3^4) - 0 \right) = 36 \left( \frac{81}{4} \right) = 9 \times 81 = 729. \] **Step 3: Combine the Results** Subtract \( I_2 \) from \( I_1 \): \[ \iint_R (2x - 4y^3) \, dA = I_1 - I_2 = (-27) - 729 = -756. \] **Final Answer:** \[ \iint_R (2x - 4y^3) \, dA = -756. \] **Answer:** \(-756\) |
-756 |
| deepseekv3 | To compute the double integral \(\iint\limits_{R}{2x - 4y^3\,dA}\) over the rectangle \(R = [-5, 4] \times [0, 3]\), we proceed as follows: The integral can be expressed as an iterated integral: \[ \iint\limits_{R}{2x - 4y^3\,dA} = \int_{-5}^{4} \int_{0}^{3} (2x - 4y^3) \, dy \, dx \] ### Step 1: Integrate with respect to \(y\) (inner integral) First, integrate the integrand \(2x - 4y^3\) with respect to \(y\) over the interval \([0, 3]\): \[ \int_{0}^{3} (2x - 4y^3) \, dy \] Since \(2x\) is treated as a constant when integrating with respect to \(y\), we get: \[ \int_{0}^{3} 2x \, dy - \int_{0}^{3} 4y^3 \, dy = 2x \cdot y \Big|_{0}^{3} - 4 \cdot \frac{y^4}{4} \Big|_{0}^{3} \] Evaluate the limits: \[ 2x \cdot (3 - 0) - \left(y^4 \Big|_{0}^{3}\right) = 6x - (3^4 - 0) = 6x - 81 \] Thus, the inner integral evaluates to: \[ 6x - 81 \] ### Step 2: Integrate with respect to \(x\) (outer integral) Now integrate \(6x - 81\) with respect to \(x\) over the interval \([-5, 4]\): \[ \int_{-5}^{4} (6x - 81) \, dx \] We compute this as: \[ 6 \cdot \frac{x^2}{2} \Big|_{-5}^{4} - 81 \cdot x \Big|_{-5}^{4} \] Simplify and evaluate the limits: \[ 3x^2 \Big|_{-5}^{4} - 81x \Big|_{-5}^{4} = 3(4^2 - (-5)^2) - 81(4 - (-5)) \] Calculate the values: \[ 3(16 - 25) - 81(9) = 3(-9) - 729 = -27 - 729 = -756 \] ### Final Answer: The value of the double integral is: \[ \boxed{-756} \] |
-756 |