| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the definite integral \(\int_{-3}^1 (7x^2 + x +1)dx\), we need to find the antiderivative of the integrand \(7x^2 + x +1\) and then apply the Fundamental Theorem of Calculus. The antiderivative \(F(x)\) of \(7x^2 + x +1\) is found by integrating each term separately: \[ \int (7x^2 + x +1)dx = \frac{7}{3}x^3 + \frac{1}{2}x^2 + x + C \] where \(C\) is the constant of integration. However, since we are dealing with a definite integral, the constant \(C\) will not affect the final result. Now, we apply the Fundamental Theorem of Calculus, which states that: \[ \int_{a}^{b} f(x)dx = F(b) - F(a) \] where \(F(x)\) is the antiderivative of \(f(x)\). Thus, \[ \int_{-3}^1 (7x^2 + x +1)dx = \left. \left( \frac{7}{3}x^3 + \frac{1}{2}x^2 + x \right) \right|_{-3}^{1} \] \[ = \left( \frac{7}{3}(1)^3 + \frac{1}{2}(1)^2 + (1) \right) - \left( \frac{7}{3}(-3)^3 + \frac{1}{2}(-3)^2 + (-3) \right) \] \[ = \left( \frac{7}{3} + \frac{1}{2} + 1 \right) - \left( \frac{7}{3}(-27) + \frac{1}{2}(9) - 3 \right) \] \[ = \left( \frac{7}{3} + \frac{1}{2} + 1 \right) - \left( -63 + \frac{9}{2} - 3 \right) \] \[ = \left( \frac{7}{3} + \frac{1}{2} + 1 \right) + \left( 63 - \frac{9}{2} + 3 \right) \] \[ = \left( \frac{7}{3} + \frac{1}{2} + 1 \right) + \left( 66 - \frac{9}{2} \right) \] \[ = \frac{7}{3} + \frac{1}{2} + 1 + 66 - \frac{9}{2} \] \[ = \frac{7}{3} + 67 - 4 \] \[ = \frac{7}{3} + 63 \] \[ = 63 + \frac{7}{3} \] \[ = 63 + 2\frac{1}{3} \] \[ = 65\frac{1}{3} \] Therefore, the solution to the definite integral \(\int_{-3}^1 (7x^2 + x +1)dx\) is \(65\frac{1}{3}\) or \(\frac{196}{3}\). |
65.333 |
| deepseekr1 | To compute the definite integral \(\int_{-3}^1 (7x^2 + x + 1) \, dx\), we first find the antiderivative of each term in the integrand: 1. The antiderivative of \(7x^2\) is \(\frac{7}{3}x^3\). 2. The antiderivative of \(x\) is \(\frac{1}{2}x^2\). 3. The antiderivative of \(1\) is \(x\). Combining these, the antiderivative \(F(x)\) is: \[ F(x) = \frac{7}{3}x^3 + \frac{1}{2}x^2 + x \] Next, we evaluate \(F(x)\) at the upper limit \(1\) and the lower limit \(-3\): 1. Evaluating \(F(1)\): \[ F(1) = \frac{7}{3}(1)^3 + \frac{1}{2}(1)^2 + 1 = \frac{7}{3} + \frac{1}{2} + 1 = \frac{14}{6} + \frac{3}{6} + \frac{6}{6} = \frac{23}{6} \] 2. Evaluating \(F(-3)\): \[ F(-3) = \frac{7}{3}(-3)^3 + \frac{1}{2}(-3)^2 + (-3) = \frac{7}{3}(-27) + \frac{1}{2}(9) - 3 = -63 + \frac{9}{2} - 3 = -\frac{126}{2} + \frac{9}{2} - \frac{6}{2} = -\frac{123}{2} \] Subtracting \(F(-3)\) from \(F(1)\): \[ F(1) - F(-3) = \frac{23}{6} - \left(-\frac{123}{2}\right) = \frac{23}{6} + \frac{123}{2} = \frac{23}{6} + \frac{369}{6} = \frac{392}{6} = \frac{196}{3} \] Thus, the value of the integral is \(\boxed{\dfrac{196}{3}}\). |
65.333 |
| gpto1 | To evaluate the definite integral: \[ \int_{-3}^{1} (7x^2 + x + 1) \, dx, \] we first find the indefinite integral (antiderivative) of the integrand: 1. **Find the antiderivative:** \[ \int (7x^2 + x + 1) \, dx = \frac{7}{3}x^3 + \frac{1}{2}x^2 + x + C, \] where \( C \) is the constant of integration. 2. **Evaluate the antiderivative at the upper and lower limits:** - **At \( x = 1 \):** \[ F(1) = \frac{7}{3}(1)^3 + \frac{1}{2}(1)^2 + 1 = \frac{7}{3} + \frac{1}{2} + 1. \] - **At \( x = -3 \):** \[ F(-3) = \frac{7}{3}(-3)^3 + \frac{1}{2}(-3)^2 + (-3) = \frac{7}{3}(-27) + \frac{1}{2}(9) - 3. \] Simplify \( F(-3) \): \[ \begin{align*} F(-3) &= \left( -63 \right) + \left( \frac{9}{2} \right) - 3 \\ &= -66 + \frac{9}{2}. \end{align*} \] 3. **Compute the definite integral:** \[ \begin{align*} \int_{-3}^{1} (7x^2 + x + 1) \, dx &= F(1) - F(-3) \\ &= \left( \frac{7}{3} + \frac{1}{2} + 1 \right) - \left( -66 + \frac{9}{2} \right) \\ &= \left( \frac{7}{3} + \frac{1}{2} + 1 \right) + 66 - \frac{9}{2}. \end{align*} \] 4. **Combine like terms:** First, find a common denominator for the fractions, which is 6. - Convert fractions: \[ \frac{7}{3} = \frac{14}{6}, \quad \frac{1}{2} = \frac{3}{6}, \quad 1 = \frac{6}{6}, \quad \frac{9}{2} = \frac{27}{6}. \] - Sum the fractions in \( F(1) \): \[ \frac{14}{6} + \frac{3}{6} + \frac{6}{6} = \frac{23}{6}. \] - Now compute: \[ \int_{-3}^{1} (7x^2 + x + 1) \, dx = \left( \frac{23}{6} \right) + 66 - \left( \frac{27}{6} \right). \] 5. **Simplify the expression:** \[ \begin{align*} \left( \frac{23}{6} + 66 \right) - \left( \frac{27}{6} \right) &= \left( \frac{23}{6} + \frac{396}{6} \right) - \frac{27}{6} \\ &= \frac{23 + 396}{6} - \frac{27}{6} \\ &= \frac{419}{6} - \frac{27}{6} \\ &= \frac{419 - 27}{6} \\ &= \frac{392}{6}. \end{align*} \] 6. **Simplify the final fraction:** \[ \frac{392}{6} = \frac{196}{3}. \] **Therefore, the value of the integral is**: \[ \int_{-3}^{1} (7x^2 + x + 1) \, dx = \frac{196}{3}. \] |
65.333 |
| GPT-4o | To solve the definite integral \(\int_{-3}^1 (7x^2 + x + 1) \, dx\), we need to follow these steps: 1. **Find the antiderivative** of the integrand \(7x^2 + x + 1\). 2. **Evaluate the antiderivative** at the upper and lower bounds of the integral. 3. **Subtract** the value of the antiderivative at the lower bound from its value at the upper bound. Let’s go through each step in detail. ### Step 1: Find the Antiderivative We need to find the antiderivative of \(7x^2 + x + 1\): \[ \int (7x^2 + x + 1) \, dx \] We can integrate each term separately: \[ \int 7x^2 \, dx + \int x \, dx + \int 1 \, dx \] Now, compute the antiderivative for each term: 1. \(\int 7x^2 \, dx\): \[ 7 \int x^2 \, dx = 7 \left( \frac{x^3}{3} \right) = \frac{7x^3}{3} \] 2. \(\int x \, dx\): \[ \int x \, dx = \frac{x^2}{2} \] 3. \(\int 1 \, dx\): \[ \int 1 \, dx = x \] Combining these, the antiderivative of \(7x^2 + x + 1\) is: \[ \frac{7x^3}{3} + \frac{x^2}{2} + x + C \] We can ignore the constant \(C\) since we are computing a definite integral. ### Step 2: Evaluate the Antiderivative at the Bounds Evaluate the antiderivative at the upper bound \(x = 1\): \[ \left. \left( \frac{7x^3}{3} + \frac{x^2}{2} + x \right) \right|_{x=1} = \frac{7(1)^3}{3} + \frac{(1)^2}{2} + 1 = \frac{7}{3} + \frac{1}{2} + 1 \] Now, compute the value: \[ \frac{7}{3} + \frac{1}{2} + 1 = \frac{7}{3} + \frac{1}{2} + \frac{2}{2} = \frac{7}{3} + \frac{3}{2} \] Find the common denominator (6) and combine: \[ = \frac{14}{6} + \frac{9}{6} = \frac{23}{6} \] Evaluate the antiderivative at the lower bound \(x = -3\): \[ \left. \left( \frac{7x^3}{3} + \frac{x^2}{2} + x \right) \right|_{x=-3} = \frac{7(-3)^3}{3} + \frac{(-3)^2}{2} + (-3) = \frac{7(-27)}{3} + \frac{9}{2} - 3 = -63 + \frac{9}{2} - 3 \] Now, compute the value: \[ -63 + \frac{9}{2} - 3 = -63 - 3 + \frac{9}{2} = -66 + \frac{9}{2} \] Find the common denominator (2) and combine: \[ = -\frac{132}{2} + \frac{9}{2} = -\frac{123}{2} \] ### Step 3: Subtract the Values Subtract the value of the antiderivative at the lower bound from its value at the upper bound: \[ \left( \frac{23}{6} \right) - \left( -\frac{123}{2} \right) = \frac{23}{6} + \frac{123}{2} \] Find the common denominator (6) and combine: \[ = \frac{23}{6} + \frac{369}{6} = \frac{392}{6} = \frac{196}{3} \] So, the value of the definite integral \(\int_{-3}^1 (7x^2 + x + 1) \, dx\) is: \[ \boxed{\frac{196}{3}} \] |
65.333 |
| llama3.1-70B-Instruct | To solve the given problem, we need to evaluate the definite integral of the function 7x^2 + x + 1 from x = -3 to x = 1. Here's the step-by-step solution: 1. First, we need to find the antiderivative of the function 7x^2 + x + 1. We can do this by integrating each term separately: ∫(7x^2 + x + 1)dx = ∫7x^2dx + ∫xdx + ∫1dx Using the power rule of integration, we get: ∫7x^2dx = (7/3)x^3 + C1 ∫xdx = (1/2)x^2 + C2 ∫1dx = x + C3 where C1, C2, and C3 are constants of integration. 2. Now, we can combine the antiderivatives to get the final antiderivative: ∫(7x^2 + x + 1)dx = (7/3)x^3 + (1/2)x^2 + x + C where C = C1 + C2 + C3. 3. Next, we need to evaluate the definite integral by applying the fundamental theorem of calculus: ∫_{-3}^1 (7x^2 + x + 1)dx = [(7/3)x^3 + (1/2)x^2 + x] from x = -3 to x = 1 = [(7/3)(1)^3 + (1/2)(1)^2 + 1] - [(7/3)(-3)^3 + (1/2)(-3)^2 + (-3)] = (7/3) + (1/2) + 1 - (-63/3) - (9/2) + 3 = 7/3 + 1/2 + 1 + 63/3 + 9/2 - 3 = 7/3 + 1/2 + 1 + 21 + 9/2 - 3 = 7/3 + 1/2 + 1 + 21 + 4.5 - 3 = 7/3 + 1/2 + 1 + 22.5 = 7/3 + 1/2 + 23.5 = 2.33 + 0.5 + 23.5 = 26.33 Therefore, the value of the definite integral is 26.33. |
65.333 |
| llama3.1-405B-Instruct | To evaluate the definite integral \int_{-3}^1 (7x^2 + x +1)dx, we can use the power rule of integration and the constant multiple rule of integration. First, let's integrate the function 7x^2 + x + 1: \int (7x^2 + x + 1) dx = \int 7x^2 dx + \int x dx + \int 1 dx Using the power rule of integration, we get: \int 7x^2 dx = \frac{7x^3}{3} \int x dx = \frac{x^2}{2} \int 1 dx = x So, the integral becomes: \int (7x^2 + x + 1) dx = \frac{7x^3}{3} + \frac{x^2}{2} + x + C where C is the constant of integration. Now, we need to evaluate the definite integral from -3 to 1. We can do this by plugging in the limits of integration into the antiderivative: \int_{-3}^1 (7x^2 + x + 1) dx = \left[\frac{7x^3}{3} + \frac{x^2}{2} + x\right]_{-3}^1 Evaluating the expression at x = 1 and x = -3, we get: \int_{-3}^1 (7x^2 + x + 1) dx = \left[\frac{7(1)^3}{3} + \frac{(1)^2}{2} + 1\right] - \left[\frac{7(-3)^3}{3} + \frac{(-3)^2}{2} + (-3)\right] Simplifying, we get: \int_{-3}^1 (7x^2 + x + 1) dx = \left[\frac{7}{3} + \frac{1}{2} + 1\right] - \left[-\frac{63}{3} + \frac{9}{2} - 3\right] Combining like terms, we get: \int_{-3}^1 (7x^2 + x + 1) dx = \frac{7}{3} + \frac{1}{2} + 1 + \frac{63}{3} - \frac{9}{2} + 3 Simplifying further, we get: \int_{-3}^1 (7x^2 + x + 1) dx = \frac{70}{3} + \frac{1}{2} + 1 - \frac{9}{2} + 3 Combining like terms, we get: \int_{-3}^1 (7x^2 + x + 1) dx = \frac{70}{3} + 1 - \frac{8}{2} + 3 Simplifying further, we get: \int_{-3}^1 (7x^2 + x + 1) dx = \frac{70}{3} + 1 - 4 + 3 Combining like terms, we get: \int_{-3}^1 (7x^2 + x + 1) dx = \frac{70}{3} + 0 Simplifying, we get: \int_{-3}^1 (7x^2 + x + 1) dx = \frac{70}{3} Therefore, the value of the definite integral \int_{-3}^1 (7x^2 + x + 1)dx is \frac{70}{3}. |
65.333 |
| DeepSeek-Coder-V2 | To solve the integral \(\int_{-3}^1 (7x^2 + x + 1)dx\), we will follow these steps: 1. **Find the antiderivative of the integrand**: The integrand is \(7x^2 + x + 1\). We need to find its antiderivative \(F(x)\). \[ F(x) = \int (7x^2 + x + 1) dx \] Using the power rule for integration, we get: \[ F(x) = 7 \int x^2 dx + \int x dx + \int 1 dx \] \[ F(x) = 7 \left( \frac{x^3}{3} \right) + \left( \frac{x^2}{2} \right) + x + C \] Simplifying, we have: \[ F(x) = \frac{7x^3}{3} + \frac{x^2}{2} + x + C \] 2. **Evaluate the definite integral using the Fundamental Theorem of Calculus**: The Fundamental Theorem of Calculus states that: \[ \int_a^b f(x) dx = F(b) - F(a) \] Here, \(a = -3\) and \(b = 1\). We need to evaluate \(F(x)\) at these bounds. First, evaluate \(F(x)\) at \(x = 1\): \[ F(1) = \frac{7(1)^3}{3} + \frac{(1)^2}{2} + 1 + C = \frac{7}{3} + \frac{1}{2} + 1 + C \] Converting to a common denominator (6): \[ F(1) = \frac{14}{6} + \frac{3}{6} + \frac{6}{6} + C = \frac{23}{6} + C \] Next, evaluate \(F(x)\) at \(x = -3\): \[ F(-3) = \frac{7(-3)^3}{3} + \frac{(-3)^2}{2} + (-3) + C = \frac{7(-27)}{3} + \frac{9}{2} - 3 + C \] Simplifying: \[ F(-3) = -63 + \frac{9}{2} - 3 + C \] Converting to a common denominator (2): \[ F(-3) = -63 + \frac{9}{2} - \frac{6}{2} + C = -63 + \frac{3}{2} + C = -63 + \frac{3}{2} + C \] Converting \(-63\) to a fraction with denominator 2: \[ F(-3) = -\frac{126}{2} + \frac{3}{2} + C = -\frac{123}{2} + C \] 3. **Subtract \(F(a)\) from \(F(b)\)**: \[ \int_{-3}^1 (7x^2 + x + 1) dx = F(1) - F(-3) \] \[ = \left( \frac{23}{6} + C \right) - \left( -\frac{123}{2} + C \right) \] Since \(C\) cancels out: \[ = \frac{23}{6} + \frac{123}{2} \] Converting to a common denominator (6): \[ = \frac{23}{6} + \frac{369}{6} = \frac{392}{6} = \frac{196}{3} \] Thus, the value of the integral is: \[ \int_{-3}^1 (7x^2 + x + 1) dx = \frac{196}{3} \] |
65.333 |
| Qwen2-72B-Instruct | To solve the definite integral \(\int_{-3}^1 (7x^2 + x +1)dx\), we need to find the antiderivative of the integrand \(7x^2 + x +1\) and then apply the Fundamental Theorem of Calculus. The antiderivative \(F(x)\) of \(7x^2 + x +1\) is found by integrating each term separately: \[ \int (7x^2 + x +1)dx = \frac{7}{3}x^3 + \frac{1}{2}x^2 + x + C \] where \(C\) is the constant of integration. However, since we are dealing with a definite integral, the constant \(C\) will not affect the final result. Now, we apply the Fundamental Theorem of Calculus, which states that: \[ \int_{a}^{b} f(x)dx = F(b) - F(a) \] where \(F(x)\) is the antiderivative of \(f(x)\). Thus, \[ \int_{-3}^1 (7x^2 + x +1)dx = \left. \left( \frac{7}{3}x^3 + \frac{1}{2}x^2 + x \right) \right|_{-3}^{1} \] \[ = \left( \frac{7}{3}(1)^3 + \frac{1}{2}(1)^2 + (1) \right) - \left( \frac{7}{3}(-3)^3 + \frac{1}{2}(-3)^2 + (-3) \right) \] \[ = \left( \frac{7}{3} + \frac{1}{2} + 1 \right) - \left( \frac{7}{3}(-27) + \frac{1}{2}(9) - 3 \right) \] \[ = \left( \frac{7}{3} + \frac{1}{2} + 1 \right) - \left( -63 + \frac{9}{2} - 3 \right) \] \[ = \left( \frac{7}{3} + \frac{1}{2} + 1 \right) + \left( 63 - \frac{9}{2} + 3 \right) \] \[ = \left( \frac{7}{3} + \frac{1}{2} + 1 \right) + \left( 66 - \frac{9}{2} \right) \] \[ = \frac{7}{3} + \frac{1}{2} + 1 + 66 - \frac{9}{2} \] \[ = \frac{7}{3} + 67 - 4 \] \[ = \frac{7}{3} + 63 \] \[ = 63 + \frac{7}{3} \] \[ = 63 + 2\frac{1}{3} \] \[ = 65\frac{1}{3} \] Therefore, the solution to the definite integral \(\int_{-3}^1 (7x^2 + x +1)dx\) is \(65\frac{1}{3}\) or \(\frac{196}{3}\). |
65.333 |
| baichuan4 | To solve the integral \(\int_{-3}^1 (7x^2 + x +1)dx\), we can integrate each term separately and then add the results together. First, let's integrate \(7x^2\): \[\int 7x^2 dx = 7 \cdot \frac{x^3}{3} = \frac{7x^3}{3}\] Next, let's integrate \(x\): \[\int x dx = \frac{x^2}{2}\] Lastly, let's integrate \(1\): \[\int 1 dx = x\] Now, we add these together and apply the integration bounds: \[\int_{-3}^1 (7x^2 + x +1)dx = \left[\frac{7x^3}{3} + \frac{x^2}{2} + x\right]_{-3}^1\] Evaluating at the upper bound (\(x=1\)): \[\frac{7(1)^3}{3} + \frac{(1)^2}{2} + 1 = \frac{7}{3} + \frac{1}{2} + 1\] Evaluating at the lower bound (\(x=-3\)): \[\frac{7(-3)^3}{3} + \frac{(-3)^2}{2} - 3 = -7(9) + \frac{9}{2} - 3 = -63 + \frac{9}{2} - 3\] Subtracting the result from the lower bound from the result of the upper bound: \[\left(\frac{7}{3} + \frac{1}{2} + 1\right) - \left(-63 + \frac{9}{2} - 3\right)\] \[= \frac{7}{3} + \frac{1}{2} + 1 + 63 - \frac{9}{2} + 3\] \[= \frac{14}{6} + \frac{3}{6} + \frac{6}{6} + \frac{378}{6} - \frac{27}{6} + \frac{18}{6}\] \[= \frac{14 + 3 + 6 + 378 - 27 + 18}{6}\] \[= \frac{402}{6}\] \[= 67\] So, the solution to the integral \(\int_{-3}^1 (7x^2 + x +1)dx\) is \(67\). |
65.333 |
| doubao | First find the antiderivative of the function 7x^2 + x + 1. The antiderivative is 7(x^3/3) + x^2/2 + x. Evaluating this from -3 to 1: [(7(1^3)/3) + (1^2/2) + 1] - [(7(-3)^3/3) + (-3)^2/2 + (-3)] = (7/3 + 1/2 + 1) - [(-63) + 9/2 - 3] = (14/6 + 3/6 + 6/6) - [(-126/2 + 9/2 - 6/2)] = (23/6) - (-123/2) = 23/6 + 123/2 = 23/6 + 369/6 = 392/6 = 196/3. |
65.333 |
| minimax | To solve the integral \(\int_{-3}^1 (7x^2 + x + 1)dx\), we integrate the function term by term: \[ \int (7x^2 + x + 1)dx = \frac{7}{3}x^3 + \frac{1}{2}x^2 + x + C \] where \(C\) is the constant of integration. Since we are evaluating a definite integral, we do not need to worry about \(C\). Now we evaluate the antiderivative at the upper and lower limits of the integral and subtract: \[ \left[\frac{7}{3}x^3 + \frac{1}{2}x^2 + x\right]_{-3}^1 = \left(\frac{7}{3}(1)^3 + \frac{1}{2}(1)^2 + (1)\right) - \left(\frac{7}{3}(-3)^3 + \frac{1}{2}(-3)^2 + (-3)\right) \] \[ = \left(\frac{7}{3} + \frac{1}{2} + 1\right) - \left(\frac{7}{3}(-27) + \frac{1}{2}(9) - 3\right) \] \[ = \left(\frac{7}{3} + \frac{3}{6} + \frac{6}{6}\right) - \left(-63 + \frac{9}{2} - 3\right) \] \[ = \left(\frac{14}{6} + \frac{3}{6} + \frac{6}{6}\right) - \left(-63 + \frac{9}{2} - 3\right) \] \[ = \frac{23}{6} - \left(-63 + \frac{9}{2} - 3\right) \] \[ = \frac{23}{6} - \left(-63 + 4.5 - 3\right) \] \[ = \frac{23}{6} - (-61.5) \] \[ = \frac{23}{6} + 61.5 \] \[ = \frac{23}{6} + \frac{369}{6} \] \[ = \frac{392}{6} \] \[ = \frac{196}{3} \] So the value of the definite integral \(\int_{-3}^1 (7x^2 + x + 1)dx\) is \(\frac{196}{3}\). |
65.333 |
| kimi | To solve the integral of the function \(7x^2 + x + 1\) from \(-3\) to \(1\), we first find the antiderivative (indefinite integral) of the function and then evaluate it at the bounds of the integral. The antiderivative \(F(x)\) of \(7x^2 + x + 1\) is found by integrating each term separately: \[ \int (7x^2 + x + 1)dx = \int 7x^2 dx + \int x dx + \int 1 dx \] \[ = 7\int x^2 dx + \int x dx + \int 1 dx \] \[ = 7\left(\frac{x^3}{3}\right) + \frac{x^2}{2} + x + C \] \[ = \frac{7x^3}{3} + \frac{x^2}{2} + x + C \] Now, we evaluate this antiderivative at the bounds \(-3\) and \(1\) and subtract the two results: \[ \left[\frac{7x^3}{3} + \frac{x^2}{2} + x\right]_{-3}^1 \] \[ = \left(\frac{7(1)^3}{3} + \frac{(1)^2}{2} + 1\right) - \left(\frac{7(-3)^3}{3} + \frac{(-3)^2}{2} + (-3)\right) \] \[ = \left(\frac{7}{3} + \frac{1}{2} + 1\right) - \left(\frac{-63}{3} + \frac{9}{2} - 3\right) \] \[ = \left(\frac{14}{6} + \frac{3}{6} + \frac{6}{6}\right) - \left(\frac{-63}{6} + \frac{27}{6} - \frac{18}{6}\right) \] \[ = \frac{23}{6} - \left(\frac{-38}{6}\right) \] \[ = \frac{23}{6} + \frac{38}{6} \] \[ = \frac{61}{6} \] So the solution to the integral is: \[ \int_{-3}^1 (7x^2 + x +1)dx = \frac{61}{6} \] |
65.333 |
| wenxinyiyan | 首先，我们需要找到7x^2 + x + 1的不定积分（原函数）。 对于7x^2，其原函数是 (7/3)x^3； 对于x，其原函数是 (1/2)x^2； 对于1，其原函数是 x。 所以，7x^2 + x + 1的原函数是 (7/3)x^3 + (1/2)x^2 + x + C，其中C是常数。 然后，我们用上下限1和-3分别代入原函数，并相减。 \int_{-3}^1 (7x^2 + x +1)dx = [(7/3)x^3 + (1/2)x^2 + x]_{-3}^1 = (7/3)(1)^3 + (1/2)(1)^2 + 1 - (7/3)(-3)^3 - (1/2)(-3)^2 - (-3) = 7/3 + 1/2 + 1 + 63/3 - 9/2 + 3 = 28 + 1/2 + 1 - 9/2 + 3 = 32 - 4 = 28 所以，\int_{-3}^1 (7x^2 + x +1)dx = 28。 |
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| zhipu | To solve the integral \(\int_{-3}^1 (7x^2 + x + 1) \, dx\), we need to find the antiderivative of the integrand \(7x^2 + x + 1\) and then evaluate it from \(x = -3\) to \(x = 1\). First, let's find the antiderivative of each term in the integrand: 1. The antiderivative of \(7x^2\) is \(\frac{7x^3}{3}\), because the derivative of \(\frac{7x^3}{3}\) is \(7x^2\). 2. The antiderivative of \(x\) is \(\frac{x^2}{2}\), because the derivative of \(\frac{x^2}{2}\) is \(x\). 3. The antiderivative of \(1\) is \(x\), because the derivative of \(x\) is \(1\). Putting these together, the antiderivative of \(7x^2 + x + 1\) is: \[ F(x) = \frac{7x^3}{3} + \frac{x^2}{2} + x \] Next, we evaluate \(F(x)\) at the upper limit \(x = 1\) and the lower limit \(x = -3\): \[ F(1) = \frac{7(1)^3}{3} + \frac{(1)^2}{2} + 1 = \frac{7}{3} + \frac{1}{2} + 1 \] \[ F(-3) = \frac{7(-3)^3}{3} + \frac{(-3)^2}{2} + (-3) = \frac{7(-27)}{3} + \frac{9}{2} - 3 \] Now, let's simplify these expressions: \[ F(1) = \frac{7}{3} + \frac{1}{2} + 1 = \frac{7}{3} + \frac{1}{2} + \frac{2}{2} = \frac{7}{3} + \frac{3}{2} = \frac{14}{6} + \frac{9}{6} = \frac{23}{6} \] \[ F(-3) = \frac{-189}{3} + \frac{9}{2} - 3 = -63 + \frac{9}{2} - 3 = -63 + 4.5 - 3 = -61.5 \] \[ F(-3) = -61.5 = -\frac{123}{2} \] Finally, we find the difference \(F(1) - F(-3)\): \[ \int_{-3}^1 (7x^2 + x + 1) \, dx = F(1) - F(-3) = \frac{23}{6} - \left(-\frac{123}{2}\right) = \frac{23}{6} + \frac{123}{2} \] To add these fractions, we need a common denominator, which is 6: \[ \frac{23}{6} + \frac{123 \times 3}{2 \times 3} = \frac{23}{6} + \frac{369}{6} = \frac{23 + 369}{6} = \frac{392}{6} = \frac{196}{3} \] So, the value of the integral is: \[ \boxed{\frac{196}{3}} \] |
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| gpto1 | To evaluate the definite integral: \[ \int_{-3}^{1} (7x^2 + x + 1) \, dx, \] we first find the indefinite integral (antiderivative) of the integrand: 1. **Find the antiderivative:** \[ \int (7x^2 + x + 1) \, dx = \frac{7}{3}x^3 + \frac{1}{2}x^2 + x + C, \] where \( C \) is the constant of integration. 2. **Evaluate the antiderivative at the upper and lower limits:** - **At \( x = 1 \):** \[ F(1) = \frac{7}{3}(1)^3 + \frac{1}{2}(1)^2 + 1 = \frac{7}{3} + \frac{1}{2} + 1. \] - **At \( x = -3 \):** \[ F(-3) = \frac{7}{3}(-3)^3 + \frac{1}{2}(-3)^2 + (-3) = \frac{7}{3}(-27) + \frac{1}{2}(9) - 3. \] Simplify \( F(-3) \): \[ \begin{align*} F(-3) &= \left( -63 \right) + \left( \frac{9}{2} \right) - 3 \\ &= -66 + \frac{9}{2}. \end{align*} \] 3. **Compute the definite integral:** \[ \begin{align*} \int_{-3}^{1} (7x^2 + x + 1) \, dx &= F(1) - F(-3) \\ &= \left( \frac{7}{3} + \frac{1}{2} + 1 \right) - \left( -66 + \frac{9}{2} \right) \\ &= \left( \frac{7}{3} + \frac{1}{2} + 1 \right) + 66 - \frac{9}{2}. \end{align*} \] 4. **Combine like terms:** First, find a common denominator for the fractions, which is 6. - Convert fractions: \[ \frac{7}{3} = \frac{14}{6}, \quad \frac{1}{2} = \frac{3}{6}, \quad 1 = \frac{6}{6}, \quad \frac{9}{2} = \frac{27}{6}. \] - Sum the fractions in \( F(1) \): \[ \frac{14}{6} + \frac{3}{6} + \frac{6}{6} = \frac{23}{6}. \] - Now compute: \[ \int_{-3}^{1} (7x^2 + x + 1) \, dx = \left( \frac{23}{6} \right) + 66 - \left( \frac{27}{6} \right). \] 5. **Simplify the expression:** \[ \begin{align*} \left( \frac{23}{6} + 66 \right) - \left( \frac{27}{6} \right) &= \left( \frac{23}{6} + \frac{396}{6} \right) - \frac{27}{6} \\ &= \frac{23 + 396}{6} - \frac{27}{6} \\ &= \frac{419}{6} - \frac{27}{6} \\ &= \frac{419 - 27}{6} \\ &= \frac{392}{6}. \end{align*} \] 6. **Simplify the final fraction:** \[ \frac{392}{6} = \frac{196}{3}. \] **Therefore, the value of the integral is**: \[ \int_{-3}^{1} (7x^2 + x + 1) \, dx = \frac{196}{3}. \] |
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| deepseekv3 | To evaluate the integral \(\int_{-3}^1 (7x^2 + x + 1) \, dx\), follow these steps: ### Step 1: Integrate the function term by term The integral of a sum is the sum of the integrals, so we integrate each term separately: \[ \int (7x^2 + x + 1) \, dx = \int 7x^2 \, dx + \int x \, dx + \int 1 \, dx \] Using the power rule for integration (\(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\)): \[ \int 7x^2 \, dx = 7 \cdot \frac{x^{2+1}}{2+1} = \frac{7x^3}{3}, \] \[ \int x \, dx = \frac{x^{1+1}}{1+1} = \frac{x^2}{2}, \] \[ \int 1 \, dx = x. \] Combine the results: \[ \int (7x^2 + x + 1) \, dx = \frac{7x^3}{3} + \frac{x^2}{2} + x + C, \] where \(C\) is the constant of integration. ### Step 2: Evaluate the definite integral from \(-3\) to \(1\) Substitute the limits of integration: \[ \int_{-3}^1 (7x^2 + x + 1) \, dx = \left[ \frac{7x^3}{3} + \frac{x^2}{2} + x \right]_{-3}^1 \] First, evaluate at \(x = 1\): \[ \frac{7(1)^3}{3} + \frac{(1)^2}{2} + (1) = \frac{7}{3} + \frac{1}{2} + 1 = \frac{14}{6} + \frac{3}{6} + \frac{6}{6} = \frac{23}{6}. \] Next, evaluate at \(x = -3\): \[ \frac{7(-3)^3}{3} + \frac{(-3)^2}{2} + (-3) = \frac{7(-27)}{3} + \frac{9}{2} - 3 = -63 + 4.5 - 3 = -61.5 = -\frac{123}{2}. \] Subtract the lower limit result from the upper limit result: \[ \frac{23}{6} - \left(-\frac{123}{2}\right) = \frac{23}{6} + \frac{123}{2} = \frac{23}{6} + \frac{369}{6} = \frac{392}{6} = \frac{196}{3}. \] ### Final Answer: \[ \boxed{\frac{196}{3}} \] |
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