| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the integral \(\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2}\), we can utilize the substitution method where \(x = \tan(\theta)\). This substitution is particularly advantageous as it simplifies the denominator. Here's a step-by-step guide to solving this integral: ### Step 1: Substitution Let \(x = \tan(\theta)\). Then, \(dx = \sec^2(\theta) d\theta\). Also, since \(x = \tan(\theta)\), we have \(1 + x^2 = 1 + \tan^2(\theta) = \sec^2(\theta)\). ### Step 2: Transform the Integral Substitute \(x = \tan(\theta)\) and \(dx = \sec^2(\theta) d\theta\) into the integral: \[ \int_{0}^{\sqrt{3}} \frac{dx}{1+x^2} = \int \frac{\sec^2(\theta) d\theta}{\sec^2(\theta)} = \int d\theta. \] ### Step 3: Transform the Limits The limits of integration also need to be transformed from \(x\) to \(\theta\). When \(x = 0\), we have \(\tan(\theta) = 0\), so \(\theta = 0\). When \(x = \sqrt{3}\), we have \(\tan(\theta) = \sqrt{3}\), so \(\theta = \frac{\pi}{3}\). ### Step 4: Evaluate the Integral Now, evaluate the integral \(\int d\theta\) from \(\theta = 0\) to \(\theta = \frac{\pi}{3}\): \[ \int_{0}^{\sqrt{3}} \frac{dx}{1+x^2} = \int_{0}^{\frac{\pi}{3}} d\theta = \left[ \theta \right]_{0}^{\frac{\pi}{3}} = \frac{\pi}{3} - 0 = \frac{\pi}{3}. \] ### Answer: The integral \(\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2}\) evaluates to \(\boxed{\frac{\pi}{3}}\). |
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| deepseekr1 | To find the integral \(\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2}\), we recognize that the integrand \(\frac{1}{1+x^2}\) is the derivative of \(\arctan(x)\). Therefore, we can use the Fundamental Theorem of Calculus to evaluate the definite integral. The integral is computed as follows: \[ \int_{0}^{\sqrt{3}} \frac{dx}{1+x^2} = \left[ \arctan(x) \right]_{0}^{\sqrt{3}} \] Evaluating the antiderivative at the bounds: \[ \arctan(\sqrt{3}) - \arctan(0) \] We know that \(\arctan(\sqrt{3}) = \frac{\pi}{3}\) because \(\tan\left(\frac{\pi}{3}\right) = \sqrt{3}\), and \(\arctan(0) = 0\). Therefore: \[ \frac{\pi}{3} - 0 = \frac{\pi}{3} \] Thus, the value of the integral is \(\boxed{\dfrac{\pi}{3}}\). |
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| gpto1 | To evaluate the integral: \[ \int_{0}^{\sqrt{3}} \frac{dx}{1 + x^2} \] we recognize that the integrand is a standard form that relates to the inverse tangent function. Recall the antiderivative of \(\frac{1}{1 + x^2}\) is \(\arctan(x)\): \[ \int \frac{dx}{1 + x^2} = \arctan(x) + C \] Applying the Fundamental Theorem of Calculus, we evaluate the definite integral by finding the antiderivative at the upper and lower limits and subtracting: \[ \int_{0}^{\sqrt{3}} \frac{dx}{1 + x^2} = \left[ \arctan(x) \right]_{0}^{\sqrt{3}} = \arctan(\sqrt{3}) - \arctan(0) \] Next, we compute the values of the inverse tangent function at the specified points. 1. **Compute \(\arctan(\sqrt{3})\):** The inverse tangent of \(\sqrt{3}\) is the angle whose tangent is \(\sqrt{3}\). We know from trigonometry that: \[ \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \] Therefore: \[ \arctan(\sqrt{3}) = \frac{\pi}{3} \] 2. **Compute \(\arctan(0)\):** The inverse tangent of \(0\) is the angle whose tangent is \(0\): \[ \tan(0) = 0 \implies \arctan(0) = 0 \] Subtracting these values: \[ \arctan(\sqrt{3}) - \arctan(0) = \frac{\pi}{3} - 0 = \frac{\pi}{3} \] **Answer:** \[ \int_{0}^{\sqrt{3}} \frac{dx}{1 + x^2} = \frac{\pi}{3} \] |
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| GPT-4o | To solve the integral \(\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2}\), we recognize that the integrand \(\frac{1}{1+x^2}\) is the derivative of the inverse tangent function, \(\arctan(x)\). Thus, we can use the fact that: \[ \frac{d}{dx} \left( \arctan(x) \right) = \frac{1}{1+x^2} \] Therefore, we have: \[ \int \frac{dx}{1+x^2} = \arctan(x) + C \] To find the definite integral from \(0\) to \(\sqrt{3}\), we evaluate: \[ \int_{0}^{\sqrt{3}} \frac{dx}{1+x^2} = \left. \arctan(x) \right|_{0}^{\sqrt{3}} \] This means we need to compute: \[ \arctan(\sqrt{3}) - \arctan(0) \] We know that: \[ \arctan(\sqrt{3}) = \frac{\pi}{3} \quad \text{and} \quad \arctan(0) = 0 \] Therefore: \[ \arctan(\sqrt{3}) - \arctan(0) = \frac{\pi}{3} - 0 = \frac{\pi}{3} \] So, the value of the integral is: \[ \int_{0}^{\sqrt{3}} \frac{dx}{1+x^2} = \frac{\pi}{3} \] Hence, the solution is: \[ \boxed{\frac{\pi}{3}} \] |
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| llama3.1-70B-Instruct | The problem is to evaluate the definite integral of $\frac{1}{1+x^2}$ from $0$ to $\sqrt{3}$. ## Step 1: Recall the antiderivative of $\frac{1}{1+x^2}$ The antiderivative of $\frac{1}{1+x^2}$ is $\arctan(x)$. ## Step 2: Apply the Fundamental Theorem of Calculus The Fundamental Theorem of Calculus states that the definite integral of a function $f(x)$ from $a$ to $b$ is equal to $F(b) - F(a)$, where $F(x)$ is the antiderivative of $f(x)$. In this case, $f(x) = \frac{1}{1+x^2}$, $F(x) = \arctan(x)$, $a = 0$, and $b = \sqrt{3}$. ## Step 3: Evaluate the antiderivative at the limits of integration We need to evaluate $\arctan(x)$ at $x = \sqrt{3}$ and $x = 0$. ## Step 4: Calculate $\arctan(\sqrt{3})$ and $\arctan(0)$ $\arctan(\sqrt{3}) = \frac{\pi}{3}$ and $\arctan(0) = 0$. ## Step 5: Subtract the values of the antiderivative at the limits of integration $\frac{\pi}{3} - 0 = \frac{\pi}{3}$. The final answer is: $\boxed{\frac{\pi}{3}}$ |
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| llama3.1-405B-Instruct | We have that $\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2} = \left[\arctan x\right]_{0}^{\sqrt{3}} = \arctan \sqrt{3} - \arctan 0 = \boxed{\frac{\pi}{3}}.$ Final Answer: The final answer is \frac{\pi}{3}. I hope it is correct. |
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| DeepSeek-Coder-V2 | To solve the integral \(\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2}\), we can use the substitution method involving trigonometric functions. Here is the step-by-step solution: 1. **Substitution**: Let \(x = \tan \theta\). This substitution is useful because \(\frac{d}{dx}(\tan \theta) = \sec^2 \theta\), which simplifies the integrand. 2. **Differential**: Compute the differential \(dx\): \[ dx = \frac{d}{d\theta}(\tan \theta) \, d\theta = \sec^2 \theta \, d\theta \] 3. **Rewrite the integral**: Substitute \(x = \tan \theta\) and \(dx = \sec^2 \theta \, d\theta\) into the integral: \[ \int_{0}^{\sqrt{3}} \frac{dx}{1+x^2} = \int_{0}^{\theta_{\sqrt{3}}} \frac{\sec^2 \theta \, d\theta}{1 + \tan^2 \theta} \] 4. **Simplify the integrand**: Use the trigonometric identity \(1 + \tan^2 \theta = \sec^2 \theta\): \[ \int_{0}^{\theta_{\sqrt{3}}} \frac{\sec^2 \theta \, d\theta}{\sec^2 \theta} = \int_{0}^{\theta_{\sqrt{3}}} d\theta \] 5. **Evaluate the integral**: The integral of \(d\theta\) is \(\theta\): \[ \int_{0}^{\theta_{\sqrt{3}}} d\theta = \theta \Big|_{0}^{\theta_{\sqrt{3}}} = \theta_{\sqrt{3}} - \theta_0 \] 6. **Determine the bounds**: We need to find the values of \(\theta\) corresponding to the bounds \(x = 0\) and \(x = \sqrt{3}\): - For \(x = 0\), \(\tan \theta = 0\) implies \(\theta = 0\). - For \(x = \sqrt{3}\), \(\tan \theta = \sqrt{3}\) implies \(\theta = \frac{\pi}{3}\). 7. **Compute the result**: Substitute the bounds into the evaluated integral: \[ \theta_{\sqrt{3}} - \theta_0 = \frac{\pi}{3} - 0 = \frac{\pi}{3} \] Thus, the value of the integral \(\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2}\) is \(\frac{\pi}{3}\). |
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| Qwen2-72B-Instruct | To solve the integral $\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2}$, we can use the substitution method where $x = \tan(\theta)$. This substitution is particularly useful because it transforms the denominator into a simpler form. Here's how to do it step-by-step: ### Step 1: Substitution Let $x = \tan(\theta)$. Then, $dx = \sec^2(\theta) d\theta$. Also, since $x = \tan(\theta)$, we have $1 + x^2 = 1 + \tan^2(\theta) = \sec^2(\theta)$. ### Step 2: Transform the Integral Substitute $x = \tan(\theta)$ and $dx = \sec^2(\theta) d\theta$ into the integral: $\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2} = \int \frac{\sec^2(\theta) d\theta}{\sec^2(\theta)} = \int d\theta$. ### Step 3: Transform the Limits The limits of integration also need to be transformed from $x$ to $\theta$. When $x = 0$, we have $\tan(\theta) = 0$, so $\theta = 0$. When $x = \sqrt{3}$, we have $\tan(\theta) = \sqrt{3}$, so $\theta = \frac{\pi}{3}$. ### Step 4: Evaluate the Integral Now, evaluate the integral $\int d\theta$ from $\theta = 0$ to $\theta = \frac{\pi}{3}$: $\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2} = \int_{0}^{\frac{\pi}{3}} d\theta = \left[ \theta \right]_{0}^{\frac{\pi}{3}} = \frac{\pi}{3} - 0 = \frac{\pi}{3}$. ### Answer: The integral $\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2}$ evaluates to $\boxed{\frac{\pi}{3}}$. |
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| baichuan4 | To solve the integral $\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2}$, we can use the substitution method. Let's set $x = \tan(u)$, which implies that $dx = \sec^2(u)du$. Now, let's apply the limits of integration: - When $x = 0$, $u = \arctan(0) = 0$. - When $x = \sqrt{3}$, $u = \arctan(\sqrt{3}) = \frac{\pi}{3}$. Substituting these into the integral, we get: $$\int_{0}^{\frac{\pi}{3}} \frac{\sec^2(u)du}{1+\tan^2(u)}$$ Since $\sec^2(u) = 1+\tan^2(u)$, the integral simplifies to: $$\int_{0}^{\frac{\pi}{3}} du$$ This is a simple integral of the identity function, which is just $u$. So, we have: $$u \Big|_{0}^{\frac{\pi}{3}} = \frac{\pi}{3} - 0 = \frac{\pi}{3}$$ Therefore, the solution to the given problem is: $$\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2} = \frac{\pi}{3}$$ |
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| doubao | We know that $\int \frac{dx}{1+x^2}=\arctan(x)+C$. Evaluating the definite integral: $\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2}=[\arctan(x)]_{0}^{\sqrt{3}}=\arctan(\sqrt{3})-\arctan(0)=\frac{\pi}{3}-0=\frac{\pi}{3}$. |
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| minimax | To solve the integral $\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2}$, we can recognize that the integrand is the derivative of the arctangent function. Specifically, the derivative of $\arctan(x)$ with respect to $x$ is $\frac{1}{1+x^2}$. Therefore, we can write the integral as: $$\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2} = \left[\arctan(x)\right]_{0}^{\sqrt{3}}$$ Now we evaluate the arctangent function at the upper and lower limits of the integral: $$\arctan(\sqrt{3}) - \arctan(0)$$ We know that $\arctan(0) = 0$ because the tangent of 0 is 0. Also, $\arctan(\sqrt{3})$ is the angle whose tangent is $\sqrt{3}$. This angle is $\frac{\pi}{3}$ radians (or 60 degrees), because $\tan(\frac{\pi}{3}) = \sqrt{3}$. So the solution to the integral is: $$\left[\arctan(\sqrt{3}) - \arctan(0)\right] = \frac{\pi}{3} - 0 = \frac{\pi}{3}$$ Therefore, $\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2} = \frac{\pi}{3}$. |
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| kimi | To solve the integral $\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2}$, we recognize that this is a standard integral involving the arctangent function. The antiderivative of $\frac{1}{1+x^2}$ is $\arctan(x)$. Therefore, we can evaluate the definite integral as follows: 1. Find the antiderivative of the integrand: $$ \int \frac{dx}{1+x^2} = \arctan(x) + C $$ 2. Evaluate the antiderivative at the upper and lower limits of the integral and subtract the results: $$ \left[ \arctan(x) \right]_{0}^{\sqrt{3}} = \arctan(\sqrt{3}) - \arctan(0) $$ 3. Calculate the values of the arctangent function at the limits: $$ \arctan(\sqrt{3}) = \frac{\pi}{3} \quad \text{(since $\tan(\frac{\pi}{3}) = \sqrt{3}$)} $$ $$ \arctan(0) = 0 \quad \text{(since $\tan(0) = 0$)} $$ 4. Subtract the lower limit from the upper limit: $$ \frac{\pi}{3} - 0 = \frac{\pi}{3} $$ So the solution to the integral is: $$ \int_{0}^{\sqrt{3}} \frac{dx}{1+x^2} = \frac{\pi}{3} $$ |
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| wenxinyiyan | 首先，我们识别到给定的积分 $\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2}$ 是一个基本积分形式的变体，它与反正切函数 $\arctan(x)$ 的导数有关。 我们知道，$\frac{d}{dx} \arctan(x) = \frac{1}{1+x^2}$。 因此，我们可以直接应用不定积分的公式，将给定的定积分转化为： $$\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2} = \left[ \arctan(x) \right]_{0}^{\sqrt{3}}$$接下来，我们计算上下限的反正切值： $$\arctan(\sqrt{3}) - \arctan(0)$$由于 $\arctan(\sqrt{3}) = \frac{\pi}{3}$（因为 $\tan(\frac{\pi}{3}) = \sqrt{3}$）且 $\arctan(0) = 0$，所以： $$\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2} = \frac{\pi}{3} - 0 = \frac{\pi}{3}$$ |
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| zhipu | To solve the integral \(\int_{0}^{\sqrt{3}} \frac{dx}{1+x^2}\), we can use the standard result for the integral of \(\frac{1}{1+x^2}\), which is \(\arctan(x)\). Here's the step-by-step solution: 1. **Identify the integral:** \[ \int_{0}^{\sqrt{3}} \frac{dx}{1+x^2} \] 2. **Use the standard integral result:** \[ \int \frac{dx}{1+x^2} = \arctan(x) + C \] 3. **Apply the limits of integration:** \[ \left[ \arctan(x) \right]_{0}^{\sqrt{3}} \] 4. **Evaluate the definite integral:** \[ \arctan(\sqrt{3}) - \arctan(0) \] 5. **Calculate the values of \(\arctan(\sqrt{3})\) and \(\arctan(0)\):** - \(\arctan(\sqrt{3})\) is the angle whose tangent is \(\sqrt{3}\). This angle is \(\frac{\pi}{3}\). - \(\arctan(0)\) is the angle whose tangent is 0. This angle is 0. 6. **Substitute these values back into the expression:** \[ \arctan(\sqrt{3}) - \arctan(0) = \frac{\pi}{3} - 0 = \frac{\pi}{3} \] Therefore, the value of the integral is: \[ \int_{0}^{\sqrt{3}} \frac{dx}{1+x^2} = \frac{\pi}{3} \] So, the solution to the problem is \(\boxed{\frac{\pi}{3}}\). |
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| gpto1 | To evaluate the integral: \[ \int_{0}^{\sqrt{3}} \frac{dx}{1 + x^2} \] we recognize that the integrand is a standard form that relates to the inverse tangent function. Recall the antiderivative of \(\frac{1}{1 + x^2}\) is \(\arctan(x)\): \[ \int \frac{dx}{1 + x^2} = \arctan(x) + C \] Applying the Fundamental Theorem of Calculus, we evaluate the definite integral by finding the antiderivative at the upper and lower limits and subtracting: \[ \int_{0}^{\sqrt{3}} \frac{dx}{1 + x^2} = \left[ \arctan(x) \right]_{0}^{\sqrt{3}} = \arctan(\sqrt{3}) - \arctan(0) \] Next, we compute the values of the inverse tangent function at the specified points. 1. **Compute \(\arctan(\sqrt{3})\):** The inverse tangent of \(\sqrt{3}\) is the angle whose tangent is \(\sqrt{3}\). We know from trigonometry that: \[ \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \] Therefore: \[ \arctan(\sqrt{3}) = \frac{\pi}{3} \] 2. **Compute \(\arctan(0)\):** The inverse tangent of \(0\) is the angle whose tangent is \(0\): \[ \tan(0) = 0 \implies \arctan(0) = 0 \] Subtracting these values: \[ \arctan(\sqrt{3}) - \arctan(0) = \frac{\pi}{3} - 0 = \frac{\pi}{3} \] **Answer:** \[ \int_{0}^{\sqrt{3}} \frac{dx}{1 + x^2} = \frac{\pi}{3} \] |
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| deepseekv3 | To evaluate the integral: \[ \int_{0}^{\sqrt{3}} \frac{dx}{1+x^2} \] we can use the fact that the antiderivative of \(\frac{1}{1+x^2}\) is \(\arctan(x)\). Therefore, the integral becomes: \[ \int \frac{dx}{1+x^2} = \arctan(x) + C \] Now, we can compute the definite integral from \(0\) to \(\sqrt{3}\): \[ \int_{0}^{\sqrt{3}} \frac{dx}{1+x^2} = \arctan(\sqrt{3}) - \arctan(0) \] We know that: \[ \arctan(\sqrt{3}) = \frac{\pi}{3} \quad \text{and} \quad \arctan(0) = 0 \] Thus, the integral evaluates to: \[ \frac{\pi}{3} - 0 = \frac{\pi}{3} \] \(\boxed{\dfrac{\pi}{3}}\) |
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