| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | The integral \(\int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}\) can be solved by using a trigonometric substitution. Let's make the substitution \(x = a + (b-a)\sin^2\theta\). Differentiating both sides with respect to \(x\) yields: \[ dx = 2(b-a)\sin\theta\cos\theta d\theta \] The integral becomes: \[ \int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}} = \int_0^{\frac{\pi}{2}} \frac{2(b-a)\sin\theta\cos\theta d\theta}{\sqrt{(b-a)^2\sin^2\theta\cos^2\theta}} = \int_0^{\frac{\pi}{2}} \frac{2(b-a)\sin\theta\cos\theta d\theta}{(b-a)\sin\theta\cos\theta} = \int_0^{\frac{\pi}{2}} 2 d\theta \] This simplifies to: \[ \int_0^{\frac{\pi}{2}} 2 d\theta = 2\theta\bigg|_0^{\frac{\pi}{2}} = 2\left(\frac{\pi}{2}\right) = \pi \] Therefore, the value of the integral is \(\pi\). Rounded to the thousands decimal, the answer is \(\boxed{3.142}\). |
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| deepseekr1 | To find the value of the integral \(\int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}\), we start by making a substitution. Let \(t = x - a\), which transforms the integral limits from \(x = a\) to \(x = b\) into \(t = 0\) to \(t = b - a\). The integral then becomes: \[ \int_0^{b-a} \frac{dt}{\sqrt{t(b-a-t)}} \] Next, we use another substitution \(u = \frac{t}{b-a}\), which normalizes the interval. This transforms \(t\) to \(u\) with \(t = (b-a)u\) and \(dt = (b-a)du\). The integral becomes: \[ \int_0^1 \frac{(b-a)du}{\sqrt{(b-a)u \cdot (b-a)(1-u)}} = \int_0^1 \frac{(b-a)du}{(b-a)\sqrt{u(1-u)}} = \int_0^1 \frac{du}{\sqrt{u(1-u)}} \] This integral is recognized as a Beta function \(B\left(\frac{1}{2}, \frac{1}{2}\right)\), which is known to be equal to \(\pi\). Therefore, the value of the original integral is \(\pi\). Rounding \(\pi\) to the thousandths decimal place, we get \(3.142\). \[ \boxed{3.142} \] |
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| gpto1 | To solve the integral: \[ I = \int_a^b \frac{dx}{\sqrt{(x - a)(b - x)}} \] We can perform a substitution to simplify the integral. **Substitution:** Let: \[ x = \frac{a + b}{2} + \frac{b - a}{2} \cos \theta \] This substitution maps the interval from \( x = a \) to \( x = b \) as \( \theta \) varies from \( \theta = \pi \) to \( \theta = 0 \). **Computing \( x - a \) and \( b - x \):** First, compute \( x - a \) and \( b - x \): \[ x - a = \frac{a + b}{2} + \frac{b - a}{2} \cos \theta - a = \frac{b - a}{2} (1 + \cos \theta) \] \[ b - x = b - \left( \frac{a + b}{2} + \frac{b - a}{2} \cos \theta \right) = \frac{b - a}{2} (1 - \cos \theta) \] **Simplifying the Square Root:** Now, compute the product: \[ (x - a)(b - x) = \left( \frac{b - a}{2} (1 + \cos \theta) \right) \left( \frac{b - a}{2} (1 - \cos \theta) \right) = \left( \frac{b - a}{2} \right)^2 (1 - \cos^2 \theta) = \left( \frac{b - a}{2} \right)^2 \sin^2 \theta \] Therefore, the square root is: \[ \sqrt{(x - a)(b - x)} = \left| \frac{b - a}{2} \sin \theta \right| \] Since \( \sin \theta \) is positive in the interval \( 0 \leq \theta \leq \pi \), we can drop the absolute value: \[ \sqrt{(x - a)(b - x)} = \frac{b - a}{2} \sin \theta \] **Computing \( dx \):** Differentiate \( x \) with respect to \( \theta \): \[ dx = -\frac{b - a}{2} \sin \theta \, d\theta \] **Setting up the Integral:** Substitute \( dx \) and \( \sqrt{(x - a)(b - x)} \) back into the integral: \[ I = \int_{\theta = \pi}^{\theta = 0} \frac{-\frac{b - a}{2} \sin \theta \, d\theta}{\frac{b - a}{2} \sin \theta} = \int_{\theta = \pi}^{\theta = 0} (-1) d\theta = \int_{0}^{\pi} d\theta \] **Evaluating the Integral:** \[ I = \int_{0}^{\pi} d\theta = \left[ \theta \right]_0^{\pi} = \pi \] **Conclusion:** Thus, the value of the integral is: \[ I = \pi \] Rounded to the thousandth decimal place: \[ I \approx 3.142 \] \(\boxed{3.142}\) |
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| GPT-4o | To solve the integral \(\int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}\), we'll use a substitution method to simplify the integral. Let's consider the substitution: \[ x = a + (b-a)\sin^2(\theta) \] First, compute \(dx\): \[ dx = \frac{d}{d\theta} \left( a + (b-a)\sin^2(\theta) \right) d\theta = 2(b-a)\sin(\theta)\cos(\theta) d\theta = (b-a)\sin(2\theta) d\theta \] Next, we need to determine the new limits of integration when \(x\) ranges from \(a\) to \(b\): - When \(x = a\): \[ a = a + (b-a)\sin^2(\theta) \implies \sin^2(\theta) = 0 \implies \theta = 0 \] - When \(x = b\): \[ b = a + (b-a)\sin^2(\theta) \implies \sin^2(\theta) = 1 \implies \theta = \frac{\pi}{2} \] Now, consider the integrand \(\frac{dx}{\sqrt{(x-a)(b-x)}}\): \[ \sqrt{(x-a)(b-x)} = \sqrt{(a + (b-a)\sin^2(\theta) - a)(b - (a + (b-a)\sin^2(\theta)))} \] \[ = \sqrt{(b-a)\sin^2(\theta) (b-a)(1-\sin^2(\theta))} = \sqrt{(b-a)^2\sin^2(\theta)\cos^2(\theta)} = (b-a)\sin(\theta)\cos(\theta) \] Thus, the integral becomes: \[ \int_0^{\frac{\pi}{2}} \frac{(b-a)\sin(2\theta)d\theta}{(b-a)\sin(\theta)\cos(\theta)} = \int_0^{\frac{\pi}{2}} \frac{(b-a)\sin(2\theta)d\theta}{(b-a)\sin(\theta)\cos(\theta)} \] \[ = \int_0^{\frac{\pi}{2}} \frac{2\sin(\theta)\cos(\theta)d\theta}{\sin(\theta)\cos(\theta)} = \int_0^{\frac{\pi}{2}} 2 d\theta \] \[ = 2 \int_0^{\frac{\pi}{2}} d\theta = 2\left[\theta\right]_0^{\frac{\pi}{2}} = 2\left(\frac{\pi}{2} - 0\right) = 2 \cdot \frac{\pi}{2} = \pi \] Thus, the value of the integral \(\int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}\) is \(\pi\). Rounded to the thousands decimal, the answer is: \[ \pi \approx 3.142 \] So, the final answer is: \[ \boxed{3.142} \] |
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| llama3.1-70B-Instruct | The given integral is \int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}. To solve this integral, we can use the substitution method. Let's substitute x = a + (b-a)sin^2(θ), where θ is a new variable. Then, dx = 2(b-a)sin(θ)cos(θ)dθ. Now, we can rewrite the integral as: \int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}} = \int_0^π \frac{2(b-a)sin(θ)cos(θ)dθ}{\sqrt{(a + (b-a)sin^2(θ) - a)(b - (a + (b-a)sin^2(θ)))}}. Simplifying the expression inside the square root, we get: \int_0^π \frac{2(b-a)sin(θ)cos(θ)dθ}{\sqrt{(b-a)sin^2(θ)(b-a)cos^2(θ)}}. Canceling out the common factors, we get: \int_0^π \frac{2sin(θ)cos(θ)dθ}{\sqrt{sin^2(θ)cos^2(θ)}}. Now, we can use the trigonometric identity sin(2θ) = 2sin(θ)cos(θ) to simplify the integral: \int_0^π \frac{sin(2θ)dθ}{\sqrt{sin^2(θ)cos^2(θ)}}. Using the substitution u = 2θ, we get: \int_0^2π \frac{sin(u)du}{\sqrt{sin^2(u/2)cos^2(u/2)}}. Simplifying the expression inside the square root, we get: \int_0^2π \frac{sin(u)du}{\sqrt{sin^2(u/2)cos^2(u/2)}} = \int_0^2π \frac{sin(u)du}{\sqrt{1/4 - (1/4)cos^2(u)}}. Now, we can use the trigonometric identity cos(2u) = 2cos^2(u) - 1 to simplify the integral: \int_0^2π \frac{sin(u)du}{\sqrt{1/4 - (1/4)cos^2(u)}} = \int_0^2π \frac{sin(u)du}{\sqrt{1/4 - (1/4)(1 + cos(2u))/2}}. Simplifying the expression inside the square root, we get: \int_0^2π \frac{sin(u)du}{\sqrt{1/4 - (1/4)(1 + cos(2u))/2}} = \int_0^2π \frac{sin(u)du}{\sqrt{1/4 - (1/8) - (1/8)cos(2u)}}. Combining the constants inside the square root, we get: \int_0^2π \frac{sin(u)du}{\sqrt{1/8 - (1/8)cos(2u)}}. Now, we can use the trigonometric identity cos(2u) = 1 - 2sin^2(u) to simplify the integral: \int_0^2π \frac{sin(u)du}{\sqrt{1/8 - (1/8)cos(2u)}} = \int_0^2π \frac{sin(u)du}{\sqrt{1/8 - (1/8)(1 - 2sin^2(u))}}. Simplifying the expression inside the square root, we get: \int_0^2π \frac{sin(u)du}{\sqrt{1/8 - (1/8) + (1/4)sin^2(u)}}. Combining the constants inside the square root, we get: \int_0^2π \frac{sin(u)du}{\sqrt{(1/4)sin^2(u)}}. Simplifying the expression inside the square root, we get: \int_0^2π \frac{sin(u)du}{(1/2)sin(u)}. Canceling out the common factors, we get: \int_0^2π du. Evaluating the integral, we get: 2π. Therefore, the value of the given integral is 2π. Rounding the answer to the thousands decimal, we get: 2.000. |
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| llama3.1-405B-Instruct | To solve the integral \(\int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}\), we can use a substitution method. Let's set \(x = a + (b-a)u\), which implies \(dx = (b-a)du\). When \(x = a\), \(u = 0\), and when \(x = b\), \(u = 1\). Substituting these into the integral, we get: \[ \int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}} = \int_0^1 \frac{(b-a)du}{\sqrt{(a + (b-a)u - a)(b - (a + (b-a)u))}} \] Simplifying the expression inside the square root: \[ = \int_0^1 \frac{(b-a)du}{\sqrt{(b-a)u(b-a)(1-u)}} \] \[ = \int_0^1 \frac{du}{\sqrt{u(1-u)}} \] Now, we can use the fact that \(\sqrt{u(1-u)} = \sqrt{u}\sqrt{1-u}\) and substitute \(u = \sin^2\theta\), which implies \(du = 2\sin\theta\cos\theta d\theta\). When \(u = 0\), \(\theta = 0\), and when \(u = 1\), \(\theta = \frac{\pi}{2}\). Substituting these into the integral, we get: \[ \int_0^1 \frac{du}{\sqrt{u(1-u)}} = \int_0^{\frac{\pi}{2}} \frac{2\sin\theta\cos\theta d\theta}{\sqrt{\sin^2\theta(1-\sin^2\theta)}} \] Simplifying the expression inside the square root: \[ = \int_0^{\frac{\pi}{2}} \frac{2\sin\theta\cos\theta d\theta}{\sqrt{\sin^2\theta\cos^2\theta}} \] \[ = \int_0^{\frac{\pi}{2}} 2 d\theta \] \[ = 2\theta \Big|_0^{\frac{\pi}{2}} \] \[ = 2\left(\frac{\pi}{2}\right) - 2(0) \] \[ = \pi \] Therefore, the value of the integral \(\int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}\) is \(\pi\), which rounded to the thousandths decimal is \(3.142\). |
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| DeepSeek-Coder-V2 | To solve the integral \(\int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}\), we can use a substitution method to simplify the integrand. Let's proceed step by step: 1. **Substitution**: Let \( y = x - a \). Then, \( dy = dx \) and \( x = y + a \). The limits of integration change accordingly: - When \( x = a \), \( y = 0 \). - When \( x = b \), \( y = b - a \). 2. **Rewrite the integral**: Substitute \( y \) into the integral: \[ \int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}} = \int_0^{b-a} \frac{dy}{\sqrt{y(b-a-y)}} \] 3. **Simplify the integrand**: Notice that \( b - x = b - (y + a) = b - a - y \). Thus, the integral becomes: \[ \int_0^{b-a} \frac{dy}{\sqrt{y(b-a-y)}} \] 4. **Another substitution**: Let \( z = \sqrt{\frac{y}{b-a-y}} \). Then, \( y = z^2 (b-a-y) \) and solving for \( y \): \[ y = \frac{z^2 (b-a)}{1 + z^2} \] Also, \( dy = \frac{2z (b-a)}{(1 + z^2)^2} dz \). 5. **Change of limits**: When \( y = 0 \), \( z = 0 \). When \( y = b - a \), \( z \to \infty \). 6. **Rewrite the integral in terms of \( z \)**: \[ \int_0^{b-a} \frac{dy}{\sqrt{y(b-a-y)}} = \int_0^\infty \frac{2z (b-a)}{(1 + z^2)^2} \cdot \frac{1}{z} dz \] Simplifying, we get: \[ \int_0^\infty \frac{2 (b-a)}{(1 + z^2)^2} dz \] 7. **Evaluate the integral**: The integral \(\int_0^\infty \frac{1}{(1 + z^2)^2} dz\) is a standard integral and evaluates to \(\frac{\pi}{2}\). Therefore: \[ \int_0^\infty \frac{2 (b-a)}{(1 + z^2)^2} dz = 2 (b-a) \cdot \frac{\pi}{2} = (b-a) \pi \] 8. **Final answer**: The value of the integral is \((b-a) \pi\). To round to the thousands decimal, we need the numerical value of \((b-a) \pi\). Assuming \( b - a \) is a specific number, we can calculate it directly. For example, if \( b - a = 1 \): \[ (b-a) \pi = 1 \cdot \pi = 3.14159 \approx 3.142 \] Thus, the rounded value to the thousands decimal is: \[ \boxed{3.142} \] |
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| Qwen2-72B-Instruct | The integral \(\int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}\) can be solved by using a trigonometric substitution. Let's make the substitution \(x = a + (b-a)\sin^2\theta\). Differentiating both sides with respect to \(x\) yields: \[ dx = 2(b-a)\sin\theta\cos\theta d\theta \] The integral becomes: \[ \int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}} = \int_0^{\frac{\pi}{2}} \frac{2(b-a)\sin\theta\cos\theta d\theta}{\sqrt{(b-a)^2\sin^2\theta\cos^2\theta}} = \int_0^{\frac{\pi}{2}} \frac{2(b-a)\sin\theta\cos\theta d\theta}{(b-a)\sin\theta\cos\theta} = \int_0^{\frac{\pi}{2}} 2 d\theta \] This simplifies to: \[ \int_0^{\frac{\pi}{2}} 2 d\theta = 2\theta\bigg|_0^{\frac{\pi}{2}} = 2\left(\frac{\pi}{2}\right) = \pi \] Therefore, the value of the integral is \(\pi\). Rounded to the thousands decimal, the answer is \(\boxed{3.142}\). |
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| baichuan4 | To solve this integral, we can use the method of substitution. Let's set $u = \sqrt{\frac{x-a}{b-x}}$, then $du = \frac{1}{2} \left( \frac{1}{\sqrt{(x-a)(b-x)}} \right) (1) dx$. From this, we can see that $dx = 2u du$. Now, let's express $x$ in terms of $u$. We have $u = \sqrt{\frac{x-a}{b-x}}$, so $u^2 = \frac{x-a}{b-x}$. Rearranging this equation, we get $x = \frac{a + bu^2}{1 + u^2}$. Now, we can substitute $u$ and $dx$ into the original integral: \[ \int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}} = \int_{0}^{\infty} \frac{2u du}{u} = 2 \int_{0}^{\infty} du \] This integral is divergent, so the original integral does not converge. Therefore, the value of $\int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}$ is not defined. |
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| doubao | Let's make the substitution \(x = a + (b - a)t\). Then \(dx = (b - a)dt\). When \(x = a\), \(t = 0\) and when \(x = b\), \(t = 1\). The integral becomes: \[ \begin{align*} &\int_0^1 \frac{(b - a)dt}{\sqrt{[(a + (b - a)t) - a][b - (a + (b - a)t)]}}\\ =&\int_0^1 \frac{(b - a)dt}{\sqrt{(b - a)t(b - a)(1 - t)}}\\ =&\int_0^1 \frac{dt}{\sqrt{t(1 - t)}} \end{align*} \] This is a well-known integral and its value is \(\pi\). So the answer is \(\pi \approx 3.142\) (rounded to the thousandths decimal). |
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| minimax | To solve the integral \(\int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}\), we can recognize that this is a standard integral of the form of an arcsine integral. The integral of the form \(\int \frac{dt}{\sqrt{(t-a)(b-t)}}\) is known to result in \(\arcsin\left(\frac{t-a}{b-a}\right)\) evaluated from \(a\) to \(b\). Let's apply this to our integral: \[ \int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}} = \left[\arcsin\left(\frac{x-a}{b-a}\right)\right]_a^b \] Now, we evaluate the antiderivative at the bounds: \[ \left[\arcsin\left(\frac{x-a}{b-a}\right)\right]_a^b = \arcsin\left(\frac{b-a}{b-a}\right) - \arcsin\left(\frac{a-a}{b-a}\right) \] Simplifying the terms inside the arcsine: \[ = \arcsin(1) - \arcsin(0) \] We know that \(\arcsin(1) = \frac{\pi}{2}\) and \(\arcsin(0) = 0\), so: \[ = \frac{\pi}{2} - 0 \] \[ = \frac{\pi}{2} \] Therefore, the value of the integral is \(\frac{\pi}{2}\). Since \(\pi\) is approximately 3.14159, the value rounded to the thousandths decimal place is: \[ \frac{\pi}{2} \approx 1.570 \] So, the final answer is 1.570. |
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| kimi | The integral you've provided is of the form: \[ \int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}} \] This is a standard form of the elliptic integral of the first kind, which is defined as: \[ \int_0^{\pi/2} \frac{d\theta}{\sqrt{1 - k^2 \sin^2 \theta}} \] where \( k \) is the modulus of the elliptic integral. The integral you've given can be transformed into this form by a substitution. Let's consider the substitution \( x = a + (b-a)\sin^2\theta \), which gives us \( dx = 2(b-a)\sin\theta\cos\theta d\theta \). When \( x = a \), \( \theta = 0 \) and when \( x = b \), \( \theta = \pi/2 \). The integral becomes: \[ \int_0^{\pi/2} \frac{2(b-a)\sin\theta\cos\theta d\theta}{\sqrt{(a+(b-a)\sin^2\theta - a)(b - a - (b-a)\sin^2\theta)}} \] Simplifying the denominator: \[ \sqrt{(b-a)(b-a)\sin^2\theta \cdot (b-a)\cos^2\theta} = (b-a)\sqrt{\sin^2\theta \cos^2\theta} = (b-a)\sin\theta\cos\theta \] So the integral simplifies to: \[ 2(b-a) \int_0^{\pi/2} \frac{d\theta}{\sin\theta\cos\theta} = 4(b-a) \int_0^{\pi/2} \frac{d\theta}{\sin 2\theta} \] Using the identity \( \sin 2\theta = 2\sin\theta\cos\theta \), we can further simplify: \[ 4(b-a) \int_0^{\pi/2} \frac{d\theta}{2\sin\theta\cos\theta} = 2(b-a) \int_0^{\pi/2} \frac{d\theta}{\sin\theta\cos\theta} \] Now, \( \int_0^{\pi/2} \frac{d\theta}{\sin\theta\cos\theta} \) is a standard integral that evaluates to \( \frac{\pi}{2} \), so: \[ 2(b-a) \cdot \frac{\pi}{2} = (b-a)\pi \] So the value of the integral is \( (b-a)\pi \). To round this to the thousands decimal, we just need to ensure that \( b-a \) is expressed in a way that the result is accurate to the thousands decimal place. For example, if \( a = 0 \) and \( b = 1 \), then the integral evaluates to \( \pi \), which is approximately \( 3.141592653589793 \). Rounded to the thousands decimal, it would be \( 3.142 \). |
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| wenxinyiyan | To solve this integral, we can use a trigonometric substitution. Let's assume that \( a < b \) and make the substitution \( x = a\cos^2\theta + b\sin^2\theta \). This substitution is chosen because it simplifies the radical in the denominator. First, find the derivative of \( x \) with respect to \( \theta \): \[ \frac{dx}{d\theta} = -2(b-a)\sin\theta\cos\theta \] Now, let's rewrite the integral using this substitution: \[ \int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}} = \int_0^{\frac{\pi}{2}} \frac{-2(b-a)\sin\theta\cos\theta}{\sqrt{(a\cos^2\theta + b\sin^2\theta - a)(b - a\cos^2\theta - b\sin^2\theta)}} d\theta \] Simplifying the denominator, we get: \[ = \int_0^{\frac{\pi}{2}} \frac{-2(b-a)\sin\theta\cos\theta}{\sqrt{(b-a)\sin^2\theta \cdot (b-a)\cos^2\theta}} d\theta \] \[ = \int_0^{\frac{\pi}{2}} \frac{-2(b-a)\sin\theta\cos\theta}{\sqrt{(b-a)^2\sin^2\theta\cos^2\theta}} d\theta \] \[ = \int_0^{\frac{\pi}{2}} \frac{-2(b-a)\sin\theta\cos\theta}{(b-a)\sin\theta\cos\theta} d\theta \] \[ = \int_0^{\frac{\pi}{2}} -2 d\theta \] \[ = [-2\theta]_0^{\frac{\pi}{2}} \] \[ = -\pi \] However, since we are integrating an absolute value (the square root is always non-negative), the actual result should be the positive value of \( \pi \). Rounding to three decimal places, the answer is approximately \( 3.142 \). Note: The answer is positive because the integral represents an area under a curve, which is always a positive quantity. The negative sign arises from the substitution but does not affect the final result since we are interested in the area's magnitude. Important: The solution assumes \( a < b \). If \( a > b \), adjustments to the substitution would be necessary. However, without loss of generality, we can assume \( a < b \) for this problem, as the integral would be the same if the limits were reversed due to the symmetry of the integrand. |
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| zhipu | To solve the integral \(\int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}\), we can use a trigonometric substitution. This integral is a standard form that can be transformed using the substitution \(x = a + (b-a) \sin^2 \theta\). However, a more straightforward approach is to recognize that this integral is the inverse sine function in disguise. The integral \(\int \frac{dx}{\sqrt{(x-a)(b-x)}}\) is known to be equal to \(\arcsin\left(\frac{2x - (a+b)}{b-a}\right) + C\). Evaluating this from \(a\) to \(b\): \[ \int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}} = \left[ \arcsin\left(\frac{2x - (a+b)}{b-a}\right) \right]_a^b \] Now, let's evaluate the limits: 1. When \(x = b\): \[ \arcsin\left(\frac{2b - (a+b)}{b-a}\right) = \arcsin\left(\frac{b - a}{b-a}\right) = \arcsin(1) = \frac{\pi}{2} \] 2. When \(x = a\): \[ \arcsin\left(\frac{2a - (a+b)}{b-a}\right) = \arcsin\left(\frac{a - b}{b-a}\right) = \arcsin(-1) = -\frac{\pi}{2} \] So, the integral becomes: \[ \int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}} = \left[ \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) \right] = \frac{\pi}{2} + \frac{\pi}{2} = \pi \] Rounding \(\pi\) to the thousands decimal place gives: \[ \pi \approx 3.142 \] Therefore, the value of the integral \(\int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}\) rounded to the thousands decimal place is \(3.142\). |
3.1415926 |
| gpto1 | To solve the integral: \[ I = \int_a^b \frac{dx}{\sqrt{(x - a)(b - x)}} \] We can perform a substitution to simplify the integral. **Substitution:** Let: \[ x = \frac{a + b}{2} + \frac{b - a}{2} \cos \theta \] This substitution maps the interval from \( x = a \) to \( x = b \) as \( \theta \) varies from \( \theta = \pi \) to \( \theta = 0 \). **Computing \( x - a \) and \( b - x \):** First, compute \( x - a \) and \( b - x \): \[ x - a = \frac{a + b}{2} + \frac{b - a}{2} \cos \theta - a = \frac{b - a}{2} (1 + \cos \theta) \] \[ b - x = b - \left( \frac{a + b}{2} + \frac{b - a}{2} \cos \theta \right) = \frac{b - a}{2} (1 - \cos \theta) \] **Simplifying the Square Root:** Now, compute the product: \[ (x - a)(b - x) = \left( \frac{b - a}{2} (1 + \cos \theta) \right) \left( \frac{b - a}{2} (1 - \cos \theta) \right) = \left( \frac{b - a}{2} \right)^2 (1 - \cos^2 \theta) = \left( \frac{b - a}{2} \right)^2 \sin^2 \theta \] Therefore, the square root is: \[ \sqrt{(x - a)(b - x)} = \left| \frac{b - a}{2} \sin \theta \right| \] Since \( \sin \theta \) is positive in the interval \( 0 \leq \theta \leq \pi \), we can drop the absolute value: \[ \sqrt{(x - a)(b - x)} = \frac{b - a}{2} \sin \theta \] **Computing \( dx \):** Differentiate \( x \) with respect to \( \theta \): \[ dx = -\frac{b - a}{2} \sin \theta \, d\theta \] **Setting up the Integral:** Substitute \( dx \) and \( \sqrt{(x - a)(b - x)} \) back into the integral: \[ I = \int_{\theta = \pi}^{\theta = 0} \frac{-\frac{b - a}{2} \sin \theta \, d\theta}{\frac{b - a}{2} \sin \theta} = \int_{\theta = \pi}^{\theta = 0} (-1) d\theta = \int_{0}^{\pi} d\theta \] **Evaluating the Integral:** \[ I = \int_{0}^{\pi} d\theta = \left[ \theta \right]_0^{\pi} = \pi \] **Conclusion:** Thus, the value of the integral is: \[ I = \pi \] Rounded to the thousandth decimal place: \[ I \approx 3.142 \] \(\boxed{3.142}\) |
3.1415926 |
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