Question

Calculate the arc length of the curve \( y = x^{2} \) from \( x = 1 \) to \( x = 3 \).

Ask by Fowler Evans. in Uzbekistan
Feb 04,2025

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Answer

Egrining arka uzunligi **8.27 birlik** ga teng.

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To calculate the arc length of the curve \( y = x^{2} \) from \( x = 1 \) to \( x = 3 \), we use the arc length formula given by: \[ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^{2}} \, dx \] First, we find the derivative \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = 2x \] Next, we compute \( \left(\frac{dy}{dx}\right)^{2} \): \[ \left(\frac{dy}{dx}\right)^{2} = (2x)^{2} = 4x^{2} \] Now, we can substitute this into the arc length formula: \[ L = \int_{1}^{3} \sqrt{1 + 4x^{2}} \, dx \] To solve this integral, we can use a trigonometric substitution or recognize it as a standard form. However, we can also evaluate it directly: 1. Let's calculate the integral \( \sqrt{1 + 4x^{2}} \): Using the substitution \( u = 2x \longrightarrow du = 2dx \longrightarrow dx = \frac{du}{2} \), we change the limits: - When \( x = 1 \), \( u = 2 \). - When \( x = 3 \), \( u = 6 \). The integral transforms to: \[ L = \int_{2}^{6} \sqrt{1 + u^{2}} \cdot \frac{du}{2} \] This gives: \[ L = \frac{1}{2} \int_{2}^{6} \sqrt{1 + u^{2}} \, du \] The integral \( \int \sqrt{1 + u^{2}} \, du = \frac{u}{2} \sqrt{1 + u^{2}} + \frac{1}{2} \ln \left( u + \sqrt{1 + u^{2}} \right) + C \). Now we evaluate it from 2 to 6: \[ L = \frac{1}{2} \left[ \left( \frac{6}{2} \sqrt{1 + 6^{2}} + \frac{1}{2} \ln(6 + \sqrt{1 + 6^{2}}) \right) - \left( \frac{2}{2} \sqrt{1 + 2^{2}} + \frac{1}{2} \ln(2 + \sqrt{1 + 2^{2}}) \right) \right] \] Calculating those: - For \( u = 6 \): \[ \frac{6}{2} \sqrt{37} + \frac{1}{2} \ln(6 + \sqrt{37}) \] - For \( u = 2 \): \[ 1 \sqrt{5} + \frac{1}{2} \ln(2 + \sqrt{5}) \] Combining these results gives: \[ L = \frac{1}{2} \left[ 3 \sqrt{37} + \frac{1}{2} \ln(6 + \sqrt{37}) - (\sqrt{5} + \frac{1}{2} \ln(2 + \sqrt{5})) \right] \] Calculating these, you will get the arc length; however, for layman's terms, you can also use numerical methods or calculators to find the final value. Thus, \(\boxed{L} \) approx will yield the actual arc length from \( x = 1 \) to \( x = 3 \).

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