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Find the arc length of y = (x+ 6) + (x + 6)- for 1 < x < 2. (Give your answer to three decimal places.) S =

Find the arc length of y = (x+ 6) + (x + 6)- for 1 < x < 2. (Give your answer to three decimal places.) S =

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Find the arc length of y = 1/12(x+6)^3+(x+6)^-1 over [1,2]

**Problem Statement:** Find the arc length of the function: \[ y = \frac{1}{12}(x + 6)^3 + (x + 6)^{-1} \] for the interval \( 1 \leq x \leq 2 \). (Give your answer to three decimal places.) **Solution:** To calculate the arc length \( s \) of the curve, use the formula for arc length from calculus, which is: \[ s = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] 1. **Differentiate the function** \( y \) with respect to \( x \) to find \(\frac{dy}{dx}\). 2. **Substitute** \(\frac{dy}{dx}\) into the arc length formula. 3. **Evaluate** the integral from \( x = 1 \) to \( x = 2 \). 4. **Provide** the answer to three decimal places.
Transcribed Image Text:**Problem Statement:** Find the arc length of the function: \[ y = \frac{1}{12}(x + 6)^3 + (x + 6)^{-1} \] for the interval \( 1 \leq x \leq 2 \). (Give your answer to three decimal places.) **Solution:** To calculate the arc length \( s \) of the curve, use the formula for arc length from calculus, which is: \[ s = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] 1. **Differentiate the function** \( y \) with respect to \( x \) to find \(\frac{dy}{dx}\). 2. **Substitute** \(\frac{dy}{dx}\) into the arc length formula. 3. **Evaluate** the integral from \( x = 1 \) to \( x = 2 \). 4. **Provide** the answer to three decimal places.
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