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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 6 - Applications of IntegrationProblem 6.R.60
Chapter 6, Problem 6.R.60

58–61. Arc length Find the length of the following curves.
y = x³/6 + 1/2x on [1,2]

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Identify the function given: \(y = \frac{x^3}{6} + \frac{1}{2}x\) and the interval \([1, 2]\) over which we want to find the arc length.
Recall the formula for the arc length \(L\) of a curve \(y = f(x)\) from \(x = a\) to \(x = b\): \(L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\).
Compute the derivative \(\frac{dy}{dx}\) of the function: \(\frac{dy}{dx} = \frac{d}{dx} \left( \frac{x^3}{6} + \frac{1}{2}x \right)\).
Square the derivative to get \(\left(\frac{dy}{dx}\right)^2\) and then add 1 inside the square root: \(1 + \left(\frac{dy}{dx}\right)^2\).
Set up the integral for the arc length: \(L = \int_1^2 \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\) and prepare to evaluate or simplify this integral.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Arc Length Formula

The arc length of a curve y = f(x) from x = a to x = b is found using the integral formula L = ∫_a^b √(1 + (dy/dx)²) dx. This formula sums the lengths of infinitesimal line segments along the curve, providing the total distance traveled along it.
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Arc Length of Parametric Curves

Derivative of the Function

To apply the arc length formula, you need the derivative dy/dx of the function y = x³/6 + 1/2 x. The derivative represents the slope of the curve at any point and is essential for calculating the integrand √(1 + (dy/dx)²).
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Derivatives of Other Trig Functions

Definite Integration

After finding the integrand, you evaluate the definite integral from x = 1 to x = 2. This process involves integrating the function √(1 + (dy/dx)²) over the interval to compute the exact length of the curve segment.
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Definition of the Definite Integral
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