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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 8 - Integration TechniquesProblem 8.R.114a
Chapter 8, Problem 8.R.114a

114. {Use of Tech} Arc length of the natural logarithm Consider the curve y = ln(x).
a. Find the length of the curve from x = 1 to x = a and call it L(a).
(Hint: The change of variables u = sqrt(x^2 + 1) allows evaluation by partial fractions.)

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1
Recall the formula for the arc length of a curve defined by y = f(x) from x = a to x = b: \[L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\]
For the curve y = \(\ln\)(x), compute the derivative: \[\frac{dy}{dx} = \frac{1}{x}\]
Substitute the derivative into the arc length formula: \[L(a) = \int_1^a \sqrt{1 + \left(\frac{1}{x}\right)^2} \, dx = \int_1^a \sqrt{1 + \frac{1}{x^2}} \, dx = \int_1^a \sqrt{\frac{x^2 + 1}{x^2}} \, dx\]
Simplify the integrand: \[L(a) = \int_1^a \frac{\sqrt{x^2 + 1}}{x} \, dx\]
Use the substitution suggested: let \[u = \sqrt{x^2 + 1}\] Then express dx and x in terms of u to rewrite the integral and evaluate it using partial fractions.

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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 given by the integral L = ∫_a^b √(1 + (dy/dx)^2) dx. This formula measures the distance along the curve by summing infinitesimal line segments, requiring the derivative of the function to account for the slope.
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Arc Length of Parametric Curves

Derivative of the Natural Logarithm Function

For y = ln(x), the derivative dy/dx = 1/x. This derivative is essential in the arc length formula to find the integrand √(1 + (1/x)^2) = √(1 + 1/x^2), which simplifies the integral needed to compute the curve's length.
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Derivative of the Natural Logarithmic Function

Integration Techniques: Substitution and Partial Fractions

To evaluate the arc length integral, substitution (u = √(x^2 + 1)) simplifies the integrand, transforming it into a rational function. Partial fraction decomposition then breaks this rational function into simpler terms, making the integral solvable using standard methods.
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Integration Using Partial Fractions
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