Find the arc length of the graph of the function over the given interval. y = In x, [1, 8] Step 1 The formula for the arc length of a curve y = f(x) over the interval [a, b] is b S = = [° √ Ja The arc length of the curve y = In S = Therefore, f(x) = In S S = 1 + [f '(x)] f'(x) = The arc length is represented as follows. 8 1 - L₁³√₁+· 1 = - £°√ 8 = 6³. X X X √ 1 + [f'(x)]² dx x² x² + 1 dx dx. + 1 dx dx over the interval [1, 8] is
Find the arc length of the graph of the function over the given interval. y = In x, [1, 8] Step 1 The formula for the arc length of a curve y = f(x) over the interval [a, b] is b S = = [° √ Ja The arc length of the curve y = In S = Therefore, f(x) = In S S = 1 + [f '(x)] f'(x) = The arc length is represented as follows. 8 1 - L₁³√₁+· 1 = - £°√ 8 = 6³. X X X √ 1 + [f'(x)]² dx x² x² + 1 dx dx. + 1 dx dx over the interval [1, 8] is
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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![### Tutorial Exercise
**Objective:** Find the arc length of the graph of the function over the given interval.
Given:
\[ y = \ln x, \quad [1, 8] \]
### Step 1
The formula for the arc length of a curve \( y = f(x) \) over the interval \([a, b]\) is:
\[ s = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} \, dx \]
For the given problem, we need to determine the arc length of the curve \( y = \ln x \) over the interval \([1, 8]\):
\[ s = \int_{1}^{8} \sqrt{1 + [f'(x)]^2} \, dx \]
To proceed, we first need to identify the derivative of the function \( f(x) \):
\[ f(x) = \ln x \]
Therefore,
\[ f'(x) = \frac{1}{x} \]
Now we substitute \( f'(x) \) into the arc length formula:
\[ s = \int_{1}^{8} \sqrt{1 + \left( \frac{1}{x} \right)^2} \, dx \]
Simplify the integrand:
\[ s = \int_{1}^{8} \sqrt{1 + \frac{1}{x^2}} \, dx \]
\[ = \int_{1}^{8} \sqrt{\frac{x^2 + 1}{x^2}} \, dx \]
\[ = \int_{1}^{8} \frac{\sqrt{x^2 + 1}}{x} \, dx \]
Thus, the arc length \( s \) is represented as follows:
\[ s = \int_{1}^{8} \sqrt{x^2 + 1} \, dx \]
This integral accounts for the arc length of \( y = \ln x \) from \( x = 1 \) to \( x = 8 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fecefb535-634b-4d72-abac-7d9b98194790%2Fe35ab992-008d-42a1-a565-c39e32055b32%2F60bdr1_processed.png&w=3840&q=75)
Transcribed Image Text:### Tutorial Exercise
**Objective:** Find the arc length of the graph of the function over the given interval.
Given:
\[ y = \ln x, \quad [1, 8] \]
### Step 1
The formula for the arc length of a curve \( y = f(x) \) over the interval \([a, b]\) is:
\[ s = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} \, dx \]
For the given problem, we need to determine the arc length of the curve \( y = \ln x \) over the interval \([1, 8]\):
\[ s = \int_{1}^{8} \sqrt{1 + [f'(x)]^2} \, dx \]
To proceed, we first need to identify the derivative of the function \( f(x) \):
\[ f(x) = \ln x \]
Therefore,
\[ f'(x) = \frac{1}{x} \]
Now we substitute \( f'(x) \) into the arc length formula:
\[ s = \int_{1}^{8} \sqrt{1 + \left( \frac{1}{x} \right)^2} \, dx \]
Simplify the integrand:
\[ s = \int_{1}^{8} \sqrt{1 + \frac{1}{x^2}} \, dx \]
\[ = \int_{1}^{8} \sqrt{\frac{x^2 + 1}{x^2}} \, dx \]
\[ = \int_{1}^{8} \frac{\sqrt{x^2 + 1}}{x} \, dx \]
Thus, the arc length \( s \) is represented as follows:
\[ s = \int_{1}^{8} \sqrt{x^2 + 1} \, dx \]
This integral accounts for the arc length of \( y = \ln x \) from \( x = 1 \) to \( x = 8 \).
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