Complete the steps needed to find the arc length of the following function. 1 y = 2² +²e=²₁ 0≤ x ≤b 0≤x≤ е 8 b L = - Lºv √ 1 + (y) ² dx =-1₁° v 1+ (2e² - e-²) ² dx 1 е 8 1 64 =√₁² √ =... 1+ (4e² - 1 2 + -e-²x) dx

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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**Complete the steps needed to find the arc length of the following function.**

Given the function:

\[ y = 2e^x + \frac{1}{8}e^{-x}, \quad 0 \leq x \leq b \]

The arc length \( L \) is calculated by the formula:

\[ L = \int_{1}^{b} \sqrt{1 + (y')^2} \, dx \]

**Steps:**

1. Substitute the derivative \( y' \):
   \[ L = \int_{1}^{b} \sqrt{1 + \left(2e^x - \frac{1}{8}e^{-x}\right)^2} \, dx \]

2. Simplify the expression inside the integral:
   \[ = \int_{1}^{b} \sqrt{1 + \left(4e^{2x} - \frac{1}{2} + \frac{1}{64}e^{-2x}\right)} \, dx \]

3. Continue with further simplification and integration as needed.
Transcribed Image Text:**Complete the steps needed to find the arc length of the following function.** Given the function: \[ y = 2e^x + \frac{1}{8}e^{-x}, \quad 0 \leq x \leq b \] The arc length \( L \) is calculated by the formula: \[ L = \int_{1}^{b} \sqrt{1 + (y')^2} \, dx \] **Steps:** 1. Substitute the derivative \( y' \): \[ L = \int_{1}^{b} \sqrt{1 + \left(2e^x - \frac{1}{8}e^{-x}\right)^2} \, dx \] 2. Simplify the expression inside the integral: \[ = \int_{1}^{b} \sqrt{1 + \left(4e^{2x} - \frac{1}{2} + \frac{1}{64}e^{-2x}\right)} \, dx \] 3. Continue with further simplification and integration as needed.
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