Set up an integral for the length of the curve. y=√₁-x²₁ - sxs ² 1/4 -1/4 5-4x4 4(1-x4) 1/4 -1/4 dx 1/4 1₁/41 4(1-x4) 1/4 4+16x6 4-4x4+16x6 dx 4-4x4+4x3 4(1-x4) dx 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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**Set up an integral for the length of the curve.**

Given the function:

\[ y = \sqrt{1-x^4}, \quad -\frac{1}{4} \leq x \leq \frac{1}{4} \]

Select the correct integral that represents the length of the curve:

1. \[ \int_{-1/4}^{1/4} \sqrt{\frac{5 - 4x^4}{4(1-x^4)}} \, dx \]

2. \[ \int_{-1/4}^{1/4} \sqrt{\frac{4 - 4x^4 + 16x^6}{4(1-x^4)}} \, dx \]

3. \[ \int_{-1/4}^{1/4} \sqrt{\frac{4 + 16x^6}{4}} \, dx \]

4. \[ \int_{-1/4}^{1/4} \sqrt{\frac{4 - 4x^4 + 4x^3}{4(1-x^4)}} \, dx \]

Each integral is structured to determine the arc length of the given curve over the specified interval. Ensure to verify the correct integral setup.
Transcribed Image Text:**Set up an integral for the length of the curve.** Given the function: \[ y = \sqrt{1-x^4}, \quad -\frac{1}{4} \leq x \leq \frac{1}{4} \] Select the correct integral that represents the length of the curve: 1. \[ \int_{-1/4}^{1/4} \sqrt{\frac{5 - 4x^4}{4(1-x^4)}} \, dx \] 2. \[ \int_{-1/4}^{1/4} \sqrt{\frac{4 - 4x^4 + 16x^6}{4(1-x^4)}} \, dx \] 3. \[ \int_{-1/4}^{1/4} \sqrt{\frac{4 + 16x^6}{4}} \, dx \] 4. \[ \int_{-1/4}^{1/4} \sqrt{\frac{4 - 4x^4 + 4x^3}{4(1-x^4)}} \, dx \] Each integral is structured to determine the arc length of the given curve over the specified interval. Ensure to verify the correct integral setup.
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