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Let C be the curve y = 2 ln(4 – x²), for – 0.3 < x < 0.8. A graph of y follows. - 2.79 2.76- 2.73 2.7 2.67 2.64- 2.61 2.58 2.55 2.52 2.49- 2.46

Let C be the curve y = 2 ln(4 – x²), for – 0.3 < x < 0.8. A graph of y follows. - 2.79 2.76- 2.73 2.7 2.67 2.64- 2.61 2.58 2.55 2.52 2.49- 2.46

Calculus: Early Transcendentals
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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Let \( C \) be the curve \( y = 2 \ln(4 - x^2) \), for \(-0.3 \leq x \leq 0.8\). A graph of \( y \) follows. ### Graph Explanation The graph illustrates the curve of the function \( y = 2 \ln(4 - x^2) \) between \( x = -0.3 \) and \( x = 0.8 \). The curve starts at \( y \approx 2.76 \) and decreases to \( y \approx 2.43 \). The endpoints of the curve are marked with blue points. ### Task Find the arc length of \( C \): \[ \int_{-0.3}^{0.8} \sqrt{1 + y'^2} \, dx \] #### Steps: 1. First find and simplify \( \sqrt{1 + y'^2} \). - Fill in: \(\sqrt{1 + y'^2} = \) [______] 2. Now integrate to find the arc length: \[ \int_{-0.3}^{0.8} \sqrt{1 + y'^2} \, dx = \) [______]
Transcribed Image Text:Let \( C \) be the curve \( y = 2 \ln(4 - x^2) \), for \(-0.3 \leq x \leq 0.8\). A graph of \( y \) follows. ### Graph Explanation The graph illustrates the curve of the function \( y = 2 \ln(4 - x^2) \) between \( x = -0.3 \) and \( x = 0.8 \). The curve starts at \( y \approx 2.76 \) and decreases to \( y \approx 2.43 \). The endpoints of the curve are marked with blue points. ### Task Find the arc length of \( C \): \[ \int_{-0.3}^{0.8} \sqrt{1 + y'^2} \, dx \] #### Steps: 1. First find and simplify \( \sqrt{1 + y'^2} \). - Fill in: \(\sqrt{1 + y'^2} = \) [______] 2. Now integrate to find the arc length: \[ \int_{-0.3}^{0.8} \sqrt{1 + y'^2} \, dx = \) [______]
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