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a) y = cos 2x on [0,1] b) x= on [1,10]

a) y = cos 2x on [0,1] b) x= on [1,10]

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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Please solve 6, a and b, thanks!
OCT 18 n[-2,6] a) x = -8y-3 on a) 8√65 b) 168 b) y = x³/2 on [0,60] 6. Set up the integral to determine the arc length of each curve, then integrate in your calculator a) y = cos 2x on [0,1] b) x == on [1,10] y a) 5.27 b) 9.15 7. Find the given definite and indefinite integrals, some of them will require substitution, integration by parts, and/or a combination of both. a) 5x²√x³-2dx b) fox√x + 1dx A t MacBook Pro W x lnx d) tv
Transcribed Image Text:OCT 18 n[-2,6] a) x = -8y-3 on a) 8√65 b) 168 b) y = x³/2 on [0,60] 6. Set up the integral to determine the arc length of each curve, then integrate in your calculator a) y = cos 2x on [0,1] b) x == on [1,10] y a) 5.27 b) 9.15 7. Find the given definite and indefinite integrals, some of them will require substitution, integration by parts, and/or a combination of both. a) 5x²√x³-2dx b) fox√x + 1dx A t MacBook Pro W x lnx d) tv
Expert Solution
Step 1

6)

a) y=cos(2x) on [0,π]

the length of the curve is given by 

L=∫ab(1+f'(x)2) dx

here f(x)=cos(2x)

then f'(x)=-2sin(2x)

then the arc length is ∫0π(1+(-2sin(2x))2 dx=∫0π(3-2cos(4x)) dx=5.27

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