Question 3 The arclength of y = 4x² − 2 from x = 0 to x = 2 is given by: 2 S₁² OL= OL= [²√1 + 64z²dz =√₁²√1 +8z²dz -1.²- OL= OL= OL Using the substitution z = 8z to re-write the integral in terms of u yields: 16 Stº f® √1 +4u²du 0 = √1+8xdx OL= 8 √1+64zdz OL= 16 1 8 0 √1 + 16 L = = √ ₁ + x² du 0 8udu -16 OL=√1+udu The value of this integral is: (You may need to use a table or technology to evaluate this integral.) Question Help: D Post to forum Submit Question
Question 3 The arclength of y = 4x² − 2 from x = 0 to x = 2 is given by: 2 S₁² OL= OL= [²√1 + 64z²dz =√₁²√1 +8z²dz -1.²- OL= OL= OL Using the substitution z = 8z to re-write the integral in terms of u yields: 16 Stº f® √1 +4u²du 0 = √1+8xdx OL= 8 √1+64zdz OL= 16 1 8 0 √1 + 16 L = = √ ₁ + x² du 0 8udu -16 OL=√1+udu The value of this integral is: (You may need to use a table or technology to evaluate this integral.) Question Help: D Post to forum Submit Question
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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Transcribed Image Text:Question 3
The arclength of y = 4x² - 2 from x = 0 to = 2 is given by:
-2
OL= √1+8xdx
-fo √1+64r²dx
-S₁²V₁-
+ -8x²dx
-1.²-
OL=
OL=
OL=
OL =
8
Using the substitution = 8x to re-write the integral in terms of u yields:
16
f® √1 +4u²du
0
OL
OL=
1
OL=
√1+64rdz
8
16
S
0
1
16
0
16
√1 +8udu
√₁ + u²du
√1+ udu
K
The value of this integral is:
(You may need to use a table or technology to evaluate this integral.)
Question Help: D Post to forum
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