The arclength of y = 32² – 5 for 0 < = < 2 is: g= 3(0)* -5 = 6-5=-5 -) (o, -5) y = 3(2)^ -5 = f(x) = $xt-5 co,z] f^ Cx) = lex (2-5= 7 → (2,2) (2,7) la JT+ (f')* dx J. JI+ (Ex)= dx When u= ax, then du -ledx, Yielding x-o, and u = 6.2 = 12 12 +u2 du

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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Hi, I tried to solve this question but got it marked wrong.  I was given no partial credit either, so I was wondering how could I solve this correctly for the next time?  My answer was initially just the (square root of 145) - 1, but I thought that was too simple.  It looks like I overcomplicated it by overthinking it matched one of the common integral formulas.  Any guidance you could provide would be helpful.  Thank you.

The arclength of y = 32² – 5 for 0 < = < 2 is:
y= 3(0)* -5 = 6-5=-5 > (6, -5)
y = 3(2)^ -5
f(x) = 8x?-5 co,2]
f^ Cx) = lex
(2-5= 7
→ (3,2)
(a,7)
Ardlayth=
la JT+ (f')* dx
J. JI+ (Ex)= dx
then du =ledx,
When u= Cex,
Yielding
u = 6.0 c 0
and
u = 6.2 = 12
12
2 du
Jitun
2.
93
as SJa?+x=dx=
T+12
t3 In 12+
2
145+
S145-[ + ln ( 12 + Je45
Transcribed Image Text:The arclength of y = 32² – 5 for 0 < = < 2 is: y= 3(0)* -5 = 6-5=-5 > (6, -5) y = 3(2)^ -5 f(x) = 8x?-5 co,2] f^ Cx) = lex (2-5= 7 → (3,2) (a,7) Ardlayth= la JT+ (f')* dx J. JI+ (Ex)= dx then du =ledx, When u= Cex, Yielding u = 6.0 c 0 and u = 6.2 = 12 12 2 du Jitun 2. 93 as SJa?+x=dx= T+12 t3 In 12+ 2 145+ S145-[ + ln ( 12 + Je45
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