What is the arc length of #f(x)= (3x-2)^2 # on #x in [1,3] #?

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Answer 1 · 2018-05-25T05:40:09

#L=1/6(7sqrt197-sqrt5)+1/12ln((14+sqrt197)/(2sqrt5))# units.

#f(x)=(3x−2)^2#

#f'(x)=6x−4#

Arc length is given by:

#L=int_1^3sqrt(1+(6x-4)^2)dx#

Apply the substitution #6x-4=u#:

#L=1/6int_2^14sqrt(1+u^2)du#

Apply the substitution #u=tantheta#:

#L=1/6intsec^3thetad theta#

This is a known integral:

#L=1/12[secthetatantheta+ln|sectheta+tantheta|]#

Reverse the last substitution:

#L=1/12[usqrt(1+u^2)+ln|u+sqrt(1+u^2)|]_2^14#

Hence

#L=1/6(7sqrt197-sqrt5)+1/12ln((14+sqrt197)/(2sqrt5))#