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

1 Answer
May 25, 2018

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

Explanation:

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))