What is the arc length of f(x)= (3x-2)^2 on x in [1,3] ?
1 Answer
May 25, 2018
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
L=1/6int_2^14sqrt(1+u^2)du
Apply the substitution
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))