What is the arclength of f(x)=3x^2-x+4 on x in [2,3]?
1 Answer
Mar 17, 2018
Explanation:
f(x)=3x^2−x+4
f'(x)=6x−1
Arclength is given by:
L=int_2^3sqrt(1+(6x-1)^2)dx
Apply the substitution
L=1/6intsec^3thetad theta
This is a known integral:
L=1/12[secthetatantheta+ln|sectheta+tantheta|]
Reverse the substitution:
L=1/12[(6x-1)sqrt(1+(6x-1)^2)+ln|(6x-1)+sqrt(1+(6x-1)^2)|]_2^3
Hence
L=1/12(17sqrt290-11sqrt122)+1/12ln((17+sqrt290)/(11+sqrt122))