What is the arclength of f(x)=3x^2-x+4 on x in [2,3]?

1 Answer
Mar 17, 2018

L=1/12(17sqrt290-11sqrt122)+1/12ln((17+sqrt290)/(11+sqrt122)) units.

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 6x-1=tantheta:

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))