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

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Answer 1 · 2018-03-17T16:23:26

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

#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))#