What is the arc length of #f(x)=x^2-2x+35# on #x in [1,7]#?

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Answer 1 · 2018-03-07T17:08:08

The arc length is #3sqrt(145)+1/4ln(12+sqrt145)# units.

#y=x^2-2x+35#

#y'=2x-2#

Arc length is given by:

#L=int_1^7sqrt(1+(2x-2)^2)dx#

Apply the substitution #2x-2=tantheta#:

#L=1/2intsec^3thetad theta#

This is a known integral. If you do not have it memorized apply integration by parts or look it up in a table of integrals:

#L=1/4[secthetatantheta+ln|sectheta+tantheta|]#

Reverse the substitution:

#L=1/4[(2x-2)sqrt(1+(2x-2)^2)+ln|(2x-2)+sqrt(1+(2x-2)^2)|]_1^7#

Insert the limits of integration:

#L=3sqrt(145)+1/4ln(12+sqrt145)#