What is the arc length of #f(x)= sqrt(5x+1) # on #x in [0,2]#?

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Answer 1 · 2018-05-28T12:25:39

#L=1/10(sqrt759-sqrt29)+5/4ln((2sqrt11+sqrt69)/(2+sqrt29))# units.

#f(x)=sqrt(5x+1)#

#f'(x)=5/(2sqrt(5x+1))#

Arc length is given by:

#L=int_0^2sqrt(1+25/(4(5x+1)))dx#

Apply the substitution #5x+1=u#:

#L=1/5int_1^11sqrt(1+25/(4u))du#

Rearrange:

#L=1/5int_1^11sqrt(4u+25)/(2sqrtu)du#

Apply the substitution #sqrtu=v#:

#L=1/5int_1^sqrt11sqrt(4v^2+25)dv#

Apply the substitution #2v=5tantheta#:

#L=5/2intsec^3thetad theta#

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

#L=5/4[secthetatantheta+ln|sectheta+tantheta|]#

Reverse the last substitution:

#L=[1/10vsqrt(4v^2+25)+5/4ln|2v+sqrt(4v^2+25)|]_1^sqrt11#

Insert the limits of integration:

#L=1/10(sqrt759-sqrt29)+5/4ln((2sqrt11+sqrt69)/(2+sqrt29))#