Arc length of integration (1+8x)^1/2 from x=0 to x=1?

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Answer 1 · 2018-06-10T11:28:28

#L=1/8(15-sqrt17)+2ln((sqrt17-1)/2)# units.

Let

#f(x)=sqrt(1+8x)#

#f'(x)=4/sqrt(1+8x)#

Arc length is given by:

#L=int_0^1sqrt(1+16/(1+8x))dx#

Apply the substitution #1+8x=16u^2#:

#L=4int_(1/4)^(3/4)sqrt(u^2+1)du#

Apply the substitution #u=tantheta#:

#L=4intsec^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=2[secthetatantheta+ln|sectheta+tantheta|]#

Reverse the last substitution:

#L=2[usqrt(u^2+1)+ln|u+sqrt(u^2+1)|]_(1/4)^(3/4)#

Hence

#L=1/8(15-sqrt17)+2ln(8/(1+sqrt17))#

Simplify:

#L=1/8(15-sqrt17)+2ln((sqrt17-1)/2)#