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

1 Answer
Jun 10, 2018

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

Explanation:

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)