Arc length of integration (1+8x)^1/2 from x=0 to x=1?
1 Answer
Jun 10, 2018
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
L=4int_(1/4)^(3/4)sqrt(u^2+1)du
Apply the substitution
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)