What is the arc length of f(x)=xsqrt(x^2-1) on x in [3,4] ?

1 Answer
May 20, 2018

L=7+1/8ln(15/8)+sum_(n=2)^oo((1/2),(n))1/4^n(1/8^(n-1)-1/15^(n-1))

Explanation:

f(x)=xsqrt(x^2-1)

f'(x)=(2x^2-1)/sqrt(x^2-1)

Arc length is given by:

L=int_3^4sqrt(1+(2x^2-1)^2/(x^2-1))dx

Simplify:

L=int_3^4xsqrt((4x^2-3)/(x^2-1))dx

Rearrange:

L=int_3^4(2x)sqrt(1+1/(4(x^2-1)))dx

Apply the substitution x^2-1=u:

L=int_8^15sqrt(1+1/(4u))du

For u in [8,15], 1/(4u)<1. Take the series expansion of the square root:

L=int_8^15sum_(n=0)^oo((1/2),(n))1/(4u)^ndu

Isolate the n=0 and n=1 terms:

L=int_8^15(1+1/(8u))du+sum_(n=2)^oo((1/2),(n))1/4^nint_8^15 1/u^ndu

Integrate term by term:

L=[u+1/8lnu]_ 8^15+sum_(n=2)^oo((1/2),(n))1/4^n[-1/u^(n-1)]_8^15

Insert the limits of integration:

L=7+1/8ln(15/8)+sum_(n=2)^oo((1/2),(n))1/4^n(1/8^(n-1)-1/15^(n-1))