What is the arc length of f(x)=2/x^4-1/x^6 on x in [3,6]?

2 Answers
Jun 23, 2018

approx 3.000205746

Explanation:

f(x)=2*x^(-4)-x^(-6)
so we get

f'(x)=8x^(-5)+6x^(-7)

and we have to solve

int_3^6sqrt(1+(8x^(-5)+6x^(-7))^2)dx
this is approx 3.0002057846

Jun 25, 2018

L=3+1/3sum_(n=1)^oosum_(m=0)^(2n)((1/2),(n))((2n),(m))1/(1+10n+2m)(8/243)^(2n)((-1)/12)^m(1-1/2^(1+10n+2m)) units.

Explanation:

f(x)=2/x^4-1/x^6

f'(x)=-8/x^5(1-3/(4x^2))

Arc length is given by:

L=int_3^6sqrt(1+64/x^10(1-3/(4x^2))^2)dx

For x in [3,6], 64/x^10(1-3/(4x^2))^2<1. Take the series expansion of the square root:

L=int_3^6sum_(n=0)^oo((1/2),(n))(64/x^10(1-3/(4x^2))^2)^ndx

Isolate the n=0 term and simplify:

L=int_3^6dx+sum_(n=1)^oo((1/2),(n))2^(6n)int_3^6 1/x^(10n)(1-3/(4x^2))^(2n)dx

Apply binomial expansion:

L=3+sum_(n=1)^oo((1/2),(n))2^(6n)int_3^6 1/x^(10n){sum_(m=0)^(2n)((2n),(m))(-3/(4x^2))^m}dx

Rearrange:

L=3+sum_(n=1)^oosum_(m=0)^(2n)((1/2),(n))((2n),(m))2^(6n)((-3)/4)^m int_3^6 1/x^(10n+2m)dx

Integrate directly:

L=3+sum_(n=1)^oosum_(m=0)^(2n)((1/2),(n))((2n),(m))2^(6n)/(1+10n+2m)((-3)/4)^m[(-1)/x^(1+10n+2m)]_3^6

Insert the limits of integration:

L=3+sum_(n=1)^oosum_(m=0)^(2n)((1/2),(n))((2n),(m))2^(6n)/(1+10n+2m)((-3)/4)^m(1/3^(1+10n+2m)-1/6^(1+10n+2m))

Simplify:

L=3+1/3sum_(n=1)^oosum_(m=0)^(2n)((1/2),(n))((2n),(m))1/(1+10n+2m)(8/243)^(2n)((-1)/12)^m(1-1/2^(1+10n+2m))