Question

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

Answer 1

`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`

Answer 2

`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))`