What is the arc length of f(x)=2/x^4-1/x^6 on x in [3,6]?
2 Answers
Explanation:
so we get
and we have to solve
this is
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
L=int_3^6sum_(n=0)^oo((1/2),(n))(64/x^10(1-3/(4x^2))^2)^ndx
Isolate the
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))