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

1 Answer
Jun 20, 2018

#L=2+2sum_(n=1)^oosum_(m=0)^oo((1/2),(n))((-4n),(m))(-1)^m/((2n+2m+1)*4^(n+m))# units.

Explanation:

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

#f'(x)=(8x)/(4-x^2)^2#

Arc length is given by:

#L=int_-1^1sqrt(1+(64x^2)/(4-x^2)^4)dx#

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

#L=int_-1^1sum_(n=0)^oo((1/2),(n))((64x^2)/(4-x^2)^4)^ndx#

Isolate the #n=0# term and simplify:

#L=int_-1^1dx+sum_(n=1)^oo((1/2),(n))4^(3n)int_-1^1x^(2n)/(4-x^2)^(4n)dx#

Rearrange:

#L=2+sum_(n=1)^oo((1/2),(n))1/4^nint_-1^1x^(2n)(1-1/4x^2)^(-4n)dx#

For #x in [-1,1]#, #1/4x^2<=1#. Take another series expansion:

#L=2+sum_(n=1)^oo((1/2),(n))1/4^nint_-1^1x^(2n){sum_(m=0)^oo((-4n),(m))(-1/4x^2)^m}dx#

Simplify:

#L=2+sum_(n=1)^oosum_(m=0)^oo((1/2),(n))((-4n),(m))(-1)^m/4^(n+m)int_-1^1x^(2n+2m)dx#

Integrate directly:

#L=2+sum_(n=1)^oosum_(m=0)^oo((1/2),(n))((-4n),(m))(-1)^m/4^(n+m)[x^(2n+2m+1)]_-1^1/(2n+2m+1)#

Insert the limits of integration and simplify:

#L=2+2sum_(n=1)^oosum_(m=0)^oo((1/2),(n))((-4n),(m))(-1)^m/((2n+2m+1)*4^(n+m))#