What is the arc length of #f(x)= 1/(2+x) # on #x in [1,2] #?
1 Answer
May 22, 2018
Explanation:
#f(x)=1/(2+x)#
#f'(x)=-1/(2+x)^2#
Arc length is given by:
#L=int_1^2sqrt(1+1/(2+x)^4)dx#
Apply the substitution
#L=int_3^4sqrt(1+1/u^4)du#
For
#L=int_3^4sum_(n=0)^oo((1/2),(n))1/u^(4n)du#
Isolate the
#L=int_3^4du+sum_(n=1)^oo((1/2),(n))int_3^4 1/u^(4n)du#
The remaining integrals are trivial:
#L=1+sum_(n=1)^oo((1/2),(n))1/(1-4n)[1/u^(4n-1)]_3^4#
Simplify:
#L=1+sum_(n=1)^oo((1/2),(n))1/(4n-1)(1/3^(4n-1)-1/4^(4n-1))#