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

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Answer 1 · 2018-05-22T16:29:25

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

#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 #2+x=u#:

#L=int_3^4sqrt(1+1/u^4)du#

For #u in [3,4]#, #1/u^4<1#. Take the series expansion of the square root:

#L=int_3^4sum_(n=0)^oo((1/2),(n))1/u^(4n)du#

Isolate the #n=0# term and simplify:

#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))#