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

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Answer 1 · 2018-06-28T05:35:51

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

#f(x)=1/x#

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

Arc length is given by:

#L=int_1^2sqrt(1+1/x^4)dx#

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

#L=int_1^2sum_(n=0)^oo((1/2),(n))1/x^(4n)dx#

Simplify:

#L=sum_(n=0)^oo((1/2),(n))int_1^2x^(-4n)dx#

Integrate directly:

#L=sum_(n=0)^oo((1/2),(n))[x^(1-4n)]_1^2/(1-4n)#

Isolate the #n=0# term and simplify:

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