What is the arclength of #f(x)=ln(x+3)# on #x in [2,3]#?

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Answer 1 · 2018-06-17T12:16:10

#L=1+sum_(n=1)^oo((1/2),(n))1/(2n-1)(1/5^(2n-1)-1/6^(2n-1))# units.

#f(x)=ln(x+3)#

#f'(x)=1/(x+3)#

Arclength is given by:

#L=int_2^3sqrt(1+1/(x+3)^2)dx#

Apply the substitution #x+3=u#:

#L=int_5^6sqrt(1+1/u^2)du#

For #u in [5,6]#, #1/u^2<1#. Take the series expansion of the square root:

#L=int_5^6sum_(n=0)^oo((1/2),(n))1/u^(2n)du#

Isolate the #n=0# term and simplify:

#L=int_5^6du+sum_(n=1)^oo((1/2),(n))int_5^6 1/u^(2n)du#

Integrate directly:

#L=1+sum_(n=1)^oo((1/2),(n))[(-1)/((2n-1)u^(2n-1))]_5^6#

Simplify:

#L=1+sum_(n=1)^oo((1/2),(n))1/(2n-1)(1/5^(2n-1)-1/6^(2n-1))#