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:

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

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