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:

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

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