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:

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

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