What is the arc length of $f(x)=lnx$ in the interval $[1,5]$ ?

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Answer 1 · 2018-03-07T17:25:47

The arc length is $sqrt26-sqrt2+ln5-ln((1+sqrt26)/(1+sqrt2))$ units.

$f(x)=lnx$

$f'(x)=1/x$

Arc length is given by:

$L=int_1^5sqrt(1+1/x^2)dx$

Rearrange:

$L=int_1^5sqrt(x^2+1)/xdx$

Apply the substitution $x=tantheta$ :

$L=intsectheta/tantheta*sec^2thetad theta$

Rewrite as:

$L=intcsctheta*(tan^2theta+1)d theta$

Hence

$L=int(secthetatantheta+csctheta)d theta$

Integrate term by term:

$L=[sectheta-ln|csctheta+cottheta|]$

Reverse the substitution:

$L=[sqrt(1+x^2)-ln|(1+sqrt(1+x^2))/x|]_1^5$

Insert the limits of integration:

$L=sqrt26-sqrt2+ln5-ln((1+sqrt26)/(1+sqrt2))$

What is the arc length of f(x)=lnx f(x)=lnx in the interval [1,5][1,5]?