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

Question

1488 views around the world

You can reuse this answer
Creative Commons License

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))#