Question

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

Answer 1

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

Explanation:

`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))`