Question

How do you find the arc length of the curve `y=lnx` over the interval [1,2]?

Answer 1

Apply the arc length formula.

Explanation:

`y=lnx`

`y'=1/x`

Arc length is given by:

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

Rearrange:

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

Multiply numerator and denominator by `sqrt(x^2+1)`:

`L=int_1^2(x^2+1)/(xsqrt(x^2+1))dx`

Integration is distributive:

`L=int_1^2x/sqrt(x^2+1)dx+int_1^2 1/(xsqrt(x^2+1))dx`

Apply the substitution `x=tantheta`:

`L=[sqrt(x^2+1)]_ 1^2+intsectheta/tanthetad theta`

Simplify:

`L=sqrt5-sqrt2+intcscthetad theta`

Integrate directly:

`L=sqrt5-sqrt2-[ln|csctheta+cottheta|]`

Rewrite in terms of `tantheta` and `sectheta`:

`L=sqrt5-sqrt2-[ln|(1+sectheta)/tantheta|]`

Reverse the substitution:

`L=sqrt5-sqrt2-[ln((1+sqrt(x^2+1))/x)]_1^2`

Insert the limits of integration:

`L=sqrt5-sqrt2-ln((1+sqrt5)/(2(1+sqrt2)))`

Rearrange for clarity:

`L=sqrt5-sqrt2+ln2-ln((1+sqrt5)/(1+sqrt2))`

Answer 2

Length = `sqrt5+1/2ln((sqrt5-1)/(sqrt5+1))-sqrt2-1/2ln((sqrt2-1)/(sqrt2+1))~~1.222`

Explanation:

The arc length of a function `y=f(x)` over the interval `[a,b]` is given by `L=int_a^bsqrt(1+(dy/dx)^2)dx`.
`y=lnx->dy/dx=1/x`
`:.L=int_1^2sqrt(1+(1/x)^2)dx`
`L=int_1^2sqrt(1+1/x^2)dx`
I will use this solution `intsqrt(1+1/x^2)dx=sqrt(x^2+1)+1/2ln|(sqrt(x^2+1)-1)/(sqrt(x^2+1)+1)|+C` courtesy of mason m (https://socratic.org/questions/how-do-you-evaluate-the-integral-int-sqrt-1-1-x-2) to find that
`L=sqrt(x^2+1)+1/2ln|(sqrt(x^2+1)-1)/(sqrt(x^2+1)+1)|]_1^2`
`:.L=sqrt5+1/2ln((sqrt5-1)/(sqrt5+1))-sqrt2-1/2ln((sqrt2-1)/(sqrt2+1))~~1.222`