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

2 Answers
Jun 28, 2018

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))

Jun 28, 2018

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