What is the arc length of f(x)= e^(4x-1) on x in [2,4] ?

1 Answer
Mar 15, 2018

L=sum_(n=0)^oo((1/2),(n))1/16^n(e^(15(1-2n))-e^(7(1-2n)))/(1-2n) units.

Explanation:

f(x)=e^(4x-1)

f'(x)=4e^(4x-1)

Arc length is given by:

L=int_2^4sqrt(1+(4e^(4x-1))^2)dx

Rearrange:

L=int_2^4(4e^(4x-1))sqrt(1+(4e^(4x-1))^-2)dx

For x in [2,4], (4e^(4x-1))^-2<1. Take the series expansion of the square root:

L=4int_2^4e^(4x-1){sum_(n=0)^oo((1/2),(n))(4e^(4x-1))^(-2n)}dx

Simplify:

L=4sum_(n=0)^oo((1/2),(n))1/16^nint_2^4e^((1-2n)(4x-1))dx

Integrate directly:

L=4sum_(n=0)^oo((1/2),(n))1/16^n[e^((1-2n)(4x-1))]_2^4/(4(1-2n))

Insert the limits of integration:

L=sum_(n=0)^oo((1/2),(n))1/16^n(e^(15(1-2n))-e^(7(1-2n)))/(1-2n)