Question

What is the arc length of `f(x)=xsqrt(x^2-1)` on `x in [3,4]`?

Answer 1

`L=7+1/8ln(15/8)+sum_(n=2)^oo((1/2),(n))1/4^n(1/8^(n-1)-1/15^(n-1))`

Explanation:

`f(x)=xsqrt(x^2-1)`

`f'(x)=(2x^2-1)/sqrt(x^2-1)`

Arc length is given by:

`L=int_3^4sqrt(1+(2x^2-1)^2/(x^2-1))dx`

Simplify:

`L=int_3^4xsqrt((4x^2-3)/(x^2-1))dx`

Rearrange:

`L=int_3^4(2x)sqrt(1+1/(4(x^2-1)))dx`

Apply the substitution `x^2-1=u`:

`L=int_8^15sqrt(1+1/(4u))du`

For `u in [8,15]`, `1/(4u)<1`. Take the series expansion of the square root:

`L=int_8^15sum_(n=0)^oo((1/2),(n))1/(4u)^ndu`

Isolate the `n=0` and `n=1` terms:

`L=int_8^15(1+1/(8u))du+sum_(n=2)^oo((1/2),(n))1/4^nint_8^15 1/u^ndu`

Integrate term by term:

`L=[u+1/8lnu]_ 8^15+sum_(n=2)^oo((1/2),(n))1/4^n[-1/u^(n-1)]_8^15`

Insert the limits of integration:

`L=7+1/8ln(15/8)+sum_(n=2)^oo((1/2),(n))1/4^n(1/8^(n-1)-1/15^(n-1))`