Question

What is the arc length of the curve given by `f(x)=x^(3/2)` in the interval `x in [0,3]`?

Answer 1

`(31sqrt31-8)/27~=6.0963`

Explanation:

`f(x)=x^(3/2)`
`f'(x)=3/2*x^(1/2)`

`L=int_a^b sqrt(1+[f'(x)]^2)*dx`

`L=int_0^3 sqrt(1+[3/2*x^(1/2)]^2)*dx`
`L=int_0^3 sqrt(1+9/4x)*dx`
`L=3/2int_0^3 sqrt(x+4/9)*dx`
`L=cancel(3/2)*cancel(2/3)(x+4/9)^(3/2)|_0^3`
`L=(3+4/9)^(3/2)-(4/9)^(3/2)=(31)^(3/2)/27-8/27`
`L=(31sqrt31-8)/27~=6.0963`