Question

What is the arclength of `f(x)=[4x^2–2ln(x)] /8` in the interval `[1,e^3]`?

Answer 1

`e^6/2+1/4~=201.964`

Explanation:

`f(x)=x^2/2-lnx/4`
`f'(x)=x-1/(4x)`

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

`L=int_1^(e^3) sqrt(1+(x-1/(4x))^2)*dx`
`L=int_1^(e^3)sqrt(1+x^2-1/2+1/(16x^2))*dx`
`L=int_1^(e^3)sqrt(x^2+1/2+1/(16x^2))*dx`
`L=int_1^(e^3)sqrt((x+1/(4x))^2)*dx`
`L=int_1^(e^3)(x+1/(4x))*dx`
`L=(x^2/2+(ln|x|)/4)|_1^(e^3)`
`L=e^6/2+(ln e^3)/4-(1/2+0)=e^6/2+3/4-1/2`
`L=e^6/2+1/4`