Improper integral 1/x^(1/3) from -1 to 8

[g-racer]

Hi there,
I am stuck on this problem: the integral of 1/x^(1/3) from -1 to 8.

I have broken it up into the integral from -1 to 0 and 0 to 8. I am confused as to how the negative values under a cubed root affect things and whether or not I need to break it up further.
I am not sure whether the limit as n goes to -1 of 3/2(x)^(2/3) is real or imaginary. As if -1 is squared first it becomes real but if not then it is imaginary. We haven't done anything on imaginary integrals in class

 

[Dick]

Hi there,
I am stuck on this problem: the integral of 1/x^(1/3) from -1 to 8.

I have broken it up into the integral from -1 to 0 and 0 to 8. I am confused as to how the negative values under a cubed root affect things and whether or not I need to break it up further.
I am not sure whether the limit as n goes to -1 of 3/2(x)^(2/3) is real or imaginary. As if -1 is squared first it becomes real but if not then it is imaginary. We haven't done anything on imaginary integrals in class

Cube roots of negative numbers are negative, not imaginary. E.g. (-2)^(1/3)=-(2)^(1/3).

 

[g-racer]

thanks so is 3/2(-1)^2/3 a cubed root squared and therefore =3/2?

 

[Dick]

thanks so is 3/2(-1)^2/3 a cubed root squared and therefore =3/2?

Yes, it is.