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

• g-racer
In summary, the problem is asking for the integral of 1/x^(1/3) from -1 to 8. The negative values under a cubed root do not affect the result and the integral can be broken up into two parts. The limit as n goes to -1 of 3/2(x)^(2/3) is real, as the negative sign is included in the cubed root.
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

g-racer said:
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).

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

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

Yes, it is.

## 1. What is an improper integral?

An improper integral is an integral where either the upper or lower bound of integration is infinite, or where the integrand is not defined at one or more points within the bounds of integration. It can also refer to an integral where the integrand tends to infinity at one or more points within the bounds of integration.

## 2. How do you evaluate an improper integral?

To evaluate an improper integral, you first rewrite it as a limit of a definite integral, where the upper or lower bound approaches infinity or a point where the integrand is undefined. Then, you evaluate the definite integral and take the limit as the bound approaches the desired value.

## 3. What is the integrand in the improper integral 1/x^(1/3) from -1 to 8?

The integrand in this improper integral is 1 divided by the cube root of x.

## 4. Can the improper integral 1/x^(1/3) from -1 to 8 be evaluated?

Yes, this improper integral can be evaluated. The lower bound of integration, -1, is finite and the integrand is defined for all values of x within the bounds of integration.

## 5. What is the value of the improper integral 1/x^(1/3) from -1 to 8?

The value of this improper integral is approximately 14.4997.

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