Solving Improper Integrals: 1/(3∙√x)dx

[mirandasatterley]

Homework Statement

Evaluate the following improper integrals of explain why they don't converge.
Integral from 0 to infinity(1/the cubed root of x)dx
I'm not sure how to make forulas, so this is the best I can do:
0∫∞ (1/(3∙√x))dx

Homework Equations

No equations

The Attempt at a Solution

I know that when there is ∞ as an upper bound, the intergration is changed to:

lim as b→∞ 0∫b (1/(3∙√x))dx
But in this form, the 0 is a problem.

and if the lower bound, 0, causes the function to be undefined, the integration is changed to:

lim as a→0+ a∫∞ (1/(3∙√x))dx
But, in this for the infinity is still a problem.


Is there any way to combine the two so I can solve this.
Any help is appreciated.

 

[HallsofIvy]

Well, the first thing you had better do is actually write out the anti- derivative!
What is $$\int \frac{1}{^3\sqrt{x}}dx= \int x^{-\frac{1}{3}}dx$$?

Does it converge as x goes to 0? What happens as x goes to infinity?

Oh, and notice that the problem specifically asks you to "explain why they don't converge". Maybe the problem you are having isn't really a problem!

 

[Hurkyl]

Is there any way to combine the two so I can solve this.

Is there anything wrong with the most obvious approach: make both changes?

 

[mirandasatterley]

Well, the first thing you had better do is actually write out the anti- derivative!
What is $$\int \frac{1}{^3\sqrt{x}}dx= \int x^{-\frac{1}{3}}dx$$?

Does it converge as x goes to 0? What happens as x goes to infinity?

Oh, and notice that the problem specifically asks you to "explain why they don't converge". Maybe the problem you are having isn't really a problem!

The anti- derivative is X^(2/3)
2/3
As x goes to infinity, the anti-derivative goes to infinity.
As x goes to 0, the anti- derivative goes to 0.

so, would I evaluate it as (infinity - 0), which is infinity, therefore it diverges.

Is this right?