# Thread: Need confirmation on an improper integral question

1. ## Need confirmation on an improper integral question

I swear I've done something wrong with this one.

$\displaystyle \int_{-1}^1 x^{-2/3} dx$ (This is an improper integral, because the integrand is an unbounded function at x=0)

Here's what I got:
$\displaystyle \int x^{-2/3} dx = 3\sqrt[3]{x}+C$
$\displaystyle \Rightarrow \int_{-1}^1 x^{-2/3} dx = 3-3\sqrt[3]{-1}\approx 1.5-2.59808i$

I'm probably missing something really basic, but I've been tangled up with other stuff and, well, when you take in new material, the other stuff either gets jumbled or pushed out.

If anyone could see if I've made a mistake somewhere, it'd be appreciated.

2. Originally Posted by Runty
$\displaystyle \int_{-1}^1 x^{-2/3} dx$
(This is an improper integral, because the integrand is an unbounded function at x=0)
Question: Does $\displaystyle \displaystyle\lim _{h \to 0^ + } \int_h^1 {x^{\frac{{ - 2}}{3}} dx}$ exist?

3. -1 has three cube roots, one of which is -1. [tex]3- 3\sqrt[3]{-1}= 6. There are two complex roots but neither of them gives what you have. How did you get that?

4. Originally Posted by HallsofIvy
-1 has three cube roots, one of which is -1. [tex]3- 3\sqrt[3]{-1}= 6. There are two complex roots but neither of them gives what you have. How did you get that?
WolframAlpha. It might have made a mistake.

I'll look into Plato's suggestion as well.

5. Originally Posted by Plato
Question: Does $\displaystyle \displaystyle\lim _{h \to 0^ + } \int_h^1 {x^{\frac{{ - 2}}{3}} dx}$ exist?
I believe it does. If I'm right, the limit comes out to 3.

EDIT: I think I have the correct answer now. The solution comes out to 6.

6. Originally Posted by Runty
I believe it does. If I'm right, the limit comes out to 3.
With computer algebra syatems it matters how one inputs the question.
Look at this way.