# Definite integral: why is this wrong?

1. Feb 1, 2013

### tahayassen

According to wolfram alpha, the correct answer is -3/4.

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2. Feb 1, 2013

### tahayassen

Never mind. It is right.

3. Feb 1, 2013

### ArcanaNoir

Wolfram isn't getting 3/4, it's getting (-3/2)(-1)^(2/3) which it is computing to be approximately equal to 0.75-1.29904i

The real question of the day is why (-1)^(2/3) doesn't evaluate to 1. :uhh:

4. Feb 1, 2013

### jbunniii

There are three valid answers. Wolfram is calculating this one:
$$(-1)^{2/3} = (e^{i\pi})^{2/3} = e^{i2\pi/3} = \cos(2\pi/3) + i\sin(2\pi/3) = -1/2 + i\sqrt{3}/2$$
These two also work:
$$(-1)^{2/3} = (e^{-i\pi})^{2/3} = e^{-i2\pi/3} = \cos(-2\pi/3) + i\sin(-2\pi/3) = -1/2 - i\sqrt{3}/2$$
$$(-1)^{2/3} = (e^{i3\pi})^{2/3} = e^{i2\pi} = 1$$

5. Feb 1, 2013

### Ray Vickson

There are other valid answers as well. If we put $\sqrt[3]{x-1} = -\sqrt[3]{1-x} \; (0 \leq x \leq 1)$ we have
$$\text{answer } = -\int_0^1 \frac{1}{\sqrt[3]{1-x}}\, dx = -3/2.$$

6. Feb 1, 2013

### tahayassen

Mind has now been totally blown.

7. Feb 2, 2013

### Dick

Oh, I wouldn't blow that hard. If you are doing this as a real integral then you want to pick the branch of cube root which is always real. Your original answer was fine. Wolfram just picked the wrong branch. You could also with equal justification claim that the integral from 0 to 1 of x^(1/3) has three possible answers. Or even more if you get more creative. But that's certainly not the usual answer. Wolfram just misinterpreted the question.

Last edited: Feb 2, 2013
8. Feb 2, 2013

### SammyS

Staff Emeritus