# Definite integral $\int_{-64}^{1}\frac{dx}{x^{1/3}}$

I am having some trouble with a problem very similar to this in my study guide, how can I start, the $-64$ is really intimidating to me.

• What is the antiderivative of $x^k$ ? – Claude Leibovici Aug 16 '14 at 4:35
• The integral is an improper integral, since our function blows up at $0$. But this improper integral converges. – André Nicolas Aug 16 '14 at 4:38
• @AndréNicolas. Could you interfere in this no end discussion ? Thanks. – Claude Leibovici Aug 16 '14 at 5:51
• What intervention do you suggest? The PV discussion is of no relevance because of the convergence. – André Nicolas Aug 16 '14 at 6:10
• @Brian. May I ask you a favor ? When you will get the "official" answer from your professor, could you add it to your post. I am really curious. Thanks and cheers :-) – Claude Leibovici Aug 16 '14 at 7:23

We have to break this up and take the limit.

$$\lim_{\epsilon\to 0} \left (\int_{-64}^{-\epsilon}x^{-\frac{1}{3}}dx+\int^{1}_{\epsilon}x^{-\frac{1}{3}}dx \right )$$

Using the power rule,

$$\lim_{\epsilon\to 0} \left (\left[\frac{3}{2}x^{\frac{2}{3}} \right ]_{-64}^{-\epsilon}+\left [\frac{3}{2}x^{\frac{2}{3}} \right ]^{1}_{\epsilon}\right )$$

This becomes

$$\lim_{\epsilon\to 0} \left (\frac{3}{2}(-\epsilon)^{\frac{2}{3}}-\frac{3}{2}(64)^{\frac{2}{3}}+\frac{3}{2}1^{\frac{2}{3}} -\frac{3}{2}\epsilon^{\frac{2}{3}} \right )$$ Notice the cancellation of the $\epsilon$ terms. So we are left with

$$=-\frac{3}{2}(4)^2+\frac{3}{2}=\frac{3}{2}(1-16)=-\frac{45}{2}$$

• This is not correct, I am afraid. – Claude Leibovici Aug 16 '14 at 5:05
• @ClaudeLeibovici I agree with you. This is Cauchy principal value. – Cortizol Aug 16 '14 at 6:00

There's really nothing to be intimidated of. Once you realize that $1/x^{1/3} = x^{-1/3}$, you can just use power rule.

Edit: I just realized that you probably want to take the riemann sum and compute it manually. In that case, I would suggest integrating the inverse.

• then would i just integrate? – brian Aug 16 '14 at 4:36
• The first step is to get the antiderivative. Then use bounds. – Claude Leibovici Aug 16 '14 at 4:38
• okay i got (3/2)-24(-1)^(2/3) thank you – brian Aug 16 '14 at 4:39
• could that be simplified in any way – brian Aug 16 '14 at 4:40
• uh.. (-1)^2/3 is 1. So you should get 3/2 - 24 = -45/2. – Tae Hyung Kim Aug 16 '14 at 4:47

It is an improper integral . one may write $\int_{-64}^0x^\frac{-1}{3}dx+\int_{0}^1x^\frac{-1}{3}dx.$

Hint

First, the antiderivative $$\int\frac{dx}{x^{\frac{1}{3}}}=\frac{3 x^{2/3}}{2}$$ Now, for the integral (taking into account André Nicolas's comment which does not need to be repeated), the result is then $$\frac{3}{2}\Big(1^{\frac{2}{3}}-(-64)^{\frac{2}{3}}\Big)=\frac{3}{2}\Big(1-(-64)^{\frac{2}{3}}\Big)$$ You were right to be "intimidated" by the $-64$ since $(-64)^{\frac{2}{3}}$ is a complex number which you need to simplify.

I suppose that this is the key part of the problem but I am sure that you can take from here.

I can accept to be totally wrong but, at least for me $$(-64)^{\frac{2}{3}}=-8+8 i \sqrt{3}$$
• I don't see how $(-64)^{\frac{2}{3}}$ is a complex number – Learning Aug 16 '14 at 5:14
• Is it not true that $(-64)^{\frac{2}{3}}$ is $16$ in this situation? It's the integral over the real number set, and it's well-defined. Of course the structure $(-64)^{\frac{2}{3}}$ may be referred to a complex number, but this is not the case I think. As I know, $(-4)^3=-64$ and $x^{\frac{1}{3}}$ in this case is the third root of a real number. No complex needed – Learning Aug 16 '14 at 5:40