Integration involving negative roots yields two answers? I have this problem that is driving me insane. 
I am asked to evaluate:
$$\int^1_{-1} \sqrt[3]{t} -2 \ dt$$
Now I evaluate this correctly to
$$[\frac{3^{4/3}t}{4}-2t]^1_{-1}$$
From here I work this problem has follows:
$$[\frac{3(1)^{4/3}}{4}-2(1)]-[\frac{3(-1)^{4/3}}{4}-2(-1)]$$ 
$\Longrightarrow$ $$[\frac{3}{4}-2]-[\frac{3}{4}+2]$$
$\Longrightarrow$ $$-4$$
Now the answer I get on online calculators(Wolfram, http://www.wolframalpha.com/widgets/view.jsp?id=8ab70731b1553f17c11a3bbc87e0b605) is 

When it comes to $(-1)^{4/3}$ it should equal to $1$, because $-1^4=1$ and$\sqrt[3]1=1$
and equivalent is the reasoning $\sqrt[3]-1=-1$ and $-1^4=1$
Is there something I am missing or another way to evaluate $(-1)^{4/3}$? And if so what makes that answer stronger than this one? In other words shouldn't both answers be valid? Thanks in advance.   

You are correct. The answer that Wolfram gives is not correct in the context.
Note that there are also two complex third roots of $-1$, they are $$\frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}$$
Wolfram is using the first of these.
