Calculating integral of signum I am supposing to calculate the following integral: $$\int _{0}^{1}\mathrm{sgn}(x-x^{3})dx.$$ I assumed that on interval $(0,1)$ signum is positive. So:$$\int _{0}^{1}\mathrm{sgn}(x-x^{3})dx=\left [  x-x^{3}\right ]_{0}^{1}=0.$$ Is it correct?

You are correct that, for $x \in (0,1)$, $x > x^3$. However, recall what the sign function $\text{sgn}( x)$ is:
$$\text{sgn}(x) =\left\{ \begin{matrix}
-1 & x<0\\ 
0 & x=0\\ 
1 & x>0
\end{matrix} \right.$$
Thus, since $\text{sgn}(x-x^3) = 1$ for all $x \in (0,1)$,
$$\int_0^1 \text{sgn}(x-x^3) dx = \int_0^1 1 \; dx$$
