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?

We have $x-x^3 >0$ for $x \in (0,1).$ Hence $\mathrm{sgn}(x-x^3)=1$ for $x \in (0,1).$
Thus $\int _{0}^{1}\mathrm{sgn}(x-x^{3})dx= \int _{0}^{1}1dx=1.$
