Mean Value of Probability Distribution Function where the Integral is equals 0 According to my studies,

The mean $μ_X$ of a continuous random variable $X$ with probability
density function $f_X(x)$ is $$μ_X=E(X)=\int_{-\infty}^{+\infty}x\cdot
f_X(x)$$ but we may, and often do, restrict the integral to points
where $f_X(x)>0$

References
The Probability Distribution Function I'm working out is
$$f_X(x)=x(x³-x)+2$$
where $f_X(x)$ is in the interval $[-1,1]$ and with Normalizing Constant of $15/56$.
Following the reference to obtain the Mean Value:
$$μ_X=E(X)=\frac{15}{56}\int_{-1}^{+1}x[\cdot x(x³-x)+2]$$
which is equal to 0.
I have for sure that's not the answer. And it is said that "we may, and often do, restrict the integral to points where $f_X(x)>0$".
So here is my QUESTION which interval may I use?
$$\int_{0}^{1}x[\cdot x(x³-x)+2]$$
which is equals to $\frac{11}{12}$ once the interval must be $<1$
or
$$\int_{0}^{2}x[\cdot x(x³-x)+2]$$
wich is equals to $\frac{20}{7}$ once I have to respect the distance in the interval $[-1,1]$.
Thankyou!
 A: The sentence "we may, and often do, restrict the integral to points where $f_X(x)>0$ simply means that we can safely ignore the parts where $f_X(x)=0$ in the integral. It does not mean that the expected value cannot be zero. For example, consider the pdf of a $U(0,1)$ distribution: $$f_X(x)=\begin{cases}
1, & x\in [0,1]\\
0, & \text{otherwise}
\end{cases}$$
In this case the expected value can be calculated as $$\int _0^1x\cdot 1\,dx=0.5$$
instead of $\int _{-\infty}^\infty f_X(x)\,dx$ because $f_X(x)=0$ for any values of $x$ not in $[0,1]$.
A: The expectation being zero is not a problem.
You have :
$$f(x)=\dfrac{15}{56}\cdot\begin{cases}(x^4-x^2+2)&:& -1\leqslant x\leqslant 1\\[1ex]0&:&\text{otherwise}\end{cases}$$
The reason why we may restrict the integral to the the region where the probability density function is non-zero (called its 'support') is because, by definition, elsewhere the pdf contributes zero to the overall integral.
We know for certain that these sections will integrate to zero so we may ignore them.   That the integration over the support turns out to evaluate to zero is not a problem; we just did not know it would do so without investigating.
$$\begin{align}\int_{-\infty}^\infty xf(x)\,\mathrm d x &= \int_{-\infty}^{-1}0\,\mathrm d x+\tfrac{15}{56}\int_{-1}^{1}x\cdot(x^4-x^2+2)\,\mathrm d x+\int_{1}^\infty 0\,\mathrm d x\\[1ex]&=\tfrac{15}{56}\int_{-1}^{1}(x^5-x^3+2x)\,\mathrm d x\\[1ex]&=0\end{align}$$
