Expected value of a random variable in a piecewise function

cumulative-distribution-functionsexpected valuepiecewise-continuity

I need to find the expected value of the following function

$$
F(x)=
\begin{cases}
0 &\text{ if }\:x < 0 \\
x^{3} &\text{ if }\: 0\leq x < 1\\
1 &\text{ if }\: x\geq 1
\end{cases}
$$

I use the formula $E[x] = \int_{-\infty}^{\infty} x\cdot f(x)dx \rightarrow \int_{0}^{1}x\cdot x^3dx\, +\, \int_{1}^{\infty} x\cdot dx$

The second integral on the interval $[1, \infty]$ is infinity, which is definitely off.

But I am not sure about the right approach here.

Best Answer

$f(x)=F'(x)=3x^2$ for $0\le x\le 1$ and $f(x)=0$ otherwise.

$E(X)=\int_0^1 3x^3dx=\frac{3}{4}$.

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