For a random variable $X$, $(X^3-1)$ is uniformly distributed in the interval $[0,7]$
I need to find the expected value and variance of $X$
and I know that:
cumulative distribution function: $$F_{X}(x)= \begin{cases} 0&;x < 1\\ \frac{x^3-1}{7}&;1\leq x \leq 2\\ 1&;x>2\\ \end{cases}$$
probability density function:
$$f_{X}(t)= \begin{cases} 0&;t \not\in (1,2)\\ \frac{3x^2}{7}&; t \in (1,2)\\ \end{cases}$$
My attempt:
$(1)$
$$E(X)=\int_{-\infty}^{\infty}t\cdot f_{_X}(t)dt$$
$$=\int_{-\infty}^{1}t\cdot 0dt+\int_{1}^{2}\frac{3t^2}{7}\cdot dt+\int_{2}^{\infty}t\cdot 0dt$$
$$=0+\frac{3}{28}x^4\bigg|_{1}^{2}+0=\boxed{\frac{45}{28}}$$
$(2)$
$$\text{Var}(X)=E(X^2)-(E(X))^2$$
$$E(X^2)=$$
$$E(X)^2=\bigg(\frac{2025}{784}\bigg)$$
Is it correct so far ? how to calculate $E(X^2)?$
$$E[g(x)]=\int_{-\infty}^{\infty}g(x)\cdot f(x)dx$$
$$E[X^2]=\int_{-\infty}^{\infty}t^2\cdot f(t)dt$$
$$=\int_{1}^{2}t^2\cdot \frac{3t^2}{7}dt$$ $$=\frac{3}{7}\int_{1}^{2}t^4dt$$ $$=\frac{3}{7}\frac{t^5}{5}\bigg|_{1}^{2}=\frac{31}{5}\frac{3}{7}=\boxed{\frac{93}{35}}$$
$$\text{Var}(X)=\frac{93}{35}-\frac{2025}{784}=\boxed{\frac{291}{3920}}$$