Given a probability space $(\Omega,\mathcal{F}, \mathbb{P})$ we define the expected value of a random variable $X:\Omega\to \mathbb{R}$ as $$ \tag{1} \mathbb{E}(X):=\int_{\Omega}X\,d\mathbb{P}. $$
Suppose that $X$ is absolutely continuous with density $f_X$, i.e., $f_X$ satisfies $$ \mathbb{P}(X\leq x)=\int_{-\infty}^{x}f_X(x)\,dx, \hspace{0.3cm} x\in \mathbb{R}. $$
Now we define the expected value of $X$ as $$ \tag{2} \mathbb{E}(X):=\int_{\mathbb{R}}xf_X(x)\,dx. $$
My question is the following.
How does one derive definition $(2)$ from the general one - definition $(1)$?
I know that, upon considering $\text{id}:\mathbb{R}\to \mathbb{R}$ we have $$ \int_{\Omega}(\text{id}\circ X)\,d\mathbb{P}=\int_{\mathbb{R}}x\,d\mathbb{P}_X(x), $$ where $\mathbb{P}_X$ is a probabilty on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ given with $\mathbb{P}_X(B):=\mathbb{P}(X\in B)$. I also notice that $$ \int_{-\infty}^{x}f_X(x)\,dx=\mathbb{P}_X((-\infty,x]), $$ which, in a very remote way, smells of $d\mathbb{P}_X=f_X$...
Obviously, non of this is formal and I am stuck. All suggestions are welcome. Thanks!