Cumulative distribution function, integration problem Given the continuous Probability density function $f(x)=\begin{cases} 2x-4, & 2\le x\le3  \\ 0 ,& \text{else}\end{cases}$
Find the cumulative distribution function $F(x)$.
The formula is $F(x)=\int _{ -\infty  }^{ x }{ f(x) } $

My Solution 
The first case is when $2\le x\le3$  then $$\int_2^x {(2u-4)} \, du=[u^2-4u]_2^x=x^2-4x-4+8=x^2-4x+4$$

so $F(x)=x^2-4x+4 , \text{ for } 2\le x\le 3$
 
The Problem
Now I have the cases where $x<2 \text{ and } x>3$ , ($\int_{-\infty}^{x}0 \, dx$) but I am not sure how to do it. I would appreciate if someone could show me the solution of this two cases.  

A general technique to deal with such type of questions is to split up the integral at the points where the density $f$ has different specifications, i.e. at $y=2$ and $y=3$. If $y<2$, then
$$
F(y)=\int_{-\infty}^yf(x)\,\mathrm dx=\int_{-\infty}^y0\,\mathrm dx=0,
$$
and if $2\leq y\leq 3$, then
$$
\begin{align}
F(y)&=\int_{-\infty}^2f(x)\,\mathrm dx+\int_2^yf(x)\,\mathrm dx=\int_{-\infty}^20\,\mathrm dx+\int_2^y(2x-4)\,\mathrm dx\\
&=0+\left[x^2-4x\right]_2^y=y^2-4y+4.
\end{align}
$$
Lastly, if $y>3$ then
$$
\begin{align}
F(y)&=\int_{-\infty}^2 f(x)\,\mathrm dx+\int_2^3 f(x)\,\mathrm dx+\int_3^y f(x)\,\mathrm dx\\
&=\int_{-\infty}^2 0\,\mathrm dx+\int_2^3 (2x-4)\,\mathrm dx+\int_3^y 0\,\mathrm dx\\
&=0+1+0=1.
\end{align}
$$
