c.d.f. of a random variable, given pdf

pdf is given by

$\displaystyle f(x) = 2-4|x| if -1/2 < x < 1/2$ and $\displaystyle 0$ otherwise

I cant do the integral of $\displaystyle 2-4|x|$ so I am going to make 2 cases, $\displaystyle x < 0$ and $\displaystyle x > 0.$

$\displaystyle If x < 0 , f(x) = 2+4x$ for $\displaystyle -1/2 < x < 0$

So we look at $\displaystyle \int_{x}^{0} 2+4t \text{d}t = -2x(x+1)$

$\displaystyle If x > 0 , f(x) = 2x-4x $for $\displaystyle 0 < x < 1/2$

So we look at $\displaystyle \int_{0}^{x} 2-4t \text{d}t = -2x(x-1)$

So I figure the cdf should be

$\displaystyle F_X(x) = $

$\displaystyle 0 $ if $\displaystyle x \leq -1/2$

$\displaystyle -2x(x+1) $ if $\displaystyle -1/2 < x < 0$

$\displaystyle -2x(x-1) + \int_{-1/2}^{0} 2+4t \text{d}t = -2x(x-1) + 1/2 $

if $\displaystyle 0 \leq x < 1/2$

$\displaystyle 1$ if $\displaystyle x \geq 1/2$

So i think is make sense to put $\displaystyle 0$ with the positive case. Since the adding of the $\displaystyle 1/2$ captures all the probabilities for negative values. However I think the function for $\displaystyle -1/2 < x < 0$ is wrong. Lower numbers are giving me higher probabilities than numbers close to $\displaystyle 0$. So something must be wrong.