Probability density function that integrates to 0 I'm running into a situation where I am calculating the definite integral for a probability density function as the first step in calculating probabilities with the CDF.
PDF graph image
$$\ f(x)= k |x| $$ 
$$\ -3 \le x \le 3 $$
$k$ has been evaluated as $1/9$ which you can tell from the graph
$$1/9\int_{-3}^3 x  = x^2/2 = 0$$
Now I understand that for f(x) to be a valid pdf then it must integrate to 1.
Where am I going wrong?
(Edited to reflect correct k value)
 A: $$|x| = \begin{cases}
-x,& x < 0 \\
x,& x \geq 0
\end{cases}$$
If you don't see why this is, think of it this way: you have a negatively-sloped line for all negative numbers, and a positively-sloped line for all positive numbers. This can be seen from the graph.
Hence 
$$\int_{-3}^{3}|x|\text{ d}x = \int_{-3}^{0}-x\text{ d}x + \int_{0}^{3}x\text{ d}x$$
Note that intuitively, it doesn't make sense for $\int_{-3}^{3}|x|\text{ d}x$ to equal $0$ since $|x| \geq 0$ and is continuous, so that the only way that $\int_{-3}^{3}|x| \text{ d}x = 0$ would be if $|x| = 0$ in $(-3, 3)$ (which is obviously false).
For this particular problem, it would be easier to take the areas of the two triangles to find $\int_{-3}^{3}|x|\text{ d}x$ going from the line to the $x$-axis than to compute the integral directly. You have one triangle with vertices $(-3, 3)$, $(0, 0)$, and $(-3, 0)$ and another one with vertices $(3, 3)$, $(0, 0)$, and $(3, 0)$. Both have height $3$ and base $3$, so they both have areas $$\dfrac{1}{2}(3)(3) = 4.5\text{.}$$
You have two of these triangles, so the resulting integral is $4.5+4.5=9$.
