Producing a CDF from a given PDF So I have this PDF: 
$$
f(x)=
\begin{cases}
x + 3 & \text{ for } -3 \leq x < -2\\ 
3 - x & \text{ for }  2 \leq x < 3\\
0 & \text{ otherwise}
\end{cases}
$$
To make this a CDF, I have integrated the PDF from $-\infty$ to some value, $x$.
$$
F(x)= 
\begin{cases}
\frac{x^2}{2} + 3x + \frac{9}{2} & \text{ for } -3 \leq x <-2\\
 \frac{1}{2} & \text{ for } -2 \leq x<2\\
\frac{-x^2}{2} + 3x + \frac{7}{2} & \text{ for } 2 \leq x<3
\end{cases}
$$
My friend argues that the first term in this CDF which is $(x^2/2 + 3x + 9/2)$ should actually be $(x^2/2 + 3x)$. But isn't this impossible? At $x = -3$, the CDF must be $0$, am I correct?. This is only true in the case where the first term is $(x^2/2 + 3x + 9/2)$.
If someone could shed light on this topic, that would be much appreciated.

Hint: integrate $ 3-x$. Say, integral is $f(x)$. Plug in $f(3)=1$ to get the value of constant
