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.
 A: $ f(x) = \begin{cases} 0 &, \phantom{-3 \leq {}}x< -3 \\ x+3 &, -3 \leq x \leq -2 \\ 0 &, -2 \leq x \leq \phantom{-{}}2 \\ 3-x &, \phantom{-{}}2 \leq x \leq \phantom{-{}}3 \\ 0 &, \phantom{-{}}3 \leq x\end{cases}$
\begin{align}
F(x) = \int_{-\infty}^x f(t) \,\mathrm{d}t &= \begin{cases} 0 &, \phantom{-3 \leq {}}x< -3 \\ 0+\int_{-3}^x t+3 \,\mathrm{d}t &, -3 \leq x \leq -2 \\ 0+\int_{-3}^{-2} t+3 \,\mathrm{d}t + 0&, -2 \leq x \leq \phantom{-{}}2 \\ 0+\int_{-3}^{-2} t+3 \,\mathrm{d}t + \int_{2}^x 3-t \,\mathrm{d}t&, \phantom{-{}}2 \leq x \leq \phantom{-{}}3 \\ 0+\int_{-3}^{-2} t+3 \,\mathrm{d}t + \int_{2}^3 3-t \,\mathrm{d}t + 0 &, \phantom{-{}}3 \leq x\end{cases} \\
&= \begin{cases} 0 &, \phantom{-3 \leq {}}x< -3 \\ 0+\frac{x^2+6x+9}{2} &, -3 \leq x \leq -2 \\ 0+\frac{1}{2} &, -2 \leq x \leq \phantom{-{}}2 \\ 0+\frac{1}{2} + \frac{-x^2+6x-8}{2}&, \phantom{-{}}2 \leq x \leq \phantom{-{}}3 \\ 0+\frac{1}{2} + \frac{1}{2} + 0 &, \phantom{-{}}3 \leq x\end{cases}
\end{align}
As you can see $F(-3) = 0$ because $\frac{9-18+9}{2} = 0$.  Your argument is correct that the CDF must increase from zero (starting) at $-3$.
A: The cdf $F(x)$ is given by
$$F(x)
=\begin{cases}
0 & \text{ for } x <-3\\
\int_{-3}^{x}(t+3) \, dt & \text{ for } -3 \leq x <-2\\
\frac{1}{2}  & \text{ for } -2 \leq x <2\\
\frac{1}{2}+\int_{-2}^{x}(3-t) \, dt & \text{ for } 2 \leq x <3\\
1 & \text{ otherwsie}
\end{cases}
$$
My suggestion is to plot the pdf $f(x)$ and then think in terms of the area of triangle(s) being swept as you move along the $x-$axis. 
A: Strictly speaking the CDF should be defined for all $x\in\Bbb R$.  If $x\le-3$ we have
$$F(x)=\int_{-\infty}^x f(t)\,dt=\int_{-\infty}^x 0\,dt=0\ .$$
If $-3<x<-2$ then
$$F(x)=\int_{-\infty}^x f(t)\,dt
  =\int_{-\infty}^{-3} 0\,dt+\int_{-3}^x t+3\,dt
  =\frac{1}{2}(x+3)^2$$
(which in fact is the answer you have already, but it's simpler this way).  For $-2\le x\le2$ we get $F(x)=\frac{1}{2}$ as you have already.  For $2<x<3$ we obtain
$$F(x)=\int_{-\infty}^x f(t)\,dt
  =\int_{-\infty}^2 f(t)\,dt+\int_2^x f(t)\,dt
  =\frac{1}{2}+\int_2^x (3-t)\,dt=1-\frac{1}{2}(3-x)^2\ .$$
Finally, if $x\ge 3$ then $F(x)=1$.
As a check, note that you should have $F(x)$ always increasing and
$$\lim_{x\to-\infty} F(x)=0\quad\hbox{and}\quad \lim_{x\to\infty} F(x)=1\ .$$
A: Hint: integrate $ 3-x$. Say, integral is $f(x)$. Plug in $f(3)=1$ to get the value of constant
A: Let's put it this way
$$\int_{-\infty}^Y f(x)=
\begin{cases}
0 & -\infty < Y < -3 \\ 
Y^2+3Y+{9\over 2} & -3\le Y <-2 \\
{1\over 2} & -2\le Y <2 \\
{1\over 2} +\int_2^Y(3-x)\, dx & 2\le Y < 3 \\
1 & Y\ge 3
\end{cases}$$
The first term is INDEED including the 9/2, this is because it is
$$\int_{-3}^Y f(x)\,dx\int_{-3}^Y (x+3)\,dx$$
By the fundamental theorem of calculus, this is the antiderivative at the top limit, Y, minus the antiderivative at the bottom limit, -3. The value of $x^2/2+3x$ at $x=-3$ is $-9/2$ so you get the result with the +9/2 because you subtract the bottom limit and subtracting a negative gives a positive.
A: Note first of all that the cdf is defined for all $x$, so a complete answer should take care of all $x$.
If $x\lt -3$, everything is easy. We have $\int_{-\infty}^x f_X(t)\,dt=0$. So $F_X(x)=0$ if $x\lt -3$. 
Now for $-3\le x\lt -2$, in effect we are integrating $t+3$ from $-3$ to $x$, since there is no "mass" before $-3$.  An antiderivative is $t^2/2+3t$. Plug in. We get $\frac{x^2}{2}+3x+\frac{9}{2}$. 
Note that this is $\frac{1}{2}$ at $x=-2$. The density function is then $0$ for a while, so exactly as you had it, if $-2\le x\lt 2$ then $F_X(x)=\frac{1}{2}$.
Now for $2\le x\lt 3$, we want the integral up to $2$, plus the integral from $2$ to $x$ of $3-t$. We get 
$$\frac{1}{2}+(3x-\frac{x^2}{2})- 4.$$
Finally, for $x\ge 3$, the cdf is $1$. 
