Cumulative Probability Functions for probability I'm having a really hard time trying to grasp the concept of Cumulative Distribution functions. I have this example: 
The cumulative distribution function of the random variable $X$ is given by
 $$F(x) = \left\{\begin{array}{rl} 0, & x<0, \\ \frac{x}{2}, & 0 \leq x < 1, \\ \frac{2}{3}, & 1\leq x < 2 \\ \frac{11}{12}, & 2 \leq x < 3, \\ 1, & 3 \leq x.\end{array}\right.$$
(a) Plot this cumulative distribution function. 
(b) What is $P(X > 1/2)$?
(c) What is $P(2 < X ≤ 4)$? 
(d) What is $P(X \leq  3)$? 
(e) What is $P(X = 1)$?
Now the problem I'm having is I've tried looking at my teachers slides and doing some of my own digging online but I can't seem to understand how it works necessarily, for example for a) the answer i know is $3/4$ and I'm assuming it's because the integral of $x/2$ is $x^2/4$, following that it's $1/4$, then $1 - 1/4 = 3/4$ Now my questions is why am I subtracting the value by $1$, and not only that why does $x$ magically become $1$ in this scenario? 
Thanks tremendously 
 A: The key to understanding everything is carefully to plot the cdf. Because plotting on this site is not pleasant, we will leave that to you, but it is critical that you do it before reading on. 
The first interesting thing happens at $x=1$. As $x$ approaches $1$ from the left, $F(x)$, that is, $x/2$, approaches $1/2$.  However, $F(1)$, the probability that $X\le 1$, is $2/3$. 
So there must be a point mass of $\frac{2}{3}-\frac{1}{2}$ at $x=1$.
Note that $\Pr(X\le x)=\frac{2}{3}$ for $1\le x\lt 2$. So the cdf is unchanging in this interval. That means that $\Pr(1\lt X\lt 2)=0$. 
There is a jump to $11/12$ at $2$, so there is a point mass of $\frac{11}{12}-\frac{2}{3}=\frac{1}{4}$ at $x=2$. So there is a point mass of $\frac{1}{4}$ at $x=2$.
Finally, there is a point mass of $\frac{1}{12}$ at $x=3$.
Our distribution has some features of a continuous distribution, and some features of a discrete distribution, it is a hybrid. 
Now we can answer the questions.
b) $\Pr(X\gt 1/2)=1=\Pr(X\le 1/2)=F(1/2)$. We can read off $F(1/2)$ from the given information. Since at $x=1/2$ we are in the interval $0\le x\lt 1$, where the cdf is $x/2$, we have $F(1/20=(1/2)/2=\frac{1}{4}$. And $1-\frac{1}{4}=\frac{3}{4}$.
c) This is $\Pr(X\le 4)-\Pr(X\le 2\lt 1$, that is, $F(4)-F(2)$.
d) Too simple!
e) Recall the discussion of the point mass at $x=1$.  If you want to use fancier language, $\Pr(X=1)=F(1)-\lim_{x\to 1^-}F(x)$.  
