How do I add two independent probability functions? (Edited this to try to be clearer, again, apologies if I'm not fully making sense here.)
Let's say that I have a task that takes a random amount of time, somewhere between 1 and 2 hours, evenly distributed. It seems that the simplest way to represent this is with this function, given a value $p$ that is a random number between 0 and 1.
$$
 f_1(p_1) = p_1 + 1
$$
I can figure out intuitively that if I wanted to flip this around and get a probability density function
(I think that's what it's called?) It'd be this:
$$
     pd_1(x) = \begin{cases}
         0, & \text{if $x$ < 1} \\
         x, & \text{if 1 >= $x$ >= 2 } \\
         0, & \text{if $x$ > 2} \\
     \end{cases}
$$
First of all, is there a name for this "flipping around" operation so that I can look it up and learn more about it?
Now, let's say that I have a second, independent task, that takes between 2 and 4 hours:
$$
 f_2(p_2) = 2 p_2 + 2
$$
Probability density (again, I figured this out by guessing:)
$$
     pd_2(x) = \begin{cases}
         0, & \text{if $x$ < 2} \\
         x/2, & \text{if 2 >= $x$ >= 4 } \\
         0, & \text{if $x$ > 4} \\
     \end{cases}
$$
Now, I know that if I perform these tasks sequentially, I'll get them done somewhere between 2 and 5 hours. How can I create a single function of the form $f(p) = ?$ to tell me this (where $0 >= $p$ >= 1$)?
And how can I create a probability density function that tells me the odds of getting the task done in $x$ time?
 A: I guess by 'probability functions' you meant 'probability distributions'. In that case, $f_1$ and $f_2$ cannot be probability distributions, since the integral of a probability distribution in their domain must be 1 and in this case the integral is just 1/2. Besides that, your question doesn't make much sense.
I guess you might be trying to ask: If I have two random variables $x$ and $y$ with given probability distributions $P_x$ and $P_y$ and define a new variable $z$ as $z=x+y$, what is the probabiliy distribution of $z$? 
A: You seem to be picking a point $(X,Y)$ uniformly randomly in the rectangle $R=(1,2)\times(2,4)$ and to ask for the distribution of $Z=X+Y$. Drawing the rectangle $R$ and some regions of equations $x+y\leqslant z$ in $R$, one sees that $3\leqslant Z\leqslant6$ with full probability and that each probability $P(Z\leqslant z)$ is proportional to the area of some specific polygon included in $R$. 
This yields the density $f_Z$ of $Z$ as the classical "tent" function, that is,
$$
f_Z(z)=\left\{\begin{array}{ccc}\frac12(z-3)& \text{if} & 3\lt z\lt 4\\ \frac12& \text{if} & 4\lt z\lt 5\\ \frac12(6-z)& \text{if} & 5\lt z\lt 6\\ 0&\text{otherwise}&\end{array}\right.
$$
Recall that this means that, for every $z$,
$$
P(Z\leqslant z)=\int_{-\infty}^zf_Z(u)\,\mathrm du,
$$
and that, for every suitable function $A$,
$$
E(A(Z))=\int_{-\infty}^\infty A(u)\,f_Z(u)\,\mathrm du.
$$
