Distribution function of $X_1 + X_2$ given probability distributions of $X_1$ and $X_2$ Suppose I have a probability distribution for random variable $X_1$ given by:
$$
P(x) =
\begin{cases}
1/a, & \text{if} \;\; 0 \leq x \leq a\\
0, & \text{otherwise}
\end{cases}
$$
and $X_2$ with probability distribution:
$$
P(x) =
\begin{cases}
1/b, & \text{if} \;\; 0 \leq x \leq b\\
0, & \text{otherwise}
\end{cases}
$$
What is the probability distribution function of $X_1 + X_2$.

These are both uniform random variables if they independent we can write it like this. First $X_{1} \sim U(0,a)$ $X_{2} \sim U(0,b)$ . Also note the proof assumes that $ a \leq b$ The pdf of a uniform random variable $X \sim U(a,b)$ is given by 
$$ f_{X}(x) =\begin{align}\begin{cases} \frac{1}{b-a} &  \textrm{ for  }  x \in [a,b] \\ \\ 0  &  \textrm{ for  everywhere else } \end{cases} \end{align} \tag{1}$$
then the pdf of $X_{1}$ is given by 
$$f_{X_{1}}(x_{1}) =\begin{align}\begin{cases} \frac{1}{a} &  \textrm{ for  }  x \in [0,a] \\ \\ 0  &  \textrm{ for  everywhere else } \end{cases} \end{align}\tag{2}$$
so the pdf of $X_{2}$ is
$$f_{X_{2}}(x_{2}) =\begin{align}\begin{cases} \frac{1}{b} &  \textrm{ for  }  x \in [0,b] \\ \\ 0  &  \textrm{ for  everywhere else } \end{cases} \end{align} \tag{3}$$
Let the density $S = X_{1} + X_{2}$ be given by
$$ f_{S}(s) = \int_{\mathbb{R}} f_{X_{1}}(s-x_{2})f_{X_{2}}(x_{2}) dx_{2} \tag{4}$$
$$f_{X_{1}}(s-x_{2}) =\begin{align}\begin{cases} \frac{1}{a} &  0 \leq s-x_{2} \leq a \\ \\ 0  &  \textrm{ for  everywhere else } \end{cases} \end{align} \tag{5}$$
$$f_{X_{2}}(x_{2}) =\begin{align}\begin{cases} \frac{1}{b} &  0 \leq x_{2} \leq b \\ \\ 0  &  \textrm{ for  everywhere else } \end{cases} \end{align} \tag{6}$$
when $ 0 \leq x_{2} \leq a$ 
$$ f_{S}(s) = \int_{0}^{s} \frac{1}{ab} dx_{2} = \frac{x_{2}}{ab}\Big|_{0}^{s} = \frac{s}{ab} \tag{7}$$
when $ a \leq x_{2} < b $
$$ f_{S}(s) =\int_{s-a}^{s} \frac{1}{ab} dx_{2} = \frac{1}{ab}\Big|_{s-a}^{s} = \frac{s}{ab} - \frac{s-a}{ab}= \frac{1}{ab}(s-(s-a)) = \frac{a}{ab} =\frac{1}{b} 
 \tag{8} $$
when $ b \leq  x_{2} \leq a+b $
$$ f_{S}(s) = \int_{s-a}^{b} \frac{1}{ab} dx_{2} =\frac{1}{ab}\Big|_{s-a}^{b} = \frac{b}{ab} - \frac{s-a}{ab} = \frac{1}{a} - \frac{s-a}{ab} = \frac{1}{a} - \frac{s}{ab}+\frac{1}{b} \tag{9}$$
$$f_{S}(s) =\begin{align}\begin{cases} \frac{1}{ab}s &  0 \leq s < a \\ \\ \frac{1}{b}  &   a \leq s < b  \\ \frac{1}{a} -\frac{s}{ab}+\frac{1}{b} &b\leq s < a+b \\ 0  & \textrm{ otherwise} \end{cases} \end{align} \tag{10}$$
In order to visualize this here is some code
import numpy as np
import matplotlib.pyplot as plt


def sum_uni(a,b,n):

    z1 = np.linspace(0,a,n)
    z2 = np.ones(n)   
    z3 = np.linspace(b,a+b,n) 

    f1 = (1/(a*b))*z1
    f2 = (1/b)*z2
    f3 = (1/a)*z2-(1/(a*b))*z3 +(1/b)*z2
    g1 = np.linspace(0,a,n)
    g2 = np.linspace(a,b,n)
    g3 = np.linspace(b,a+b,n)
    my_plot = np.concatenate((g1,g2,g3),axis=0)
    my_distribution = np.concatenate((f1,f2,f3),axis=0)

    return my_plot, my_distribution

my_plot, my_dist = sum_uni(2,3,100)

plt.plot(my_plot,my_dist)


