I have the cdf of the maximum of two iid uniformly distributed random variables $x,y \sim U[a,b]$ (if that's incorrect, please let me know why). $X = \max (x,y)$ :
$$ F_X(x)= \begin{cases} 1 & x \ge b \\ (\frac{x-a}{b-a})^2 & x \in (a,b) \\ 0 & x \le a \\ \end{cases} $$
Now, I determine the pdf by taking the derivative:
$$ f_X(x)= \begin{cases} 0 & x \ge b \\ \frac{2(x-a)}{(b-a)^2} & x \in (a,b) \\ 0 & x \le a \\ \end{cases} $$
Now I'd like to get the pdf of $X_1 + X_2$. $X_1, X_2 = \max (x,y) $. So I do convolution:
$$f_{X_1 + X_2}(t)=\int_{2a}^{2b} f(x)*f(t-x) dx = \int_{2a}^{2b} \frac{2(x-a)}{(b-a)^2} \frac{2(((t-x)-a)}{(b-a)^2} dx = \frac{(8 ((a + 2 b)^2 - 3 b t))}{(3 (a - b)^3)}$$
The above result does not integrate to $1$. $$\int_{2a}^{2b} \frac{(8 ((a + 2 b)^2 - 3 b t))}{(3 (a - b)^3)} dt \ne 1$$ So, I must have done something incorrect here. What's wrong about this?
I have a feeling something is wrong with my integration limits in one of the above integrals. Is that what I'm doing incorrectly?