There is much information about the probability density function of the sum of two iid variables which are uniform on [0,1]. I am trying to derive a general probability density formula for the sum of two iid variables, $X_1$ and $X_2$, which are uniform on $(a,b)$. For $Y = X_1 + X_2$ I was able to obtain the following formula geometrically: $$f_Y(y) = \begin{cases} \frac{y-2a}{(b-a)^2} & \text{for $2a < y < a+b$} \\ \frac{2b-y}{(b-a)^2} & \text{for $a+b \le y < 2b$} \\ 0 & \text{otherwise.} \end{cases}$$
However, I am trying to arrive at this formula using convolution. I know we start with $$f_Y(y)=\int_{-\infty}^\infty f_{X_1}(x)f_{X_2}(y-x)\,dx.$$ Then because $f_X(x) = 1$ if $a\le x\le b$ we have $\int_a^b f_{X_2}(y-x)\,dx$. I don't really understand where I'm supposed to go from here. Any help or explanation is greatly appreciated.
It looks very much like your answer follows along from here. However, you missed a crucial part when generalizing.
$f_{X_1}(x) = \dfrac{1}{b-a}$ when $a \leq x \leq b$ and $0$ otherwise so your convolution becomes
$$f_Y(y) = \int_a^b f_{X_2}(y-x) \dfrac{1}{b-a} dx$$
Now this is $0$ unless $a \leq y -x \leq b$ or stated otherwise, when $y-b \leq x \leq y-a$ in which case $f_Y(\cdot) = 1$. Can you finish it from here?