I'm trying to find the CDF the sum of $X$ and $Y$ (which are independent). $X$ is uniform distributed over $[0,1]$ and $Y$ over $[0,2]$. I've seen some similar questions which explain the situations when both distributions are identical. I tried using the PDF of $Z =X+Y$ which results in $$\int_{-\infty}^\infty f_X(z-y)f_Y(y)\,\mathrm{d}y$$ out of which I get $$\int_0^2 f_X(z-y)\,\mathrm{d}y$$ (convolution). I'm somehow stuck here. For identical distributions I would make two calculations, for $0\le z\le1$ and $1\le z\le2$.
Should I split up in three possibilities here or is there an easier/other way to find the CDF of $X+Y$ (without the pdf?)?
Let $W = X + Y$. $$\begin{align*} \Pr[W \le w] &= \Pr[X + Y \le w] \\ &= \int_{y=0}^2 \Pr[X \le w - y \mid Y = y] f_Y(y) \, dy. \end{align*}$$ Here, we see we need to deal with different cases for the conditional probability. If $w-y < 0$, then the conditional probability is obviously zero; similarly, if $w-y > 1$, then the conditional probability is one. But in this latter case, the integrand is not zero, so we can't simply ignore it. That is to say, if $0 \le y < w-1$, $$\Pr[X \le w - y \mid Y = y] = 1.$$ But if $w < 1$, this inequality is never satisfied. So we will consider two cases: $w \le 1$, for which it suffices to write $$\Pr[W \le w] = \frac{1}{2} \int_{y = 0}^{w} \Pr[X \le w-y \mid Y = y] \, dy = \frac{1}{2} \int_{y=0}^w w-y \, dy = \frac{w^2}{4}.$$ In the case where $1 < w \le 2$, we then have $$\Pr[W \le w] = \frac{1}{2} \int_{y=0}^{w-1} \, dy + \frac{1}{2} \int_{y=w-1}^w w-y \, dy = \frac{2w-1}{4}.$$ And finally, when $2 < w \le 3$, the upper limit of the second integral will be $2$ because $Y$ cannot exceed $2$. This gives $$\Pr[W \le w] = \frac{(w-1)(5-w)}{4}.$$ So all told, the CDF of $W$ is $$F_W(w) = \begin{cases} 0, & w \le 0 \\ \frac{w^2}{4}, & 0 < w \le 1 \\ \frac{2w-1}{4}, & 1 < w \le 2 \\ \frac{(w-1)(5-w)}{4}, & 2 < w \le 3 \\ 1, & 3 < w. \end{cases}$$
To understand why there are five cases, consider the diagram below, which shows the joint support of $X$ and $Y$ in a Cartesian coordinate plane (light blue rectangle). The diagonal line $X + Y = w$ and its inequality $X + Y \le w$ are drawn for varying $w \in [0,3]$. When $w \le 0$, the line fails to intersect the rectangle at all, thus there is no area and zero probability. When $0 < w \le 1$, the line cuts off exactly one corner, and the area cut off is simply the area of the triangle, which is a quadratic function of its side length; i.e., $w^2/2$, and the probability (because the area of the underlying rectangle is $2$) is $w^2/4$. When $1 < w \le 2$, the line cuts off two corners, forming a trapezoid, and the area is a linear function of $w$. When $2 < w \le 3$, the line cuts off three corners, and it is perhaps best to calculate its area as the complement of the area not cut off. When $w > 3$, the entire rectangle is shaded.