Suppose $X ∼ Beta(a = 3; b = 1; θ = 1)$ and $Y ∼ U (−2, 2)$ are independent. Derive an expression for the cumulative distribution function of $X + Y$.
I am trying to do this by a convolution but I am having trouble to find the right boundaries.
As a density function, I found $f(x)= x^3$. Therefore I end up with this integral from -2 to 2: $\displaystyle\int fX(s-y)\dfrac14 dy$.
However working this out, don’t result in a valid cdf.
Rather than a convolution, draw a picture of the support of $X$ and $Y$ in the plane. Add a line cutting through it, $w = x + y$. Shading the support below this line represents the probability $P(W\le w)$. To compute this value, of course, you'd simply integrate $f(x,y)$ over the shaded region.
Note, however, that while there are a lot of options for where to draw the line $w = x+y$ -- I don't think I'm giving anything away by saying $w$ can be any number between $-2$ and $3$ -- they're not all the same for our purposes. Where the line gets drawn may alter the shape of the shaded region associated with $P(W\le w)$. If you figure out where these changes occur and break the cases up accordingly, you should be able to find a piecewise distribution function $F_W(w)$.
If you have questions on how to proceed, feel free to ask.
I had to sort out some thoughts as to why your convolution wasn't working. One reason is that your integral isn't supposed to lead to a cdf but rather a pdf. (I'm wondering if you have your terms swapped, as $f(x) = x^3$ is one of your cdfs, not a pdf.) However, that's not the only issue.
Technically your marginal pdfs are not $f_X(x) = 3x^2$ and $f_Y(y) = \frac{1}{4}$. That's the way you'll often see them written, but then there will be a note nearby saying, for example, $0\le x \le 1$ or $-2 \le y \le 2$. The way to incorporate such information into the functions themselves is to write $f_X(x) = 3x^2 \cdot \mathbf{1}_{\{0\le x \le 1\}}$ and $f_Y(y) = \frac{1}{4} \cdot \mathbf{1}_{\{-2 \le y \le 2\}}$, where $\mathbf{1}_{\{\}}$ is the usual indicator function. So if we wanted to set up a convolution, we would write
\begin{align} f_W(w) &= \int_\mathbb{R} f_X(w-y) f_Y(y) \; dy \\ &= \int_\mathbb{R} 3(w-y)^2 \cdot \mathbf{1}_{\{0\le w-y \le 1\}} \frac{1}{4} \cdot \mathbf{1}_{\{-2 \le y \le 2\}} \; dy \\ &= \frac{1}{4} \int_\mathbb{R} 3(w-y)^2 \cdot \mathbf{1}_{\{w - 1 \le y \le w\}} \cdot \mathbf{1}_{\{-2 \le y \le 2\}} \; dy \end{align}
Now, the product of these two indicator functions is $0$ unless both conditions they encode are true -- in which case the product evaluates to $1$. (More succinctly, $\mathbf{1}_A \cdot \mathbf{1}_B = \mathbf{1}_{A\cap B}$.) So on one end, we need both $y \ge w - 1$ and $y \ge -2$ -- that is, $y \ge \max \{w-1, -2 \}$. Similarly, $y \le \min\{w, 2 \}$. These are precisely the values of $y$ we need to integrate over -- and note that one, they change depending on $w$, and two, they never run all the way from $-2$ to $2$. So then we have
\begin{align} f_W(w) &= \frac{1}{4} \int_{\max \{w-1, -2 \}} ^{\min\{w, 2 \}} 3(w-y)^2 \; dy \\ &= \left[ \frac{(w-y)^3}{4} \right]_{\min\{w, 2 \}} ^{\max \{w-1, -2 \}} \end{align}
So that right there is your pdf. It can be rewritten (and arguably simplifed) as a piecewise function, as suggested above. If you want a cdf, you just have to compute $F_W(w) = \int_{-2}^w f_W(t) \; dt$.
The latter solution may end up a bit more succinct, but I think the first one -- drawing a graph to determine your separate definite integrals -- is a bit more intuitive -- and so perhaps nicer if you're trying to understand what's going on with the problem.