<s>[INST] Compute convolution $f*g(x)$ such that : $f,g\in L^{p}$ Question : 
Compute convolution : $f*g(x)$
$f(x)=\begin{cases}3x^2 & \text{ if } |x|\leq4 \\0 & \text{ otherwise}\end{cases}$
$g(x)=\begin{cases}1  &\text{ if }|x|\leq 2 \\0 & \text{ otherwise}\end{cases}$
My try : 
\begin{align}
f*g(x)&=\int f(x-y)g(y)dy\\
&=\int_{[-4,4]}3(x-y)^{2}1_{[-2,2]}(x-y)dy\\
&=\int_{[-4,4]∩[x-2,x+2]}3(x-y)^{2}dy
\end{align}
Now, how I find or discussed with this?
$[-4,4]\cap[x-2,x+2]=?$
Please, give me ideas and method to approach it.
 [/INST] Your bounds aren't quite right. We have $$f(x-y) = 3(x-y)^2\cdot\mathsf 1_{[-4,4]}(x-y) $$
and $g(y) = \mathsf 1_{[-2,2]}(y)$, so we have the following inequalities for $y$:
\begin{align}
x-4&\leqslant y\leqslant x+4\\
-2&\leqslant y\leqslant 2.
\end{align}
For $-6\leqslant x\leqslant -2$ we have
$$
f\star g(x) = \int_{-2}^{x+4} 3(x-y)^2\ \mathsf dy = (x+6)(x^2+12).
$$
For $-2\leqslant x\leqslant 2$ we have
$$
f\star g(x) = \int_{-2}^2 3(x-y)^2\ \mathsf dy = 12x^2+16.
$$
For $2\leqslant x\leqslant 6$ we have
$$
f\star g(x) = \int_{x-4}^{2} 3(x-y)^2\ \mathsf dy = (-x+10)(x^2-8x+28).
$$
</s>