Okay so for this question: Convolution of a function with itself
The answer stated that in the case of $x\le 0$: the integral bounds are from 0 to x. Why is this?
I also don't understand why from $1\le x\le2$ the bounds are from $x-1$ to $1$. If somebody could clarify that'd be much appreciated.
Write $\phi = 1_{[0,1]}$.
We want to find $\phi *\phi $.
So $\phi*\phi(x) = \int_{-\infty}^\infty \phi(y)\phi(x-y)dy$.
Now we want to find out when $\phi(y)\phi(x-y)\ne 0$. By the definition of this functions, this happens iff $y,x-y\in[0,1]$.
Now, remember that we can think of $x$ as fixed and $y$ as a running variable.
Thus, we can first only take $y\in [0,1]$ , $\phi*\phi(x) = \int_0^1 \phi(y)\phi(x-y)dy$
And now we know $\phi(y) = 1$ on that segment, so $\phi*\phi(x) = \int_0^1 \phi(x-y)dy$.
Now we need to treat $\phi(x-y)$. we can do that as follows:
First, if $x>2$ then for every $y\in[0,1]$ , $x-y\notin [0,1]$ so $\phi(x-y) =0$ and $\phi*\phi(x) = 0$. Same reasoning shows that $\phi*\phi(x) =0$ for $x<0$.
Now if $1\le x \le 2$ then $ 0\le x-y \le 1$ is equivalent to $y\ge x-1$ and this is possible since $x\in [1,2] $ so $x-1 \in [0,1]$ and the bounds are $x-1 $ and $1$ so we get $\phi*\phi(x) = \int_{x-1}^1 1 dy$.
Last case is $x\in[0,1]$ then $x-y\in [0,1] \iff y\le x$ so the bounds are $0$ and $x$ and we get $\phi*\phi(x) = \int_0^x 1 dy$.
Does this makes the answer you saw clearer?