Convolution integrals finding pdf question My question is if $X$ and $Y$ are independent random variables, each with pdf $f(x)=2x$, $0 < x < 1$, how I would find the pdf of $Z=X+Y$?
I understand that I need to use the convolution integral to help me with this answer I have obtained:
$$f_Z(z)=\int_{S_X}f_X(x)f_Y(z-x)\,dx=\int_0^14x(z-x)I(0<z-x<1)\,dx$$
The problem is that whilst I know that I have to include $I(0 < z-x < 1)$, I don't fully understand its purpose. My current understanding is that the value of the pdf depends on this $I$ --- if $z-x$ ends up being in this interval, $I = 1$ and if not, then $I = 0$, is that correct?
The answer my book gives me is
\begin{cases}
\displaystyle\int_0^z4x(z-x)\,dx=2x^3/3&0\leq z < 1\\
\displaystyle\int_{\color{red}{z-1}}^14x(z-x)\,dx=-2z^3/3+4z-8/3\quad&1\leq z < 2\\
\\
\displaystyle 0&\text{elsewhere.}
\end{cases}
However, what I'm not clear about is why the integral for $1 < z < 2$ is over $z-1$ to $1$. I understand that the maximum value $x$ can take is 1, so the upper bound makes sense, but why is the lower bound $z-1$ and not $z$?
Thanks in advance!

The convolution is
$$
(f*g)(x) = \int_{-\infty}^\infty g(w)f(x-w)\,dw,
$$
but in this case both $f$ and $g$ are the same function and it is given by
$$
f(x) = \begin{cases} 2x & \text{if }0<x<1, \\ 0 & \text{if }x>1\text{ or } x<0. \end{cases}
$$
(Its value at the isolated points $0$ and $1$ doesn't matter.)
Within the integral you have the expression $f(w)f(x-w)$.  That is $0$ if $w>1$ or $w<0$.  It is also $0$ if $x-w>1$ or $x-w<0$.  That would mean $w<x-1$ or $w>x$.
Thus $f(w)f(x-w)=0$ if $w<\max\{0,x-1\}$ and also if $w>\min\{1,x\}$.  Hence the convolution is
$$
\int_{\max\{0,x-1\}}^{\min\{1,x\}} f(w)f(x-w)\,dw.
$$
If $0<x<1$ then $\max\{0,x-1\} =0$.  If $1<x<2$ then $\max\{0,x-1\}=x-1$.
If $0<x<1$ then $\min\{1,x\} = x$.  If $1<x<2$ then $\min\{1,x\}=1$.
