Sum of two uniform random variables

Hint: Split the calculation into two cases: (i) $0\le z\le 1$ and (ii) $1\lt z\le 2$.

Added: (i) if $0\le z\le 1$, then $f_X(z-y)=1$ if $0\le y\le z$, and $f_X(z-y)=0$ if $y\gt z$. It follows that $$\int_0^1 f_X(z-y)\,dy=\int_0^z 1\cdot dy=z$$.

(ii) If $1\lt z\le 2$, then $f_X(z-y)=1$ if $z-1\le y \le 1$, and $f_X(z-y)=0$ elsewhere. It follows that $$\int_0^1 f_X(z-y)\,dy=\int_{z-1}^1 1\cdot dy=2-z.$$ Thus $f_Z(z)=z$ if $0\le z\le 1$, and $f_Z(z)=2-z$ if $1\le z\le 2$. And for completeness, $f_Z(z)=0$ if $z$ is outside the interval $[0,2]$.

Remark: I suspect that the convolution way is in this case effectively no faster than the "slow" way of finding first the cumulative distribution function $F_Z(z)$, and differentiating.

hint: the integrand is zero unless $0 \le z-y \le 1$