So we get \begin{equation} f(b-z) = \begin{cases} b-z & b-z\in (0,1)\\ 2-(b-z) & b-z\in[1,2)\\ 0 & \text{ otherwise} \end{cases}. \end{equation} We have that \begin{equation} 0<z<1 \end{equation} or \begin{equation} b-1<b-z<b \end{equation} If $b < 0$, then $b - z < 0$, so $h(b) = 0$ according to the boundaries we have. On the other hand if $b - 1> 2$ (or $b>3$), that is is $b-z>2$, we also get $f(b) = 0$. Other than that, we can distinguish three cases:
Case 1: If $0<b<1$ \begin{equation} h(b) = \int_{0}^{1}f(b-z)dz = \int_0^{b} b-z \ dz = \frac{b^2}{2} \end{equation}
Case 2: If $1<b<2$ \begin{equation} h(b) = \int_{0}^{1}f(b-z)dz = \int_{b-1}^{1} b-z \ dz + \int_{0}^{b-1} 2-(b-z) \ dz = \frac{b(2-b)}{2}-\dfrac{b^2-4b+3}{2} \end{equation}
Case 3: If $2<b<3$, we have \begin{equation} h(b) = \int_{0}^{1}f(b-z)dz = \int_{b-2}^{1} 2-(b-z) \ dz = \frac{b^2-6b+9}{2} \end{equation}
Distribution of B \begin{equation} h(b) = \begin{cases} \frac{b^2}{2} & b\in (0,1)\\ \frac{b(2-b)}{2}-\frac{b^2-4b+3}{2} & b\in(1,2)\\ \frac{b^2-6b+9}{2} & b\in(2,3)\\ 0 & \text{ otherwise} \end{cases}. \end{equation}
