Sum of three uniformly distributed random variables. Let $X,Y,Z$ be uniformly distributed $U(0,1)$. Then I know that the density function for the random variable $A= X+Y$ is 
$$f(a) = \begin{cases} 
      a & a\in (0,1)\\
      2-a & a\in[1,2)\\
      0 & \text{ otherwise} 
   \end{cases}.
$$
and 
$$
g(z)=\begin{cases}
   1  & z\in (0,1)\\
   0   & \text{otherwise}
\end{cases}
$$
My goal is to find the $B = A+Z.$ By the convolution theorem:
$$h(b) = \int_{-\infty}^{\infty}f(b-z)g(z)dz=\int_{0}^{1}f(b-z)dz.$$
After this I am not sure how to proceed. Any help will be much appreciated. 
 A: 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}

