= Solution
For <independent random variables>, the <probability density function> of a sum is the <convolution of probability densities>. Since each density is one on $[0,1]$ and zero elsewhere,
$$
f_{X+Y}(u)=\int_{\mathbb R}\mathbf1_{[0,1]}(x)\mathbf1_{[0,1]}(u-x)\,dx.
$$
This is the length of $[0,1]\cap[u-1,u]$. Therefore \b[the density is]
$$
\boxed{f_{X+Y}(u)=\begin{cases}u&0\le u\le1,\\2-u&1\le u\le2,\\0&\text{otherwise}.\end{cases}}
$$
The two triangular areas sum to one, as a <probability density function> must.
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