Given $X$ and $Y$ are R.V.s, are indepent and each uniformly distributed over $(0,1)$. How does the book go from here: $\int_0^1 f_X(a-y) \ dy$ to (for $(0 \leq a \leq 1)$): $\int_0^a dy$ ? They are trying to find the p.d.f. of $X+Y$ there.
EDIT:
As you know: $F_{X+Y}(a) = \int_{-\infty}^{-\infty} F_X (a-y)f_Y(y) dy$. Therefore in our case, $f_Y(y)$ is just $1$. Thus, $f_{X+Y}(a) = \int_0^1 \frac{d}{da}F_X(a-y)dy=\int_0^1f_X(a-y)dy$. And then they basically jump from this result to the one I identified above. And that would be the case for $0\leq a\leq 1$, they then have a case for $1 < a \leq 2$, and the result for that case is $\int_{a-1}^1 dy = 2-a$. No idea how they do that.