To Jan: It can also be done without ever knowing that you're essentially doing a convolution. To simplify the notation, I'll use S = U + V.

Since U and V are independent, their joint pdf is g(u, v) = (1)(1) = 1 for

,

, and zero elsewhere.

The cdf of S is given by

.

There are two cases to consider:

and

.

**Case 1:** .

.

Therefore

for

.

**Case 2:** .

I'll leave the mopping up to you. You should get

for

.

And of course f(s) = 0 elsewhere. These are the same results as CaptainB got using the convolution theorem.