Well, going with the hint you could argue that since f(x) is continuous in , for any , then f is uniformly continuous there, so , and then: taking a partition P with n points of s.t. , we get .

Of course, we still need to deal with , but this is not problem: the difference can be made as little as wanted.

Now, not going with the hint is much simpler and shorter, but perhaps you guys haven't yet studied this: f is Riemann integrable there since it is bounded and the set of points of discontinuities of f in [0,1] (which is one single point) has (Borel-Lebesgue) measure zero.

Tonio