It's the case no ? The step is 1/n for u_n and 1/(n+1) for u_(n+1) (finer subdivision than for u_n).
In my exercice I have f(x)=xexp(x), but I suppose this is true for all increasing function.
I verify with Maple for this f.
Consider the following example: if , if (yes, I know, it is not continuous). Then for the Riemann sum is . And for it is . To get a continuous example, just consider going quickly from 0 to 1 just before .
In your example, f is convex; this could help.