I see what you mean. They give $\displaystyle f(x)=x^2\sin(1/x^3)$ for x in (0,1], f(0)=0.

It's easy to check f'(x) is unbounded in this case. And then they conclude f'(x) is not integrable.....

It wouldnt be too hard to make such function with f(1)=1 and f(0)=0.

I'm confused right now. I think you may be right after all.