I cannot solve the following problem:
Show that $$\left|\int_a^bf(u)\,du-f(a)(b-a)-\frac{1}{2}f'(a)(b-a)^2\right|\le\frac{C}{2}|b-a|^{2+\delta},$$ where $0\le a\le b \le 1$, $f$ is continuous, $f'$ is Lipschitz of order $\delta$ on $[0,1]$ and $C$ is the Lipschitz constant.
The only result I've got so far is that $$\left|\int_a^bf(u)\,du-f(a)(b-a)\right|\le\frac{C}{2}|b-a|^{2},$$ and I have no idea how to prove the initial problem.
By Taylor's Theorem with the remainder term (or by the mean value theorem): $$ f(u) = f(a) + (u-a) f'(\xi(u)) ,$$ where $a \le \xi(u) \le u$. Hence $$ f(u) - f(a) - (u-a) f'(a) = (u-a) (f'(\xi(u)) - f'(a)) \tag 1.$$ Also $$ |f'(\xi(u)) - f'(a)| \le C |\xi(u) - a|^\delta \le C |b-a|^\delta .$$ Now integrate both sides of (1) with respect to $u$ from $a$ to $b$.