all. I'm unfortunately stuck trying to prove the following result:
Let $$E(h) = \int_0^1f(x)\,\text{d}x - h\sum_{i=1}^Nf(ih-h/2),\quad h={1\over N}.$$ If $f\in C^1([0,1])$, prove that there exists a constant $c$ such that $|E(h)| \le ch$.
I approach this problem similar to the way we do it for the case that $f\in C^2([0,1])$: Take the Taylor expansion of the function over each of the subintervals up to the first derivative: $$\begin{align*}\underbrace{\int_0^1f(x)\,\text{d}x}_{I(f)} &= \sum_{i=1}^N\int_{(i-1)h}^{ih}\left[f(ih-h/2) + f'(ih-h/2)(x-ih+h/2)\right]\,\text{d}x\\&= \sum_{i=1}^N\int_{(i-1)h}^{ih}f(ih-h/2)\,\text{d}x + \sum_{i=1}^N\int_{(i-1)h}^{ih}f'(ih-h/2)(x-ih+h/2)\,\text{d}x\\ &=\underbrace{h\sum_{i=1}^Nf(ih-h/2)}_{Q(f)} + \sum_{i=1}^N\int_{(i-1)h}^{ih}f'(ih-h/2)(x-ih+h/2)\,\text{d}x\end{align*}$$ Hence $$\begin{align*}|E(h)| &= |I(f) - Q(f)|\\ &= \left|\sum_{i=1}^N\int_{(i-1)h}^{ih}f'(ih-h/2)(x-ih+h/2)\,\text{d}x\right|\\ &\le \sum_{i=1}^N\int_{(i-1)h}^{ih}|f'(ih-h/2)(x-ih+h/2)|\,\text{d}x\\ &\le \sum_{i=1}^N\|f'\|_\infty{h^2\over4}\\ &\le {h^2\over4}\|f'\|_\infty\left(N(N+1)\over2\right)\\ & = {1+h\over 8}\|f'\|_\infty,\end{align*}$$ but it's not clear what I could've done to have $h$ by itself in the numerator. Does anyone know what I did wrong in terms of estimation?
You apparently replaced $\sum_{i=1}^N i$ by $N(N+1)/2$, which isn't correct. The sum is simply $N$.
I'd approach the problem a little differently. Focus on one subinterval: we want to estimate $$ E(f) = \int_a^b f(x)\,dx - (b-a) f(c), \quad c = \frac{a+b}{2} $$ For any linear function $g$ we have $E(g)=0$, so we can subtract such a function from $f$ without changing $E(f)$. By subtracting a suitable constant function we can get $f(c) = 0$. By the mean value theorem, $$ |f(x)| \le |x-c| \sup_{[a,b]}|f'| $$ and integrating both sides yields $$ |E(f)|\le \frac{(b-a)^2}{4} \sup|f'| $$
So, the error on a subinterval of size $h$ is at most $\frac{h^2}{4}\|f'\|_\infty$. Summing over $N$ such subintervals and using $hN=1$, we get $$ \frac{h}{4} \|f'\|_\infty $$ as an error bound.
The above is still inefficient in that I could have subtracted a linear function from $f$ to make $f(c)=f'(c)=0$. Doing so results in replacing $\|f'\|_\infty$ by $$ \omega(h) = \sup \{|f'(t)-f'(s)| : |t-s|\le h/2 \} $$ which tends to $0$ as $h\to 0$. So the error is actually $o(h)$.