limit of the error in approximating definite integral with midpoint rule I want to calculate $\lim_{n \rightarrow \infty} n^2 |\int_{[0,1]}f(x)-I_n(x)|$ where $I_n$ is the integral approximation by midpoint rule:
$I_n=\frac{1}{n}\sum_{k=1}^nf(c_k)$ and $c_k$ is the point in the middle of $k^{th}$ interval.
My attempt:
I know the error is bounded by $\frac{max_{[0,1]} f''(x)}{24n^2}$.
Also by Lagrange theorem, we know that the remainder at point $x$ is $\frac{f^{(n+1)}(\xi)(x-x_\circ)^{n+1}}{(n+1)!}$. 
Now if I want to use the Lagrange formula I'll have to integrate over the whole $[0,1]$ interval. This is what confuses me, because if $n$ is even the integral will become zero and if it odd nonzero.
Even if I can integrate, $\lim_{n \rightarrow \infty} n^2 \frac{f^{(n+1)}(\xi)(x-x_\circ)^{n+1}}{(n+1)!}$ is $\infty \times 0$ and I don't know how to deal with it.
I appreciate any help.

Assume $f''$ is continuous.  The error in a single Midpoint Rule interval of length $h$ is
$$ \int_a^{a+h} f(x)\ dx - h f(a+h/2) = -\frac{h^3}{24} f''(\xi)$$ 
for some $\xi \in [a,a+h]$.  The error for $n$ equal subintervals of $[0,1]$ is the sum of the errors in each: if $h = 1/n$ and $x_j = j h$ we get
$$\int_0^1 f(x)\ dx - I_n = \sum_{j=0}^{n-1} \left(\int_{x_j}^{x_{j+1}} f(x)\ dx - \frac{1}{n} f(x_j + h/2)\right) = - \frac{1}{24n^3} \sum_{j=0}^{n-1} f''(\xi_j) $$
where $\xi_j \in [x_j, x_{j+1}]$.
Now notice that $\displaystyle \frac{1}{n} \sum_{j=0}^{n-1} f''(\xi_j)$ is a Riemann sum
for $\int_0^1 f''(x)\ dx = f'(1) - f'(0)$, so  ...
