# 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.

-
No, that "Lagrange theorem" result is not true. You are thinking of the remainder in Taylor's theorem for a degree $n$ approximation, which is completely different. – Robert Israel Feb 20 '13 at 5:37
Yes, you are right. But I thought there should be a way to connect them. Because this is the way we get the upper bound on the error. Isn't it true that $|\int_{[0,1]}f(x)-I_n(x)|=\int_{[0,1]}\frac{f^{(n+1)}(\xi)(x-x_\circ)^{n+1}}{(n‌​+1)!}$? – Bunny Feb 20 '13 at 5:48
No, it isn't. For example, if $f(x) = x^2$ and $n > 1$, $f^{(n+1)} = 0$ but $\int_0^1 f(x)\ dx > I_n(x)$. – Robert Israel Feb 20 '13 at 7:35

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 ...