Limit of the Trapezoidal Rule 
Let $f$ be Riemann-integrable and $\zeta=\{x_0,...x_n\}$ the most equidistant pratition. Show that $$\lim_{n\to\infty} \frac{b-a}{n} (\sum_{k=1}^n \frac{f(x_{k-1})+f(x_k)}{2})=\int_a^b f(x)dx$$

My try (but this is obviously wrong)
We define the Riemann-integral as (short) $$\lim_{n\to\infty} \sum_{k=1}^n f(\xi)\Delta x_k$$
Hence $f(\xi)=\frac{f(x_{k-1})+f(x_k)}{2}\in f([x_{k-1},x_k])$
(Note, this is only true if $f$ is increasing).
Could anyone give me a hint on how to continue, without the need for an increasing-function assumption?
 A: The sum you wrote down is a Riemann sum. Let $$I_0=\left[a,a+\frac{(b-a)}{2n}\right]$$
$$I_k=\left[a+\left(k-\frac12\right)\frac{(b-a)}n,a+\left(k+\frac12\right)\frac{(b-a)}n\right]$$
For $1\le k\le n-1$, and
$$I_n=\left[b-\frac{(b-a)}{2n},b\right]$$
The intervals cover $[a,b]$ and the width $\Delta x_k$ of each interval containing $x_k$ is the weight given to $f(x_k)$ in the trapezoidal rule.
A: Clearly if $f$ is Riemann integrable on $[a, b]$ and if $\zeta = \{x_{0}, x_{1}, x_{2}, \ldots, x_{n}\}$ is a partition of $[a, b]$ with $x_{k} = a + kh$ and $h = (b - a)/n$ then by the definition of Riemann integration it follows that $$\lim_{n \to \infty}\frac{b - a}{n}\sum_{k = 1}^{n}f(\xi_{k}) = \int_{a}^{b}f(x)\,dx\tag{1}$$ where $\xi_{k}$ is any arbitrary point in interval $[x_{k - 1}, x_{k}]$.
Now we need to take $\xi_{k} = x_{k - 1}$ in $(1)$ and again take $\xi_{k} = x_{k}$ in $(1)$ and adding the resulting equations we get $$\lim_{n \to \infty}\frac{b - a}{n}\sum_{k = 1}^{n}\{f(x_{k - 1}) + f(x_{k})\} = 2\int_{a}^{b}f(x)\,dx$$ and dividing by $2$ we get the desired equation. All we need to assume is that $f$ is Riemann integrable on $[a, b]$.
