approximate error between integral an sum I am new here. My problem: There is an integral $I:=\int_0^1 f(x)\,dx$ for $f\colon [0,1]\to\mathbb{R}$ and I want to compute it by $H_n:=\frac{f\left(0\right)}{2n}+\frac{1}{n}\left[f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+\dotsm + f\left(\frac{n-1}{n}\right)\right]+\frac{f\left(1\right)}{2n}$
Whats an easy way to prove the error $|I-H_n|\leq \frac{L}{4n}$?
The function $f$ suffices the lipsch. condition $\Vert f(x_1)-f(x_2) \Vert\leq L \vert x_1-x_2\Vert$?

I didn't attend any lecture about numerical analysis yet.
I know how it looks like. I take equidistant steps an only compute the mean of these values. From the lip. condition I know that $\max_{x\in [0,1]} \frac{d}{dx}f(x)\leq L$.
Is there some literature or an EASY way to see this inequality?
I think induction in $n$ won't make sense. I don't even know if this approximation has a name in the literature.

This approximation is called the trapezoid rule.   Write
$I - H_n = \sum_{j=1}^n E_j$ where $$E_j = \int_{(j-1)/n}^{j/n} f(x)\ dx - \frac{f((j-1)/n) + f(j/n)}{2n} $$
For convenience I'll write $(j-1)/n = a$, $j/n = b$, $1/n = b - a = \delta$, so
$$E_j = \int_a^b f(x)\ dx - \frac{\delta}{2}(f(a) + f(b))$$
Given the Lipschitz condition and the values $f(a)$ and $f(b)$, for $a \le x \le b$ we have $$f(x) \le g(x) = \cases{f(a) + L(x-a) & for $a \le x \le (a+b)/2$\cr f(b) + L (b-x)) &for $(a+b)/2 < x \le b$\cr}$$ so $\int_a^b f(x)\ dx \le \int_a^b g(x)\ dx$, which works out to  $$\frac{\delta^2}{4} L + \frac{\delta}{2} (f(a) + f(b))$$ 
Thus $E_j \le \dfrac{\delta^2}{4} L$.  Similarly, we get a lower bound $E_j \ge -\dfrac{\delta^2}{4} L$.
So $|E_j| \le \dfrac{\delta^2}{4} L = \dfrac{L}{4n^2}$
and $|I - H_n| \le \sum_{j=1}^n |E_j| \le \dfrac{L}{4n}$.
