How did Wikipedia derive this inequality for increasing functions: $\int_{a-1}^{b} f(s)\ ds \le \sum_{i=a}^{b} f(i) \le \int_{a}^{b+1} f(s)\ ds$ The inequality itself is listed under summation approximations via integrals. It is being applied to a monotonically increasing function $f:\mathbb{R} \rightarrow \mathbb{R}$:
$$\int_{a-1}^{b} f(s)\ ds \le \sum_{i=a}^{b} f(i) \le \int_{a}^{b+1} f(s)\ ds$$
After looking at the Euler–Maclaurin formula article, I believe that the inequality is some result of the approximation of the integral $$\int_{a}^{b} f(s)\ ds$$ by $f(a) + f(a+1) + \ldots + f(b)$. Therefore, it would make sense that for a monotonically increasing function $f$,
$$
\int_{a-1}^{b} f(s)\ ds \le \int_{a}^{b} f(s)\ ds \le \int_{a}^{b+1} f(s)\ ds  \\
\Rightarrow \quad \int_{a-1}^{b} f(s)\ ds \le \sum_{i=a}^{b} f(i) \le \int_{a}^{b+1} f(s)\ ds.
$$
This, however, is a very sloppy justification. Any direction towards a more formal approach would be greatly appreciated!
 A: You have $$\int_{a-1}^b f(s) ds = \sum_{i=a}^{b} \int_{i-1}^{i} f(s) ds$$
But, as $f$ is increasing, 
$$\forall i, \ \int_{i-1}^{i} f(s) ds \leq \int_{i-1}^{i} f(i) ds = f(i)$$
So, 
$$\int_{a-1}^b f(s) ds \leq  \sum_{i=a}^{b} f(i)$$
A: 
This diagram illustrates that
$$
\int_0^9f(x)\,\mathrm{d}x\le\sum_{n=1}^9f(n)\tag1
$$
which follows from summing
$$
\int_{n-1}^nf(x)\,\mathrm{d}x\le f(n)\tag2
$$
which is true because $f(x)\le f(n)$ for $x\in[n-1,n]$.


This diagram illustrates that
$$
\sum_{n=1}^9f(n)\le\int_1^{10}f(x)\,\mathrm{d}x\tag3
$$
which follows from summing
$$
f(n)\le\int_n^{n+1}f(x)\,\mathrm{d}x\tag4
$$
which is true because $f(n)\le f(x)$ for $x\in[n,n+1]$.
A: For example
$$
\int_a^{b+1} f(x)\, dx=\sum_{i=a}^b\int_{i}^{i+1}f(x)\, dx\ge \sum_{i=a}^b f(i)
$$
as
$$
\int_{i}^{i+1}f(x)\ge \int_{i}^{i+1}f(i)=f(i)
$$
since $f$ is increasing.
A: Note that $$f(i)=\int_{i}^{i+1} f(i)\ ds\leq \int_{i}^{i+1} f(s)\ ds$$ and that $$f(i)=\int_{i-1}^{i} f(i)\ ds\geq \int_{i-1}^{i} f(s)\ ds$$  
Using these inequalities in the sum, you will find the desired result.
A: $\sum _{i=a} ^bf(i)$ is an inferior sum of the integral $\int _a ^{b+1}f(s)ds$ and a superior sum of $\int _{a-1} ^b f(s)ds$. 
This is a consecuence of the monotony of $f$. If you partition the interval $[a-1,b]$ taking $t_i = (a-1)+1$ we obtain 
$$\qquad t_{i+1} - t_i = 1 \qquad \text{ and } \qquad M_i = \sup \{f(x) : x\in [t_i,t_{i+1}] \} = f(t_i)$$
Then, taking $n=b-(a-1) = b-a +1$
$$\int _ {a-1} ^b f(s)ds \leq \sum _{i=1} ^n M_i(t_{i+1} - t_{i}) = \sum _{i=1} ^n f(t_{i+1}) = \sum _{i=1} ^nf(a+i) = \sum _{j=a} ^b f(j)$$
For the last equality you make the change $j = (a-1) + i$ so for $i=1,\; j=a$ and for $i=b-a+1$ you get $j=b$
The other inequality is analogus considering that $f(t_{i-1})$ is the infimum of the $f(x)$ for $x\in [t_i, t_{i+1}]$ and taking the interval $[a,b+1]$ in consideration.
A: Note that
$f(i)-\int_i^{i+1} f(x) dx
=\int_i^{i+1} (f(i)-f(x)) dx
$
and
$f(i+1)-\int_i^{i+1} f(x) dx
=\int_i^{i+1} (f(i+1)-f(x)) dx
$.
If $f$ is increasing then
$f(i)-f(x) \le 0$
and
$f(i+1)-f(x) \ge 0$
so
$f(i)-\int_i^{i+1} f(x) dx
\le 0$
and
$f(i+1)-\int_i^{i+1} f(x) dx
\ge 0$.
Similarly,
if $f$ is decreasing then
$f(i)-f(x) \ge 0$
and
$f(i+1)-f(x) \le 0$
so
$f(i)-\int_i^{i+1} f(x) dx
\ge 0$
and
$f(i+1)-\int_i^{i+1} f(x) dx
\le 0$.
A unified way to write these is
$\min(f(x))_{x=a}^b
\le \dfrac1{b-a}\int_a^b f(x) dx
\le \max(f(x))_{x=a}^b
$.
In words,
the average is between the min and the max.
