How to demonstrate $ \int\limits_1^{N+1}f(x)dx\leq \sum\limits_{k=1}^{N}f(k)\leq \int\limits_0^N f(x)dx $ for $f$ decreasing?

Question

I found this inequality in an exercises here, on this site (see the inequality below). It applies to a decreasing function. I used this for my exercise and it worked. The problem is i cannot find from where this comes. Can someone explain it to me ? Evaluating $ \lim\limits_{n\to\infty} \sum_{k=1}^{n^2} \frac{n}{n^2+k^2} $ the inequality is in the first answear

Added

For any decreasing function $f:\mathbb{R}\to\mathbb{R}$ and any $N>1$ we have $$ \int\limits_1^{N+1}f(x)dx\leq \sum\limits_{k=1}^{N}f(k)\leq \int\limits_0^N f(x)dx. $$

Answer 1

If $$K-1\leq x\leq K\leq y\leq K+1,$$ since $f$ is decreasing, $$f(y)\leq f(K)\leq f(x),$$ and thus $$\int_{K}^{K+1} f(y)\,\mathrm dy\leq f(K)\leq \int_{K-1}^{K}f(x)\,\mathrm dx.$$

Summing $K$ from $1$ to $N$ gives you the result.