# Prove that if $f(x)$ is an increasing function then $f(i) < \int_{i}^{i+1} f(x)dx.$ where $i$ is a positive integer

Prove that if $f(x)$ is an increasing function then $$f(i) < \int_{i}^{i+1} f(x)dx.$$ where $i$ is a positive integer.

I can see why $f(i+1) > \int_{i}^{i+1} f(x)dx$, but I don't see an easy way of thinking about the above inequality.

$$f(i)=\int_i^{i+1}f(i)\ \mathrm dx<\int_i^{i+1}f(x)\ \mathrm dx$$

Consider the average value of $f$ on the interval $[i,i+1]$, $$f(x_i)=\frac{1}{i+1-i}\int_i^{i+1} f(x) dx =\int_i^{i+1} f(x) dx$$

for some $x_i$, $i\leq x_i\leq i+1$

Since $f$ is increasing, $f(i)\leq f(x_i)$ (strict when $i<x_i$), which was to be shown.

A more general way of stating this which works for all continuous functions, whether increasing or decreasing, is

$\min_{i \le x \le i+1} f(x) \le \int_{i}^{i+1} f(x) dx \le \max_{i \le x \le i+1} f(x)$.

If you want to have something that works for any interval, you can use this:

$\min_{a \le x \le b} f(x) \le \frac1{b-a}\int_{a}^{b} f(x) dx \le \max_{a \le x \le b} f(x)$.