Approximating the integral $\int_t^{t+h} f(s)\,ds$ by $f(t)h$ Suppose $f$ is a continuous real-valued function, and $h>0$. 
Does the following condition hold 
$$
\int_t^{t+h} f(s)\,ds\sim f(t)h
$$
and if so is this due to mid value rule?

The basic idea here is that if $h$ is sufficiently small, then, on the interval $[t,t+h]$, $f(s)$ is essentially constant. Treating $f(s)$ as a constant, you end up with $$\displaystyle\int_{t}^{t+h} f(s) ds \approx f(s) \displaystyle\int_{t}^{t+h} ds = f(s) \left( (t+h) - t \right) \Longrightarrow \displaystyle\int_{t}^{t+h} f(s) ds  \approx f(s) h$$
I'm not sure if there's a more formal way of reaching this approximation, but I have seen this type of derivation used in examples from physics. 
