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?

You could also see this as a first-order Taylor's approximation. Let $F(t)=\int_c^tf(s)ds$ be an antiderivative of $f$, then
$$
\int_{t}^{t+h}f(s)ds=F(t+h)-F(t)\approx hF'(t)=hf(t).
$$
One advantage of this is that it makes it clear on how to improve the approximation. For example, if $f'$ exists at $t$, then the second-order approximation is 
$$
hF'(t)+\frac{1}{2}F''(t)h^2=hf(t)+\frac{1}{2}f'(t)h^2.
$$
