Explanation of this Integral of Probability Density Function

Question

I don't understand one part of my notes:

My question is, how do we go from $\int_{x}^{x+\Delta}f(t)dt$ to $f(x)\Delta + o(\Delta)$, and what is $o(\Delta)$ in this case?

Answer 1

Let $g(x,\Delta)=\frac 1{\Delta}\int_x^{x+\Delta} [f(t)-f(x)]dt \equiv \frac 1{\Delta}\int_x^{x+\Delta} f(t)dt-f(x)$ assuming that $f$ is continuous at $x$ it is easy to see that $g(x,\Delta) \to 0$ as $\delta \to 0$. We can write $\int_x^{x+\Delta} f(t)dt=\Delta g(x,\Delta)+f(x)\Delta$. So we get $\int_x^{x+\Delta} f(t)dt=f(x)\Delta+\Delta g(x,\Delta)$ and $o(\Delta)$ stands for $\Delta g(x,\Delta)$