Dose $F' > 0$ imply $F$ is stirctly increasing?

Question

$F'(x)$ is positive and integrable on $[a,b]$. Does that imply $F$ is strictly increasing?

Intuitively I think it's true, but don't know how to prove.

Answer 1

By the fact that $F'(x)$ exists we can conclude that $F$ is continuous.

Hence by the mean value theorem for all $a_0,b_0$ such that $b\ge b_0>a_0\ge a$ exists $c$ such that $a_0<c<b_0$ and $F'(c)=\frac{F(b_0)-F(a_0)}{b_0-a_0}$. Because $b_0-a_0$ and $F'(c)$ are positive from assumption $F(b_0)-F(a_0)>0\implies F(b_0)>F(a_0)$ for all $b_0>a_0$

Answer 2

Note: $$\int_a^x f(t)dt=\lim_\limits{n\to\infty} \sum_{k=0}^n \frac1n \cdot f\left(a+k\frac{x-a}{n}\right)>0,$$ because $f(x)>0, x\in (a,b]$.