fundamental theorem of calculus

Question

Let $f: [a, b] \to \mathbb R$ be integrable function and $F: [a,b] \to \mathbb R$, $F(x) = \int_a^x f(t) dt$

a) Show that if $f(x) \ge 0$ for all $x \in [a, b]$ then $F$ is increasing

b) If $F$ is increasing, can you conclude that $f(x) \ge 0$ for all $x \in [a, b]$?

So, I can write down the proof of fundamental theorem of calculus, and that will prove it, but is there a neater way to prove it, just for the increasing part?

Answer 1

a) If $a \le x <y \le b$, then $F(y)-F(x)=\int_x^y f(t) dt \ge 0$, hence $F(x) \le F(y)$

b) Let $a=0,b=1$ and

$f(0)=-1$ and $f(x)=0$ for $x \in (0,1]$. Then $F$ is constant, hence increasing, but $f(0)<0$.