How to show that $f(x)\ge 0 \implies \int_a^x fdt \ge 0$? This is clear to me thinking about integration like area under a curve, but how can it be proven? (assuming $f$ is continuous and using the fundamental theorem of calculus, and $\forall x\in[a,b]$)
Assume that $x\ge a$ in the integral $\int_a^x f(t)dt$

Let $f(x)$ be the derivative of a function $F(x)$

Since $f(x)\ge0$, the function $F(x)$ must be weakly increasing, and since $x\ge a$ we have $F(x)\ge F(a)$

We know from the fundamental theorem of calculus that
$$
\int_a^x f(t)\,\mathrm dt=F(x)-F(a)
$$
Can you see it now?
Simply use $\int_{a}^{x} (f-0)dt \geq0$. Now split the integrals and take the second one to the RHS