No, there is no pathological case. If $f(x)$ is integrable, and $m\leq f(x)\leq M$ for all $x\in[a,b]$, then $m(b-a) \leq \int_a^b f(x)\,dx \leq M(b-a).$ This follows easily from the definition of the integral as a limit of Riemann sums. If $a=x_0\lt x_1\lt\cdots \lt x_n = b$ is a partition of $[a,b]$, $\Delta x_i = x_i - x_{i-1}$, and $x_i^*$ is a point in $[x_{i-1},x_i]$, then $m\leq f(x_i^*)\leq M$ for each $i$, so the Riemann sum satisfies $ m(b-a) = \sum_{i=1}^n m\Delta x_i \leq \sum_{i=1}^n f(x_i^*)\Delta x_i \leq \sum_{i=1}^n M\Delta x_i = M(b-a),$ so taking limits you get $m(b-a) \leq \lim_{||P||\to 0}\sum_{i=1}^n f(x_i^*)\Delta x_i \leq M(b-a)$ or $m(b-a) \leq \int_a^b f(x)\,dx \leq M(b-a)$ (using the assumption that $f$ is integrable on $[a,b]$.
In particular, if $f(x)\geq 0$ for all $x\in [a,b]$, then $0 = 0(b-a) \leq \int_a^b f(x)\,dx = F.$ Redefining a function at a finite number of points on $[a,b]$ does not change its integrability, nor the value of the integral, so if all you know is that $f(x)\geq 0$ on $(a,b)$, you can always redefine it at $a$ and $b$ so that you get $f(x)\geq 0$ on $[a,b]$ without changing the value of the integral.
(In fact, you only need $f(x)\geq 0$ "almost everywhere on $[a,b]$", meaning at all $x$ except perhaps for a set of measure $0$ that is contained on $[a,b]$).