Let f be continuous on [a,b] such that \int_A^B f(x)\,dx = 0 for every a \leq A < B \leq b.
We show, by contradiction, that under these conditions it cannot be the case that f(c) \neq 0 for any c \in (a,b):
Suppose c \in (a,b) is such that f(c) > 0. Then, by continuity of f at c, there exists a \delta > 0 such that if A=\max\{a, c - \delta\} < x < \min\{b, c + \delta\}=B then f(x) > 0. But then by a basic property of integrals we have
\int_A^B f(x)\,dx > \int_A^B 0\,dx = 0, which is a contradiction. Thus such a c cannot exist. A similar argument shows that there cannot exist any c \in (a,b) such that f(c) < 0.
Hence f is constantly zero on (a,b) and by continuity f(a) = f(b) = 0 also.
This is a more "conceptual" proof than one using the fundamental theorem, and one can rephrase it thus:
"\int_a^b \frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x}\,dx = 0 says only that, on average, cars are neither created nor destroyed on the stretch of highway between a and b. But suppose there exists a c \in [a,b] where \frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x} \neq 0. Then, if \frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x} is continuous, there is around c a stretch of highway where cars are being spontaneously created if \frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x} > 0 or destroyed if \frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x} < 0. Both of these we reject as unphysical."