Prove that $\int_a^b f(x)\,dx \gt 0$ if $f \geq 0$ for all $x \in [a,b]$ and $f$ is continuous at $x_0 \in [a,b]$ and $f(x_0) \gt 0$
EDIT. Please ignore below. It is very confusing actually -.-
Note: After typing this all out, I think I realized that my proof is complete, but I'll just post it to make sure :)
Attempt:
Find a partition $P = \{t_0,\ldots,t_n\}$ of $[a,b]$ s.t. $f(x)\gt f(x_0)/2$ for any $x \in [t_{i-1},t_i]$
Thus, the lower sum, $L(f,P)\geq x_0 (t_{i}-t_{i-1})/2 > 0$
[Basic Idea: since f is continuous at $x_0$, in the worst case scenario (i.e. $f=0$ at all points except in a nbhd of $x_0$) there must be some "bump" at $x_0$, which prevents the integral from actually equaling $0$. If we can find just one lower sum of a partition to be $> 0$, we will be done (I think...)]