My solution to this problem is the following:
Solution: $[a,b]$ is closed interval, so $f$ is uniformly continuous on the interval $[a,b]$. So, we know that there exists $\delta >0$ such that for all $\epsilon > 0$, if $|x-c| < \delta$ then $|f(x) - f(c)| < \epsilon$. In the above definition of uniform continuity, let $x=b$ and $c>b-\delta$, and notice that $x-c<\delta$. Then applying the definition of continuity, we see that this implies $|f(b) - f(b-\delta)| < \epsilon$. Expanding this inequality we see that we have $-\epsilon < f(b) - f(b-\delta) < \epsilon$. Since this is true for any $\epsilon > 0$, then we can choose $\epsilon >0$ as we wish. Choose $\epsilon < -f(b)$, and since $f(b) < 0$, then this means we still have $\epsilon > 0$. Notice this implies that for all points $x_0$ in the domain strictly between $x$ and $c$, we have $f(x_0) < 0$.
Then observe that in the definition of integral, we have $\int_{c=b-\delta}^{b} f(x) dx = \inf\{U(P,f,\alpha): P\}.$ And since by definition $U = \sum_{i=1}^n M_{f_i} \Delta \alpha_i$.
Then obviously since $f(x_0) < 0$ for all $x_0$ in this summation, then $\int_{c=b-\delta}^{b} f(x) dx < 0$.
I have a few questions about this solution:
1) Is uniform continuity necessary? Or is just continuity sufficient and if so, why?
2) Is there an easier way to prove this than going to the definition of a Riemann integral as a sum?