Continuous function and positive integral

Question

I am currently working on this problem:

Let $f: [a, b] \to \mathbb{R}$ be an integrable function such that $f(x) \ge 0$ for all $x \in \mathbb{R}$.

1. Prove that $\int_a^bf(t)dt \ge 0$
2. Assume that there exists $c \in (a,b)$ such that $f$ is continuous at $c$ and $f(c) > 0$. Prove that $\int_a^bf(t)dt > 0$

My solution for the first part is that $\int_a^bf(t)dt$ is always greater than or equal to the lower Darboux sum $L(f)$, and since $f(x) \ge 0$ we have $L(f) \ge 0$, hence the proof is done.

I am currently stuck on the second part. This is my current progress:

Since $f$ is continuous at $c$, $\forall \epsilon >0$, $\exists \delta > 0$ such that if $|x - c| < \delta$ implies $|f(x) - f(c)| < \epsilon$. Choose $\epsilon = f(c)$ we have $f(x) > 0$ if $x$ is in the interval $(c- \delta, c+\delta)$. Same analogy as the above part shows that $\int_a^bf(t)dt > 0$ for $x \in (c-\delta, c + \delta)$.

I haven't found the solution for the remaining part. Can someone give hints?

Answer 1

1. It is a direct consequence of a definition of the integral.

2. It is not true without the assumption $f(c)>0.$ Indeed, the zero function is a counterexample. If we assume what I wrote, $f$ is positive on a neighbourhood of $c$, then the integral is necessarily positive.