Prove: Let $f$ be continuous on $[a,b]$, then suppose $f(x) \geq 0$ for all $x \in [a,b]$. if there exists a point $c \in [a,b]$, such that $f(c) > 0. then $$\int_{a}^{b} f(x) dx>0$.
This is what i have so far, since $f(x)$ is continuous and $f(c)>0$ $\exists [t,s]$ such that $f(x) > f(x)/2$, $x \in [t,s]$ ... and I have no idea how to continuous to prove that, anyone help me? Thanks
First prove that $f(x) \geqslant 0$ implies $\displaystyle \int_a^bf(x) \, dx \geqslant 0$.
This follows because for any partition $Q$ and lower Darboux sum $L(Q,f)$
$$0 \leqslant L(Q,f) \leqslant \sup_{P} L(P,f) = \int_a^bf(x) \, dx.$$
As you observed, if there is at least one point $c \in [a,b]$ where $f$ is continuous and $f(c) > 0$, then by continuity there exists a subinterval $[\alpha,\beta]$ with $c \in (\alpha, \beta)$ and where $f(x) > f(c)/2 > 0$ for all $x \in [\alpha,\beta]$. The infimum of $f$ on $[\alpha,\beta]$ must, therefore, be strictly greater than $0$.
Hence,
$$\int_a^b f(x) \, dx \geqslant \int_\alpha^\beta f(x) \, dx\geqslant \inf_{x \in [\alpha,\beta]}f(x)(\beta - \alpha) > 0.$$
Here we are using $\displaystyle f(x) \geqslant g(x) \implies \int_a^b f(x) \, dx \geqslant \int_a^b g(x) \, dx$ which follows from the first part of the proof using $f(x) - g(x) \geqslant 0$ as the integrand. In particular,
$$f(x) \geqslant f(x) \chi_{[\alpha,\beta]} \implies \int_a^b f(x) \, dx \geqslant \int_\alpha^\beta f(x) \, dx, $$
and on $[\alpha,\beta]$,
$$f(x) \geqslant \inf_{x \in [\alpha,\beta]}f(x) \implies \int_\alpha^\beta f(x) \, dx\geqslant \inf_{x \in [\alpha,\beta]}f(x)(\beta - \alpha) .$$
