True or False. If $g \ge 0$ is continuous on $[a,b]$ and $g(x_0) > 0$ for $\ge 1$ point $x_0 \in [a,b]$,
then $\int^{b}_a g > 0.$
1. Need determine if true or false. Ergo do we need intuition? I know $g \ge 0 \implies \int^{b}_a g \ge \quad \int^{b}_a 0 = 0$. But what's the intuition of the issues of $g(x_0) > 0$ ?
True. Since g is continuous and $g(x_0) > 0$, there exists $d > 0$ such that $|y − x_0| < d$ implies that $|g(y) − g(x0)| < g(x0)/2 \iff \color{seagreen}{-\frac{g(x_0) }{2 } < g(y) − g(x_0)} < \frac{ g(x_0) }{2 } \implies \not\Leftarrow \color{seagreen}{\frac{g(x_0) }{2 } < g(y) }$.
2. I know definition of continuity warrants any $e > 0$. But how can we presage $e = \frac{g(x_0) }{2 }$?
3. Where did this function emanate from? How can we presage this? I know the rest of the proof behaves by dint of this function, but it feels fey. I believe $\int^{b}_{a} h dx = \int^{\delta + x_o}_{-\delta + x_0} \dfrac{ g(x_0) }{ 2 } dx = \dfrac{ g(x_0) }{ 2 } \int^{\delta + x_o}_{-\delta + x_0} 1 \, dx$ but this h(x) fazes me.
The intuition for one is that,
if a continuous function take positive values at a point, it is also positive for a while. You can think of it like when you drive your car, you can't stop suddenly! You need a small piece of time to stop and in that small piece of time, your car would be still going.
Formally: Let $f$ be a continuous function from $\Bbb R$ to $\Bbb R$ s.t. $f(x_0)>0$, then there is an open interval $U$ containing $x_0$ s.t. $f(x)>0 \ \forall x\in U$. Its proof is as you write but you can take $\epsilon$ any positive number smaller than $f(x_0)$ to be sure that images of corresponding $U$ is positive. You can also take it as $\epsilon =f(x_0)/2014$.
Since your car is still going while you are trying to stop, the result of the integral is positive!!!
Formally: Since every open interval contains a closed interval, (if you do not like this sentence, take $C=[x_0-\delta/2,x_0+\delta/2]$ where $\delta$ corresponds to $\epsilon$) $f$ has min value $m$ on $C$. Thus, it is easy to see that the integral of $f$ on $C$ is bigger or equal than $m\delta$ which is positive. We are done.