Prove $\int_{a}^{b}{f(x)dx}>0$. Let $f\ge0$ be integrable on $[a,b]$. Let $f$ be continuous on $x_0\in (a,b)$ and let $f(x_0)>0$. 
Prove $$\int_{a}^{b}{f(x)dx}>0$$
It is easy to see it, graphically, because $$f\ge 0 \Rightarrow\int_{a}^{b}{f(x)dx}\ge0$$
and if $f>0$ somewhere between $a$ and $b$, then the zone of $F$ between those point is "real" (positive) and $\ne 0$.
My problem is, how do I show it formally?
I would appreciate your help.
 A: Hint: 
Let $m = \frac{f(x_0)}{2}$. There exists $\delta > 0$ such that $f(x) > m$ for all $x \in [x_0 - \delta, x_0 +\delta]$. Take any partition $P$ of $[a,b]$ that contains the points $x_0 - \delta$ and $x_0 +\delta$, then we have 
$$L(f:P) > 2m\delta \tag {*}$$
Thus as $f$ is integrable
$$\int_a^bf(x) \,dx \geq L(f:P) > 2m\delta$$
Edit:
$(*)$

 Notice that for some $s \in \{0,1,\ldots, n\}$ we have that $$m_s = \inf _{f \in [x_0 -\delta, x_0 + \delta ]} f \geq m > 0$$Now if you split 
 $$L(f;P) = \sum_{i=1}^{s-1} \underbrace{m_i}_{\geq 0}\underbrace{\Delta_{i-1}}_{>0} + \underbrace{m_s}_{\geq m}\underbrace{\Delta_{s-1}}_{\geq 2m\delta} + \sum_{i=s+1}^{n} \underbrace{m_i}_{\geq 0}\underbrace{\Delta_{i-1}}_{>0} \geq m(x_0 - \delta , x_0 + \delta) > 2m\delta $$

A: Look at the point $x_0$. Positivity and continuity imply that $f(x) > 0$ for a lot of points near $x_0$. Fix some $0 < \varepsilon < f(x_0)$. By the definition of continuity, there exists a $\delta > 0$ such that $$x \in [a, b]\, \text{ and }\,0 < |x - x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon.$$
Of course, $x = x_0$ also implies $|f(x) - f(x_0)| < \varepsilon$, and we can shrink $\delta$ (if necessary) to get the following stronger statement: $$x \in [x_0-\delta, x_0+\delta] \implies |f(x) - f(x_0)| < \varepsilon.$$
Now, $|f(x) - f(x_0)| < \varepsilon$ implies $f(x) > f(x_0) - \varepsilon$, and that lower bound is positive by choice of $\varepsilon$. This means that under the curve $y = f(x)$ we have a little solid rectangle with base $[x_0-\delta, x_0+\delta]$ reaching up to the line $y = f(x_0) - \varepsilon$. So, $\int_a^b f(x)\,dx$ is at least the area of this rectangle, namely $2\delta(f(x_0)-\varepsilon)$, and this is positive.
A: An idea:
We're given $\;f\;$ is continuous and non-negative on $\;(a,b)\;$ and such that for some $\;x_0\in (a,b)\;,\;\;f(x_0)>0\;$ .
From continuity we get that there exists $\;\epsilon>0\;$ s.t. $\;f(x)>0\;\;\;\forall\;x\in (x_0-\epsilon\,,\,\,x_0+\epsilon)\;$ . 
Since we're given $\;f\;$ is integrable and non-negative in $\;[a,b]\;$ , we then get
$$\int_a^b f(x)\,dx\ge\int_{x_0-\epsilon}^{x_0+\epsilon}f(x)\,dx>0$$
and we're done.
A: Bound the integral below by the usual lower-sum approximation, choosing the left endpoint of each interval for the height. Then you will have a non-negative lower bound.
A: All the proofs I have seen use the fact that if $f$ is continuous and $f(x_0) > 0$, then $f$ is positive on a small interval. Here is a proof that does not use such an argument.
By contrapositive, let's suppose that $f : [a,b] \rightarrow \mathbb{R}$ is continuous, non-negative, and satisfies $\displaystyle\int_a^b f(t) dt = 0$.
Let $F$ denote a primitive function of $F$ on $[a,b]$ (which exists because $f$ is continuous). Then $0=\displaystyle\int_a^b f(t) dt = F(b)-F(a)$, so $F(a)=F(b)$. But $F'=f \geq 0$, so $F$ is increasing.
This implies directly that $F$ is constant, hence $f=0$.
