The properties of integral Assume $f(x)$ is continuous on $[a, b]$.
(a) Prove that if $f(x) > 0$ for all $x \in [a, b]$, then the integral of 
$f(x)$ from $a$ to $b > 0$. I tried to apply the Extreme Value Theorem and
into it, but I can not deal  with it  successfully.
(b) Prove that if $f(x) \ge 0$ for all $x \in [a, b]$, and that $f(c) > 0$ for some $c \in [a, b]$, then the integral of
$f(x)$ from $a$ to $b> 0$. I tried to use the
continuity assumption to get that $f(x) > 0$ for some sub-interval $[s, t]$ of $[a, b]$, but I have no idea about the rest of processes.
Could someone help me to finish it? I'd appreciate it!
 A: Observe that part (a) follows from part (b), so you only need to prove (b). You have the right idea: continuity implies there exists $\delta$ with $f(x) > 0$ for all $x \in (c-\delta, c+\delta$).
Now, what can you say about the integral of $f$ on $(c-\delta, c+\delta$)? To finish, what can you say about $\int_{a}^{b}f$ compared to $\int_{c-\delta}^{c+\delta}f$?
Note: you will have to make sure to choose $\delta$ sufficiently small that $(c-\delta, c+\delta) \subset [a,b]$. 
A: Consider the given claim:
For, $m\leq f(x)\leq M$ $\forall$ $x\in [a,b]$, then:
$$m(b-a)\leq\int_a^b f\leq M(b-a)$$
The proof is as follows:
Suppose $f$ is integrable on $[a,b]$, and $m\leq f(x)\leq M$
Then given some partition $P=\left\{a,b\right\}$,
$$m(b-a)\leq L(P,f)\leq \underline {\int_a^b}f=\int_a^bf=\overline {\int_a^b}f\leq U(P,f)\leq M(b-a)$$
Now consider choosing $m=0$, then we have:
$$0\leq\int_a^b f\leq M(b-a)\Rightarrow \int_a^bf\geq0$$
So, given $f(x)>0$, and $\int_a^bf \ \exists$, then indeed:
$$0\leq\int_a^bf$$
$\mathbf{EDIT}$:
Here is another approach:
We note if $\int_{a}^{b}f \ \exists$, and $\exists \ c\in[a,b]$ such that $f$ is continuous, then $f$ has an antiderivative $F(x)=\int_a^cf$.
Note a separate proof is required for the above claim.
Furthermore, by taking the continuous extension of $f$ onto $\mathbb{R}$, we can thus note that if $f$ is continuous, then infact $f$ has an antiderivative $F(x)=\int_a^bf \ \forall \ x\in[a,b]$.
So let's prove your question, but in a different way:
If $f(x)\geq0 \ \forall \ x\in[a,b]$, and continuous on $[a,b]$, then its antiderivative $F(x) \ \exists$ on $[a,b]$.
Furthermore, by the definition of an antiderivative, $F'(x)=f(x)\geq0\Rightarrow F(x)$ is an increasing (monotone) function on $[a,b]$.
By the fundamental theorem of calculus, assuming $b>a\Rightarrow \int_a^bf=F(b)-F(a)\geq0$.
Can you prove the case where $f(x)=0$? In fact, by doing so, this restricts $\int_a^bf$ to be $>0$.  
