Suppose $f \geq 0$, if continuous on [a,b] and $\int_{a}^{b} f(x)dx=0$. Prove that $f(x)=0$ for all $x\in [a,b]$. My attempt:
Let P the partition of $[a,b]$. Let $x_{i}^{*} \in [x_{i-1},x_{i}]$ and f is non negative in $[x_{i-1},x_{i}]$. 
Since f is continuous on $[a,b]$, then $f$ is R-integrable with $\int_{a}^{b} f(x)dx=0$.
$$\displaystyle \lim_{n\rightarrow \infty} \sum_{i=1}^{n}f(x_{i}^{*}) \delta x_{i}=0$$
This implies $f(x_{i})=0$ for all i. i.e $f(x)=0$.
Is this proof is correct? 
 A: I would have used the Mean value theorem to solve this.
Take a generic point $c \in (a ,b).$ You know that:
$$\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx = 0.$$
Since $f$ is continuous in $[a, b]$ (with $b > a$), then there exist $d \in (a, c)$
and $e \in (c, b)$ such that:
$$\int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) d = f(d)(c-a) + f(e)(b-c) = 0.$$
This can be rewritten as follows:
$$f(d)(c-a) = -f(e)(b-c).$$
We know that $a < c < b$. As a consequence, $(c-a) > 0$ and $(b-c) > 0$.
Therefore:
$$f(d) = -f(e) \frac{c-a}{b-c}.$$
Since $f \geq 0$ in $[a, b]$, the previous equation is only satisfied by $f(d) = f(e) = 0.$
This means that $f(x) = 0$ for all $x \in [a, b].$
A: Assume by contradiction that $f(x_0)>0$ for some $x\in [a,b]$. Since $f$ is continuous, then there is a closed ball$^*$ around $x_0$ of radius $\epsilon$ s.t $f(x)>0$ for all points inside this ball. Since the closed ball is compact, $f$ has a (positive) minimum value inside the ball - $m$. What does this tell you about the Lower Darboux sum of $f$ in a partition which involves this ball? Try reaching a contradiction to the fact that the integral of $f$ is $0$.
$*$Note that the assumption about the closed ball is not accurate if $x_0$ is on the boundary of $[a,b]$, but this can easily be fixed - try to think how.
A: Overkill?
$f$ is continuous on $[a,b]$.
$F(x):=\displaystyle{\int_{a}^{x}}f(x)dx$, $x \in [a,b]$; and 
$F$ is differentiable: $F'(x)=f(x) \ge 0$, 
i.e. $F$ is an increasing function of $x$.
$\rightarrow F(x)\equiv 0$ , for $x \in [a,b]$.
(Since $0 \le F(x) \le F(a)=0$)
Then $F'(x)=f(x)=0$, for $x \in (a,b)$.
Since $f$ is continuous on [a,b] : $f(x)=0$ , $x \in [a,b].$
(Consider $\lim_{x \rightarrow a^+}f(x)=f(a)$, similarly for $b$)
