Riemann Integration and continuity Let $f$ be continuous and Riemann integrable on $[a, b]$ and $f(x) \geq 0$ for all $x \in [a,b]$. 
I'm trying to show that if $\int^b_a f(x) \ dx = 0$ implies that $f(x) = 0$ for all $x \in [a,b]$.
Could someone give me a hint? I really do not know where to begin.
 A: Otherwise, there exists $x_0\in [a,b]$, such than $f(x_0)>0$. Without loss any generality , we may assmume that $x_0\in (a,b).$ Due to $\lim\limits_{x\to x_0}f(x)=f(x_0)>0$, there exists $\delta>0$, such that $(x_0-\delta,x_0+\delta)\subset(a,b)$ and 
$f(x)\geq\frac{1}{2}f(x_0).$
So 
$$\int_{a}^bf(x) dx\geq\int_{x_0-\delta}^{x_0+\delta}\frac{1}{2}f(x_0) dx=\delta f(x_0)>0.$$
And here comes the contradiction. 
A: First,I want to say that the condition "$f$ is Riemann integrable on $[a,b]$ " can be deleted because "$f$ is continuous" implies "$f$ is Rieman integrable".
Prove by contradiction.If not,then $\exists x_0\in[a,b]$ such that $f(x_0)> 0$.$f$ is continuous on $[a,b]$,which means $\exists \varepsilon>0$ such that $\forall t\in (x_0-\varepsilon,x_0+\varepsilon)$,$|f(t)-f(x_0)|\leq \frac{f(x_0)}{2}$.So $\forall t\in (x_0-\varepsilon,x_0+\varepsilon)$,$\frac{f(x_0)}{2}\leq f(t)\leq \frac{3f(x_0)}{2}$.Now set a [partition] $P$ of $[a,b]$ such that $x_0-\varepsilon,x_0+\varepsilon\in P$.Then $L(f,P)\geq \varepsilon f(x_0)$(Why?).Because $\int_a^bf(x)dx\geq L(f,P)$(Why?),so $\int_a^bf(x)dx\geq \varepsilon f(x_0)>0$,this contradicts "$\int^b_a f(x) \ dx = 0$".So $\forall x\in [a,b]$,$f(x)=0$.
