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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.

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    @Jon: The $\epsilon$ determines a "horizontal band" around $f(c)$ ( with $\epsilon$ being half the width of the band). Make sure the entire band stays above the $x$-axis.2012-04-18

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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.

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    yes,youcan understand in this way. But I thiank the following is easy to understand :$\int_{x_0-\delta}^{x_0+\delta}\frac{1}{2}f(x_0)dx=\frac{1}{2}f(x_0)\int_{x_0-\delta}^{x_0+\delta}dx= 2\delta\frac{1}{2}f(x_0)=\delta f(x_0).$2012-04-18
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First,I want to say that the condition "$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 $\forall x\in [a,b]$,$f(x)=0$.