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Infinite Series $1+\frac12-\frac23+\frac14+\frac15-\frac26+\cdots$

Let $f$ be a continues function in $[a,b]$

$\forall x \in [a,b] \ \ \ f(x)\geq 0$

$ \int_{a}^{b}f(x)dx \ = 0$

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Proof that $ \forall x \in [a,b] \ \ f(x) = 0 $

So how do I do that ?

What I know is:

because $f$ is continues function in $[a,b]$ I know it bounded and because $\forall x \in [a,b] \ \ \ f(x)\geq 0$ I know that $ \exists M \in \mathbb{N} \ \forall x \in [a,b] \ \ \ M \geq f(x)\geq 0$.

I also know that because $f$ is continues function in $[a,b]$ that $f \in R[a,b]$ meaning that there exists $F$ so that $F(a) = F(b)$, but how can I show that $ \forall x \in [a,b] \ \ F(b) = F(a) = F(x)$ ?

So how do I continue from here ?

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Suppose not. There exists a $x$ such that $f(x) > 0$. Let $r = f(x)$. Then by continuity, there exists a $\delta$ such that for all $y$ with $|x – y| < \delta$, $|f(x) – f(y)| = |r – f(y)| < \frac{r}{2}$. Hence for all $y \in (x – \delta, x + \delta)$, $f(y) > \frac{r}{2}$. Hence $\int_a^b f \ dx \geq \int_{x – \delta}^{x + \delta} f \ dx \geq \frac{r}{2} \cdot 2 \delta > 0$. This contradicts $\int_a^b f \ dx = 0$.

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