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Let $f(x)\in\mathbb{R}$, $x\in\mathbb{R}^{+}$, with $f(x)$ continuous. Is the following statement true?

$\exists M, \forall y> x\geq M$ such that $\lvert\int_{x}^{y}{f(s)ds}\rvert\leq \varepsilon$ for all $\varepsilon>0$.

81235
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    As currently written, If I pick $y = x$, then $|\int_{x}^{y}f(s),ds| = 0 \le \epsilon$ for all $\epsilon > 0$. So the statement is true. However, I suspect that you meant to have the conditionals in a different order. – JimmyK4542 Dec 15 '15 at 03:41
  • @JimmyK4542. revised. – 81235 Dec 15 '15 at 04:07

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The only way "$|\int_x^y f(s)\,ds| \le \epsilon$ for all $\epsilon > 0$" can be true is if $\int_x^y f(s)\,ds = 0$.

Hence, the statement is equivalent to "$\exists M, \forall y > x \ge M, \int_x^y f(s)\,ds = 0$"

This clearly fails if we take $f(x) = 1$. Hence, the statement "$\exists M, \forall y> x\geq M$ such that $|\int_{x}^{y}{f(s)ds}|\leq \varepsilon$ for all $\varepsilon>0$" is false.

JimmyK4542
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