integral of continuous function can be very small 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 $\varepsilon>0$.


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.
