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I think that the title of the thread summarises all I would like to know about an integral of one variable only. The question may appear silly (maybe it actually is), but it stems from the idea of the raw approximation of an integral provided by calculating the area of a rectangle over the domain of integration.

Bounded
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    Welcome to MSE. Please clarify your question. What do you mean by “depends at least linearly”? – José Carlos Santos Feb 22 '18 at 11:40
  • Hi thanks. I think that it is better-off to tell the whole story. I need to deal with some definite integrals and the only thing I know is that the integrand function, say $f$, is bounded over the integration interval, which is of the form $(-\epsilon, \epsilon)$ and $\epsilon=O(n^{-1/2})$, where $n$ is a quantity that potentially grows to infinity. The goal is to say that the integral is just $O(n^{-1/2})$, not interested on its actual value at all. (continued below) – Bounded Feb 22 '18 at 13:51
  • Because I have not much information to try to figure out how the integral would look like, I will be just happy to say that "always" the integral can be thought as a linear function of integration extremes. In writing the reply, I came out with the following argument: $\int_{-\epsilon}^\epsilon f(x) d(x)=g(\epsilon)-g(-\epsilon)$. Taylor expanding $g$ about $0$ gives $g(0)-g(0)+\epsilon g'(0) -\epsilon g'(0) + O(\epsilon^2)=O(\epsilon)$, as long as $g'(0)$ and $g''(0)$ are bounded. Does it makes sense? – Bounded Feb 22 '18 at 13:51

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Let's try to answer your question: Let $f:[-\varepsilon,\varepsilon]\rightarrow\mathbb{R}$ be bounded by a constant $C>0$ and integrable. Then $$|\int_{-\varepsilon}^\varepsilon f(t)\, dt|\leq \int_{-\varepsilon}^\varepsilon |f(t)|\, dt \leq C\int_{-\varepsilon}^{\varepsilon}dt=2C\varepsilon$$ hence the integral is in $O(\varepsilon)$.

humanStampedist
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