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Suppose I have an integral:

$F \equiv \int_{a}^{b} f(x) dx$

I know that $f(x)$ is real-valued, finite and non-negative everywhere on the interval $(a, b)$ where $a$, $b$ are real numbers or $\pm \infty$ and $a \leq b$. I know that in the obvious, well-behaved cases this implies $F \geq 0$.

Is there any obscure pathological case where $F < 0$? You may generalize to cases that involve double, triple or higher order integrals. However, for each integral the property that the upper bound is greater than or equal to the lower bound and the bounds are always real numbers or $\pm \infty$ must hold.

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    No. Integrals are monotone. If they weren't, they wouldn't be integrals.2011-01-21
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    I think you mean $a\leq b$. Otherwise your integral will be negative unless $f(x)$ is zero almost everywhere.2011-01-21
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    @Christian: Exactly right. Fixed.2011-01-21
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    A geometrical way to think about it is that the value of $F$ is the area **under** the graph of $f$ and **above** the $x$-axis, and so if $f$ is non-negative this area cannot be negative.2011-01-21
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    @Asaf: I thought of the geometric interpretation. It's just that this assumption is an intermediate step in a proof I'm working on, the proof needs to work even in pathological cases, and in my experience such geometric, intuitive arguments often fall apart in pathological cases.2011-01-21
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    @dsimcha: this is not a matter of intuition, it is a matter of what integrals _mean._ If this were true, then the appropriate response wouldn't be to gaze in awe at how non-intuitive mathematics can be but to change the definition of integration. (Admittedly, erring on the side of caution is far preferable to the alternative.)2011-01-21

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