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Here's the definition of an integral from Wikipedia:

Given a function $f$ of a real variable $x$ and an interval $[a, b]$ of the real line, the definite integral

$$\int_a^b f(x) \, dx$$

is defined informally to be the area of the region in the $xy$-plane bounded by the graph of $f$, the $x$-axis, and the vertical lines $x = a$ and $x = b$, such that area above the $x$-axis adds to the total, and that below the $x$-axis subtracts from the total.

Why do they specify a closed interval? Wouldn't using $(a,b)$ make no difference as the contribution to the integral from the endpoints $a$ and $b$ is zero as points have no width?

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    Why does the definition of an integral specify a closed interval since it makes no difference? Because it makes no difference.2012-09-24
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    It may be motivated by the fact that $\int_a^b f(x) dx$ exists if $f$ is continuous on $[a,b]$, but not necessarily if $f$ is merely continuous on $(a,b)$.2012-09-24
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    @did Well by specifying a closed interval it implies that a closed interval is necessary. Wouldn't it be better to use both an open and closed interval in the definition?2012-09-24
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    At the level of informality of the passage you reproduced, this makes no difference since every function on (a,b) can be extended to some function(s) on [a,b]. If one starts defining things rigorously, @HagenvonEitzen's remark applies.2012-09-24
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    @did Going by what Hagen said it seems it is a necessity to have a closed interval, I presume because f(x) could approach infinity as x goes to 0 or 1 if we took an open interval.2012-09-24
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    @Jim_CS: You might as well ask, why not allow finitely many point gaps in the definition. And for example, with Riemann integral, you want to evaluate $f$ at *arbitrary* points in the sub-intervals, which might be the left or right ends. Starting with open intervals (or gaps), you have to consider too many stupid special cases. On the other hand, You can show that the "open" integral exists iff the "closed" integral exists for $f$ extended aith zwero (or arbitrary values, cf. what did said)2012-09-24
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    @Did Even at that level you'd expect not to be able to define the area under 1/x over [0,1] in the same way you define it for x.2013-03-31
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    @LokiClock You do not have to make a, mostly irrelevant, comment to support your newly posted answer. By the way, Riemann integrals on closed intervals are not restricted to continuous functions, at all.2013-03-31
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    @Did That wasn't the point of my comment. It was to illustrate how discontinuities affect an intuition about taking areas under curves over closed intervals, how indefinite integrals come about. My answer is related because after I replied to you I thought to mention how the area not being affected by taking the open interval instead isn't special about open intervals.2013-04-01
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    We had this question again here: http://math.stackexchange.com/questions/461459/is-there-any-notable-difference-between-studying-the-riemann-integral-over-open/464512#4645122013-08-22

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