- #71

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Wasn't he the first who came up with this idea?Why would it be called that?

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- Thread starter martinbn
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- #71

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Wasn't he the first who came up with this idea?Why would it be called that?

- #72

martinbn

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Aristoteles! Do you mean Archimedes?Wasn't he the first who came up with this idea?

- #73

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Sure. How embarrassing ...Aristoteles! Do you mean Archimedes?

- #74

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Not if you are a physicist with an intuitive feeling for limitations of any heuristic rules.But if you are going to use ##\int df=f## and ##\int dx = x## in that line, then you may as well say that ##\int fdx=\int dxf=xf=fx## with the appropriate ##\cong##'s.

- #75

martinbn

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That is a good question on its own.Sure.

- #76

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Due to the Stigler's law: https://en.wikipedia.org/wiki/Stigler's_law_of_eponymyBtw., does anybody know why the Riemann integral isn't called Aristoteles integral?

- #77

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Which doesn't say anything about the notation ##\int fdx##.

But it does seem to write the Lebesgue integral in terms of infinitesimals.

- #78

martinbn

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But those infinitesimals are for another function. The point was to be able to make sense,at least intuitively, of ##\int dxf(x)## as an infinite sum of infinitesimals.But it does seem to write the Lebesgue integral in terms of infinitesimals.

- #79

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$$\int\int dxdy$$

of a region on a 2-dimensional plane?

- #80

martinbn

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- #81

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Let ##f## be a function from reals to reals (for simplicity, let it be nonnegative). Let ##S(A,B)## be the set of points ##(x,y)## such that ##A \leq x \leq B## and ##0 \leq y \leq f(x)##. Then if the area of set ##S(A,B)## is defined, it will be equal to ##\int_A^B f(x) dx##. Right? By "area" I mean a measure on ##R^2## such that every set of the form ##\{ (x,y) | x_0 \leq x \leq x_1 \wedge y_0 \leq y \leq y_1\}## has measure ##|(x_1 - x_0) \cdot (y_1 - y_0)|##

- #82

martinbn

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May be, it depends on the unstated assumptions. Say you take area to mean the Lebesgue measure in the plane. It satisfies your requirement for area. Take a function ##f(x)## so that ##S(A,B)## is Lebesgue measurable and ##f(x)## is not Riemann integrable. Then if the integral you wrote is the Riemann integral, then the area is not it, the integral doesn't exists. I assume that is not what you mean, but you didn't say anything about the relationship between the measure in the plane and the integral on the line in you statement. But let's say that it is all fine. Then what? Where are you going with this?Let ##f## be a function from reals to reals (for simplicity, let it be nonnegative). Let ##S(A,B)## be the set of points ##(x,y)## such that ##A \leq x \leq B## and ##0 \leq y \leq f(x)##. Then if the area of set ##S(A,B)## is defined, it will be equal to ##\int_A^B f(x) dx##. Right? By "area" I mean a measure on ##R^2## such that every set of the form ##\{ (x,y) | x_0 \leq x \leq x_1 \wedge y_0 \leq y \leq y_1\}## has measure ##|(x_1 - x_0) \cdot (y_1 - y_0)|##

- #83

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But let's say that it is all fine. Then what? Where are you going with this?

I was just trying to answer the question "What do you mean by area"?

- #84

nuuskur

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##\int dx f(x) = (x+C)f(x) ##

The expression ##\int\int dx\,dy ## is just some function ##f:\mathbb R ^2\to\mathbb R## (hopefully).

I'm having difficulty in understanding how either of them would be special cases of finding an 'area'. An application of either theory would be finding the area/volume/measure of an ##n##-dimensional, pre-determined to be measurable, set.

$$\int\int dxdy$$

of a region on a 2-dimensional plane?

The expression ##\int\int dx\,dy ## is just some function ##f:\mathbb R ^2\to\mathbb R## (hopefully).

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