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How does the integral of f(x) with respect to x in $R^2$ change if a non-Euclidean metric is used? For instance, let f(x)=$x^2$. Would the value of a definite integral along an interval change?

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    on what space is your distance defined ? (by the way I think you're missing a square somewhere). also integrals are defined with respect to a measure, not a distance...2011-11-24
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    I think it is more a matter of integrating against a measure, which may end up taking the metric into account, not the metric itself.2011-11-24
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    and also if as I think you mean to define a distance on $R^2$ then for some $f$ what you said is not a distance (if there is a nonzero $x$ such that $f(x)=0$ for example)2011-11-24
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    One thing that may be related here is that of fundamental forms, which are a variant of the Pythagorean thm. These forms depend on the parametrization of the surface: http://en.wikipedia.org/wiki/First_fundamental_form2011-11-24
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    I meant, since the norm (and so the metric) in $\mathbb R^n$ is generated by an inner-product, a different metric may ( or may not) be generated by a different inner-product.2011-11-24
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    Thank you @gary. I would appreciate an explanation from the downvoter.2011-11-24
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    @analysij, no problem; sorry to tell you : don't count on an explanation, I had the same problem, brought up the issue in "meta", arguing against undexplained downvotes, and I was strongly downvoted there. I can only say I did not do it myself.2011-11-25
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    @analysisj I downvoted before the edit of the question and now I cannot un-downvote. As it was formulated, the question did not make any sense, $f$ was not clearly defined, nor was the metric.2011-11-25

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