Integration in Various Metric Spaces 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?
 A: This may not be 100% of an answer, but I think it is close: If you can find a reparametrization (x(t),y(t)) of $\mathbb R^2$ that gives rise to a different choice of distance, then the first fundamental form: http://en.wikipedia.org/wiki/Line_element for lengths, and http://en.wikipedia.org/wiki/First_fundamental_form for area
will describe how length and area of objects respectively. 
This will give you a rescaling of the standard dx's dy's you use to find the value of the integral. Note that for the case of Euclidean space, the coefficients of the second fundamental form are all 1, which return to us the standard Pythagorean theorem. This new parametrization gives us the form that the Pythagorean theorem equivalent would take in our space.
We could think, e.g., of $\mathbb R^2$ in polar coordinates, in which case the Pythagorean theorem would be given by:
$x=rcos \theta$ so that $dx=-rsin\theta d\theta$ , 
$y=rsin \theta$ , so dy=$rcos\theta d\theta$
allows us to construct a new Pythagorean theorem for a new surface, where we can then define the length of paths and areas in our new space.
In more general cases, you can think of Finsler manifolds http://en.wikipedia.org/wiki/Finsler_manifold
A: The answer is: the integral is not affected by considering other metric in $\mathbb{R}^2$. 
Think about this: the Lebesgue measure is build from the volume of intervals (boxes) in $\mathbb{R}^n$. The volume of an interval $I=[a_1,b_1]\times\ldots,\times [a_n,b_n]$ is $V(I)=\prod_{i=1}^n b_i-a_i$. Ultimately, the lebesgue measure depends on how the intervals are measured in $\mathbb{R}$. If you are considering the Riemann integral, it doesn't matter because, again, the Riemann integral depends on how the intervals  are measured.
So, the integral may change if we change the way of measuring intervals in $\mathbb{R}$. Note that usual measure of an interval $[a,b]$ in $\mathbb{R}$ agrees with $d(a,b)$ with the metric given by the absolute value. Now, consider $f:\mathbb{R}\to [0,\infty[$ strictly increasing. If we define $$l_f([a,b])=f(b)-f(a),$$
for closed bounded intervals, $$l_f(I)=l_f(\bar{I}),$$ for finite intervals (the bar is for closure), and $l_f(I)=\infty$ for infinite intervals, and we continues with all the constructions for the Riemann or Lebesgue integrals, based in the lengths of the intervals of $\mathbb{R}$, this probably affect the value of the integrals of integrable functions over subsets of $\mathbb{R}^n$. Moreover $d(x,y)=|f(x)-f(y)|$ is a metric in $\mathbb{R}$, for any strictly monotonic $f:\mathbb{R}\to[0,\infty[$, but note that the change was made in $\mathbb{R}$ not in $\mathbb{R}^n$
