Differential forms and order of integration I don't understand how $$
  \int_{a_2}^{b_2} \int_{a_1}^{b_1} f(t_1,t_2) dt_1 dt_2 =
    \int_{a_1}^{b_1} \int_{a_2}^{b_2} f(t_1,t_2) dt_2 dt_1
$$ can agree with the fact that $dt_1 \wedge dt_2 = -dt_2 \wedge dt_1$.
I tried to work with pullbacks, but I must be doing something wrong. I'd really like to see a very low-level step-by-step derivation of the equality above from the point of view of differential forms.
 A: From the point of view of differential forms what you have is:
$$
\int_{[a_1,b_1]\times[a_2,b_2]}
f~dt_1\wedge dt_2$$
without order.
What I think you are looking for is Fubini's theorem.
Edit:
To add a bit more information, in general, we start by thinking $\int_\Omega
f~dt_1\wedge dt_2$ as an integration over chains, and after dealing with orientation we move to compute it with integration with respect to some standard measure as $\int_{[a_1,b_1]\times[a_2,b_2]}
f~dt_1dt_2$.
A: This formula is about integration of a real-valued function $f:\Bbb R^n\to\Bbb R$, not about integration of differential forms. And this formula should have been introduced way before you learn differential forms.
You should interpret the integration as doing two Riemann integrations. You read it as $$
  \int_{a_2}^{b_2} \left(\int_{a_1}^{b_1} f(t_1,t_2) dt_1\right) dt_2 =
    \int_{a_1}^{b_1} \left(\int_{a_2}^{b_2} f(t_1,t_2) dt_2\right) dt_1.
$$
Edit: Also, $dt_1 dt_2$ is not to be interpreted as $dt_1\wedge dt_2$. The reason is simple: this formula is introduced before differential forms.
Edit 2: I should write down the whole settings of Fubini's theorem.
First of all, assume you do NOT know anything about differential forms and their integrations. This is important, because Fubini's theorem is stated and proved without the knowledge of differential forms.
Let $f:[a_1,b_1]\times[a_2,b_2]\to\Bbb R$ be a continuous function. We could define the Riemann integral of $f$ over the rectangle $[a_1,b_1]\times[a_2,b_2]$, denoted $\int_{[a_1,b_1]\times[a_2,b_2]} f$. You could Riemann-integrate $f$ with respect to one variable and regard it as a function of another variable:
$$g(t_2):=\int_{a_1}^{b_1}f(t_1,t_2)dt_1, \ \ \ \ \ \ h(t_1):=\int_{a_2}^{b_2}f(t_1,t_2)dt_2.$$ So you obtain two functions $g:[a_2,b_2]\to\Bbb R,\ h:[a_1,b_1]\to\Bbb R$, both being single-variable functions.
Fubini's theorem tells you that both $g,h$ are also Riemann-integrable, and $$\int_{[a_1,b_1]\times[a_2,b_2]} f=\int_{a_2}^{b_2}g=\int_{a_1}^{b_1}h.$$
There should be no differential forms mentioned in the settings of Fubini's theorem, for they are unhelpful.
As for why the author writes $f(t_1,t_2)dt_1dt_2=-f(t_1,t_2)dt_2dt_1$, it is that his own abuse of notation caused himself trouble. This formula is in fact incorrect. What is correct is $f(t_1,t_2)dt_1\wedge dt_2=-f(t_1,t_2)dt_2\wedge dt_1$.
