# Integral of $f dx_1 \wedge dx_2$ zero over $\mathbb{R}^2$ and bump functions

Give an example of compactly supported smooth function $$f$$ on $$\mathbb{R}^2$$ such that $$\int_{\mathbb{R}^2} f dx_1 \wedge dx_2 = 0$$, but one of the functions $$f_1(x_1, x_2) = \int_{-\infty}^{x_1} f(t,x_2) dt$$ or $$f_2(x_1, x_2) = \int_{-\infty}^{x_2} f(x_1,t) dt$$ is not compactly supported where $$f_j : \mathbb{R}^2 \to \mathbb{R}$$.

Okay so I think bump functions will be the key here, but my current issue is not understanding what are these $$f_j$$'s are. What is the intuition for the defined maps?

I'm not sure how to provide useful "intuition" for the functions $$f_1$$ and $$f_2$$, but I can describe the gist of the argument.
Consider the function $$f(x_1, x_2) = \begin{cases} 1 & \text{if } (x_1, x_2) \in [-1, 0) \times [0, 1] \\ -1 & \text{if } (x_1, x_2) \in [0, 1] \times [0, 1] \\ 0 & \text{otherwise} \end{cases}.$$
This function is compactly supported, though it is not continuous (let alone smooth). By direct calculation, you should be able to see that $$f_2$$ is not compactly supported: the support of $$f_2$$ is $$[-1,1] \times [0, \infty)$$. And yet $$\int f dx_1 \wedge dx_2 = 0$$.
However, my $$f$$ doesn't solve your problem, since it's not smooth. So you need to think of an $$f$$ that has a similar profile to my $$f$$, but is smooth. Your idea of using bump functions should do the trick.