$$ \int_0^2 \int_0^{4-x^2} \frac{xe^{2y}}{4-y} \, dy\, dx $$
I'm stuck with this problem. I think I should change it so I integrate with respect to $dx \, dy$ but I'm not sure. Any help? Thanks
2026-03-26 19:03:48.1774551828
Solve double integral
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Yes, integrate first with respect to $x$, where $x$ goes from $0$ to $\sqrt{4-y}$. That will cancel the nasty $4-y$ at the bottom.