Question

Integral doble, ¿como la resuelvo?

∫⁰₋₂ ∫ ˣ₀ 2x eˣ² dx dy

Answer 1

$\displaystyle\\I=\int_{-2}^0\int_{0}^x2xe^{x^2}dx~dy=2\int_{-2}^{0}x^2e^{x^2}~dx\\\\\\I=\int_{-2}^{0}xd(e^{x^2})=\left.xe^{x^2}\right|_{-2}^0-\int_{-2}^0e^{x^2}~dx\\\\\\I=2e^{4}-\int_{-2}^0e^{x^2}~dx\\\\\\I=2e^{4}+\dfrac{\sqrt\pi i\text{ erf}(2i)}{2}$

$\displaystyle\text{Donde } \text {erf} (x)={\frac {2}{\sqrt {\pi }}}\int _{0}^{x}e^{-t^{2}}\,\mathrm {d} t$