Area Between Two Curves

first find the intersection points ...

for $x \geq 0$ ...

$4xe^{-x^2} = x$

$4xe^{-x^2} - x = 0$

$x(4e^{-x^2} - 1) = 0$

$x = 0$ , and ...

$e^{-x^2} = \frac{1}{4}$

$-x^2 = \ln\left(\frac{1}{4}\right)$

$x^2 = \ln{4}$

$x = \sqrt{\ln{4}}$

for $x < 0$ ...

$4xe^{-x^2} = -x$

$4xe^{-x^2} + x = 0$

$x(4e^{-x^2} + 1) = 0$

$x = 0$ only.

$A = \int_0^{\sqrt{\ln{4}}} 4xe^{-x^2} - x \, dx$

$A = \left[-2e^{-x^2} - \frac{x^2}{2}\right]_0^{\sqrt{\ln{4}}}$

you can evaluate the area value yourself.