$$y=x|x|$$ & $$y=x-|x|$$

$$\begin{aligned} & y=x|x|= \begin{cases}x^2, & x>0 \\ -x^2, & x<0\end{cases} \\ & y=x-|x|= \begin{cases}0, & x>0 \\ 2 x, & x<0\end{cases} \end{aligned}$$

Point of intersections are $$(0,0)$$ & $$(-2,4)$$.

$$\begin{aligned} \therefore \quad & \text { Area }=\int_\limits{-2}^0\left(-x^2-2 x\right) d x \\ & =\left[\frac{-x^3}{3}-\frac{2 x^2}{2}\right]_{-2}^0 \\ & =-\left[\frac{8}{3}-4\right]=\frac{4}{3} \text { sq. unit } \end{aligned}$$