QJEEM24Ma2132: JEE Main 2024 Mathematics - Class 12 JEEM24Ma2132 Solution: The area enclosed between the curves and is :

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

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

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

$$\begin{array}{rl}\therefore & \text{ Area }=\underset{-2}{\overset{0}{\int }}(-x^2-2x)dx\\ & ={[\frac{-x^3}{3}-\frac{2x^2}{2}]}_{-2}^0\\ & =-[\frac{8}{3}-4]=\frac{4}{3}\text{ sq. unit }\end{array}$$