The area (in sq. units) enclosed by the curve $y=\frac{1}{2}x|x|$ and the line y = x is :

$\frac{4}{3}$

The correct answer is Option (2) → $\frac{4}{3}$

$y=\frac{1}{2}x|x|=\left\{\begin{matrix}\frac{x^2}{2}&x≥0\\-\frac{x^2}{2}&x<0\end{matrix}\right.$

area is symmetric

area = $2×\int\limits_0^2x-\frac{x^2}{2}dx$

$=2\left[\frac{x^2}{2}-\frac{x^3}{6}\right]_0^2$

$=2\left[\frac{4}{2}-\frac{8}{6}\right]$

$=\frac{4}{3}$ sq. units