If the area of the region \ ( x, y ) : 0 ≤ y ≤ min \ 2x, 6x - x ^ 2...

Solving, $y = 2x\text{and}y = 6x - {x}^{2}$

$⇒ 2x = 6x - {x}^{2}$

$⇒ 4x - {x}^{2} = 0$

$⇒ x\left(4 - x\right) = 0$

$⇒ x = 0, 4$

$⇒ y = 0, 8$

$⇒ \left(x, y\right) ≡ \left(0, 0\right), \left(4, 8\right)$

So, the required area is given by,

$A = \dfrac{1}{2} \times 4 \times 8 + {∫}_{4}^{6}\left(6x - {x}^{2}\right)dx$

$⇒ A = 16 + {\left[3{x}^{2} - \dfrac{{x}^{3}}{3}\right]}_{4}^{6}$

$⇒ A = \dfrac{76}{3}$

$⇒ 12A = 304$