Answer
$$ A=36 $$
Work Step by Step
The Area enclosed between two curves $y = f(x)$ and $y = g(x)$ over $[a,b]$ such that $f(x) \geq g(x)$ is $$\displaystyle\int_{a}^{b} f(x) - g(x) dx$$
From the graph, we can see $2x \geq x^{2} -4x$ on the interval $[0,6]$
Therefore, the area between the two curves is $$\displaystyle\int_{0}^{6} 6x-x^{2}dx$$ $$\left[ 3x^{2} - \frac{x^{3}}{3} \right]^{6}_0$$
$$ = 3\times 6^{2} - \frac{6^{3}}{3} = 36$$
