The area of the region enclosed by the parabola y = 4 x − x 2 and 3 y = x − 4 2 is equal to

Given: $y=4x-x^2 \text{and} 3y={\left(x-4\right)}^2$

$\Rightarrow 4x-x^2=\frac{{\left(x-4\right)}^2}{3}$

$\Rightarrow 12x-3x^2=x^2+16-8x$

$\Rightarrow 4x^2-20x+16=0$

$\Rightarrow x^2-5x+4=0$

$\Rightarrow \left(x-4\right)\left(x-1\right)=0$

$\Rightarrow x=4, 1$

$\Rightarrow y=0, 3$

So, the intersection points are $\left(4,0\right) \text{and} \left(1,3\right)$.

Area $={\int }_1^4\left[\left(4x-x^2\right)-\frac{{\left(x-4\right)}^2}{3}\right]dx$

Area $={\left|\frac{4x^2}{2}-\frac{x^3}{3}-\frac{{\left(x-4\right)}^3}{9}\right|}_1^4$

$=\left|\left(\frac{64}{2}-\frac{64}{3}-\frac{4}{2}+\frac{1}{3}-\frac{27}{9}\right)\right|$

$=\left(27-21\right)=6$