Write the two curves in comparable form

The given parabolas are $y=4x-x^2$ and $3y=(x-4)^2 \,\Rightarrow\, y=\frac{(x-4)^2}{3}.$

Find the points of intersection

Set the two expressions for $y$ equal: $4x-x^2=\frac{(x-4)^2}{3}.$

Multiply by $3$: $12x-3x^2=x^2-8x+16.$

Rearranging, $-4x^2+20x-16=0$ $4x^2-20x+16=0$ $x^2-5x+4=0$ $(x-1)(x-4)=0.$

Hence, the curves intersect at $x=1 \quad \text{and} \quad x=4.$

Determine which curve lies above the other

Let $f(x)=4x-x^2, \qquad g(x)=\frac{(x-4)^2}{3}.$

Check at a point between $1$ and $4$, say $x=2$: $f(2)=8-4=4,$ $g(2)=\frac{(2-4)^2}{3}=\frac{4}{3}.$

So on $[1,4]$, the upper curve is $y=4x-x^2$ and the lower curve is $y=\dfrac{(x-4)^2}{3}$.

Set up the area integral

Area enclosed is $A=\int_{1}^{4}\left[\left(4x-x^2\right)-\frac{(x-4)^2}{3}\right]dx.$

Simplify the integrand

First expand: $\frac{(x-4)^2}{3}=\frac{x^2-8x+16}{3}.$

Therefore, $\left(4x-x^2\right)-\frac{x^2-8x+16}{3}$ $=\frac{12x-3x^2-(x^2-8x+16)}{3}$ $=\frac{20x-4x^2-16}{3}$ $=\frac{-4(x^2-5x+4)}{3}$ $=\frac{-4(x-1)(x-4)}{3}.$

So, $A=\int_1^4 \frac{20x-4x^2-16}{3}\,dx.$

Integrate

$A=\frac{1}{3}\int_1^4 (20x-4x^2-16)\,dx.$

Antiderivative: $\int (20x-4x^2-16)dx=10x^2-\frac{4x^3}{3}-16x.$

Hence, $A=\frac{1}{3}\left[10x^2-\frac{4x^3}{3}-16x\right]_1^4.$

At $x=4$: $10(16)-\frac{4(64)}{3}-64=160-\frac{256}{3}-64=96-\frac{256}{3}=\frac{32}{3}.$

At $x=1$: $10-\frac{4}{3}-16=-6-\frac{4}{3}=-\frac{22}{3}.$

Subtract: $\frac{32}{3}-\left(-\frac{22}{3}\right)=\frac{54}{3}=18.$

Therefore, $A=\frac{1}{3}\cdot 18=6.$

Match with the options

$A=6$

So the correct option is D.

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.