The area of the region enclosed by the parabola $$y = 4x - x^2$$ and $$3y = (x - 4)^2$$ is equal to

We need to find the area enclosed by the parabola $$y = 4x - x^2$$ and the curve $$3y = (x-4)^2$$.

Curve 1: $$y = 4x - x^2$$
Curve 2: $$y = \frac{(x-4)^2}{3} = \frac{x^2 - 8x + 16}{3}$$
Setting them equal gives $$4x - x^2 = \frac{x^2 - 8x + 16}{3}$$, which leads to $$12x - 3x^2 = x^2 - 8x + 16$$, then $$4x^2 - 20x + 16 = 0$$ and $$x^2 - 5x + 4 = 0$$, so $$(x-1)(x-4) = 0$$ and the curves intersect at $$x = 1$$ and $$x = 4$$.

To determine which curve is above the other on $$[1,4]$$, we evaluate at $$x = 2$$: Curve 1 gives $$y = 8 - 4 = 4$$ and Curve 2 gives $$y = \frac{4}{3} \approx 1.33$$, so Curve 1 lies above Curve 2 on this interval.

The area is calculated as $$A = \int_1^4 \left[(4x - x^2) - \frac{(x-4)^2}{3}\right] dx = \int_1^4 \left[4x - x^2 - \frac{x^2 - 8x + 16}{3}\right] dx = \int_1^4 \frac{12x - 3x^2 - x^2 + 8x - 16}{3}\, dx = \int_1^4 \frac{-4x^2 + 20x - 16}{3}\, dx = \frac{1}{3}\int_1^4 (-4x^2 + 20x - 16)\, dx = \frac{1}{3}\left[-\frac{4x^3}{3} + 10x^2 - 16x\right]_1^4.$$ Evaluating at $$x = 4$$ gives $$-\frac{256}{3} + 160 - 64 = -\frac{256}{3} + 96 = \frac{32}{3}$$ and at $$x = 1$$ gives $$-\frac{4}{3} + 10 - 16 = -\frac{4}{3} - 6 = \frac{-22}{3}$$, so $$A = \frac{1}{3}\left(\frac{32}{3} - \frac{-22}{3}\right) = \frac{1}{3} \cdot \frac{54}{3} = \frac{54}{9} = 6.$$

Option C) 6