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

Question

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

Accepted Answer

Setting $4x-x^2=\dfrac{(x-4)^2}{3}$: $3(4x-x^2)=(x-4)^2 \Rightarrow 4x^2-20x+16=0 \Rightarrow x^2-5x+4=0\Rightarrow x=1,4$.

$$\text{Area}=\int_1^4\left[(4x-x^2)-\frac{(x-4)^2}{3}\right]dx=\frac13\int_1^4(-4x^2+20x-16)\,dx.$$

$$=\frac13\left[-\frac{4x^3}{3}+10x^2-16x\right]_1^4=\frac13\left(\frac{32}{3}-\left(-\frac{22}{3}\right)\right)=\frac13(18)=6.$$

Solutions · 1