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

Question

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

Accepted Answer

$$\begin{aligned}
& \text { Area }=\left\lvert\, \int_\limits1^4\left[\left(4 x-x^2\right)-\frac{(x-4)^2}{3}\right] d x\right. \
& \text { Area }=\left|\frac{4 x^2}{2}-\frac{x^3}{3}-\frac{(x-4)^3}{9}\right|_1^4 \
& =\left|\left(\frac{64}{2}-\frac{64}{3}-\frac{4}{2}+\frac{1}{3}-\frac{27}{9}\right)\right| \
& \Rightarrow(27-21)=6
\end{aligned}$$

Solutions · 1