Area Under Curves Question 12
Question 12 - 2024 (31 Jan Shift 2)
The area of the region enclosed by the parabola $y=4 x-x^{2}$ and $3 y=(x-4)^{2}$ is equal to
(1) $\frac{32}{9}$
(2) 4
(3) 6
(4) $\frac{14}{3}$
Show Answer
Answer (3)
Solution
$$ \begin{aligned} & \text { Area }=\left|\int_{1}^{4}\left[\left(4 x-x^{2}\right)-\frac{(x-4)^{2}}{3}\right]\right| d x \ & \text { Area }=\left|\frac{4 x^{2}}{2}-\frac{x^{3}}{3}-\frac{(x-4)^{3}}{9}\right|_{1}^{4} \ & =\left|\left(\frac{64}{2}-\frac{64}{3}-\frac{4}{2}+\frac{1}{3}-\frac{27}{9}\right)\right| \ & \Rightarrow(27-21)=6 \end{aligned} $$
