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} $