JEE MAIN - Mathematics (2024 - 31st January Evening Shift - No. 9)
The area of the region enclosed by the parabolas $$y=4 x-x^2$$ and $$3 y=(x-4)^2$$ is equal to :
$$\frac{32}{9}$$
$$\frac{14}{3}$$
4
6
Explanation
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$$\begin{aligned} & \text { Area }=\left\lvert\, \int_\limits1^4\left[\left(4 x-x^2\right)-\frac{(x-4)^2}{3}\right] d x\right. \\ & \text { Area }=\left|\frac{4 x^2}{2}-\frac{x^3}{3}-\frac{(x-4)^3}{9}\right|_1^4 \\ & =\left|\left(\frac{64}{2}-\frac{64}{3}-\frac{4}{2}+\frac{1}{3}-\frac{27}{9}\right)\right| \\ & \Rightarrow(27-21)=6 \end{aligned}$$
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