QJEEM24Ma45d8: JEE Main 2024 Mathematics - Class 12 JEEM24Ma45d8 Solution: The area of the region enclosed by the parabolas y = 4 x − x 2…

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QJEEM24Ma45d8: JEE Main 2024 Mathematics - Class 12 JEEM24Ma45d8 Solution: The area of the region enclosed by the parabolas y = 4 x − x 2…

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

Multiple Choice Options

A 329
B 143
C 4
D 6

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Accepted Answer

JEE Main 2024 (Online) 31st January Evening Shift Mathematics - Area Under The Curves Question 27 English Explanation

$\begin{aligned} Area = | \int_{1}^{4} [(4 x - x^{2}) - \frac{(x - 4)^{2}}{3}] d x \\ Area = {| \frac{4 x^{2}}{2} - \frac{x^{3}}{3} - \frac{(x - 4)^{3}}{9} |}_{1}^{4} \\ = | (\frac{64}{2} - \frac{64}{3} - \frac{4}{2} + \frac{1}{3} - \frac{27}{9}) | \\ \Rightarrow (27 - 21) = 6 \end{aligned}$

QJEEM24Ma45d8: JEE Main 2024 Mathematics - Class 12 JEEM24Ma45d8 Solution: The area of the region enclosed by the parabolas y = 4 x − x 2…

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