If the area of the region {(x, y): 0 leq y le | vedclass

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(B) To find the area $A$,we first determine the intersection points of the curves $y = 2x$ and $y = 6x - x^2$.Setting $2x = 6x - x^2$,we get $x^2 - 4x

Vedclass · Accepted Answer

$304$

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(B) To find the area $A$,we first determine the intersection points of the curves $y = 2x$ and $y = 6x - x^2$.Setting $2x = 6x - x^2$,we get $x^2 - 4x = 0$,which implies $x(x - 4) = 0$. Thus,the curves intersect at $x = 0$ and $x = 4$.For $0 \leq x \leq 4$,$2x \leq 6x - x^2$,so $\min \{2x, 6x - x^2\} = 2x$.For $x > 4$,$6x - x^2 The curve $y = 6x - x^2$ intersects the $x$-axis at $x = 0$ and $x = 6$.Thus,the area $A$ is given by:$A = \int_0^4 2x \, dx + \int_4^6 (6x - x^2) \, dx$Calculating the first integral:$\int_0^4 2x \, dx = [x^2]_0^4 = 16 - 0 = 16$Calculating the second integral:$\int_4^6 (6x - x^2) \, dx = [3x^2 - \frac{x^3}{3}]_4^6 = (3(36) - \frac{216}{3}) - (3(16) - \frac{64}{3}) = (108 - 72) - (48 - \frac{64}{3}) = 36 - \frac{144 - 64}{3} = 36 - \frac{80}{3} = \frac{108 - 80}{3} = \frac{28}{3}$Therefore,$A = 16 + \frac{28}{3} = \frac{48 + 28}{3} = \frac{76}{3}$.Finally,$12A = 12 	imes \frac{76}{3} = 4 	imes 76 = 304$.

If the area of the region $\{(x, y): 0 \leq y \leq \min \{2x, 6x-x^2\}\}$ is $A$,then $12A$ is equal to

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