The area of the region enclosed by the parabo | vedclass

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(C) To find the area,we first determine the intersection points of the two curves:$y = 4x - x^2$ and $3y = (x - 4)^2$Substitute $y$ from the first equ

Vedclass · Accepted Answer

$6$

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(C) To find the area,we first determine the intersection points of the two curves:$y = 4x - x^2$ and $3y = (x - 4)^2$Substitute $y$ from the first equation into the second:$3(4x - x^2) = (x - 4)^2$$12x - 3x^2 = x^2 - 8x + 16$$4x^2 - 20x + 16 = 0$$x^2 - 5x + 4 = 0$$(x - 1)(x - 4) = 0$So,the curves intersect at $x = 1$ and $x = 4$.The area $A$ is given by the integral of the upper curve minus the lower curve from $x = 1$ to $x = 4$:$A = \int_1^4 \left[ (4x - x^2) - \frac{(x - 4)^2}{3} \right] dx$$A = \left[ 2x^2 - \frac{x^3}{3} - \frac{(x - 4)^3}{9} \right]_1^4$$A = \left( 2(4)^2 - \frac{(4)^3}{3} - \frac{(4 - 4)^3}{9} \right) - \left( 2(1)^2 - \frac{(1)^3}{3} - \frac{(1 - 4)^3}{9} \right)$$A = \left( 32 - \frac{64}{3} - 0 \right) - \left( 2 - \frac{1}{3} - \frac{-27}{9} \right)$$A = \left( \frac{96 - 64}{3} \right) - \left( 2 - \frac{1}{3} + 3 \right)$$A = \frac{32}{3} - \left( 5 - \frac{1}{3} \right) = \frac{32}{3} - \frac{14}{3} = \frac{18}{3} = 6$

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

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