The ratio in which the x-axis divides the are | vedclass

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(A) The curves are $y = x^2 - 4x$ and $y = 2x - x^2$. Intersection points: $x^2 - 4x = 2x - x^2 \Rightarrow 2x^2 - 6x = 0 \Rightarrow 2x(x - 3) = 0$,s

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$\frac{4}{23}$

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(A) The curves are $y = x^2 - 4x$ and $y = 2x - x^2$. Intersection points: $x^2 - 4x = 2x - x^2 \Rightarrow 2x^2 - 6x = 0 \Rightarrow 2x(x - 3) = 0$,so $x = 0$ and $x = 3$.Total area $A = \int_{0}^{3} |(2x - x^2) - (x^2 - 4x)| \, dx = \int_{0}^{3} |6x - 2x^2| \, dx = [3x^2 - \frac{2}{3}x^3]_{0}^{3} = (27 - 18) = 9$.The $x$-axis divides the region into two parts: one above the $x$-axis and one below.The part above the $x$-axis is bounded by $y = 2x - x^2$ from $x = 0$ to $x = 2$. Area $A_1 = \int_{0}^{2} (2x - x^2) \, dx = [x^2 - \frac{x^3}{3}]_{0}^{2} = 4 - \frac{8}{3} = \frac{4}{3}$.The part below the $x$-axis is bounded by $y = x^2 - 4x$ from $x = 0$ to $x = 3$ (partially) and $y = 2x - x^2$ from $x = 2$ to $x = 3$. The total area below the $x$-axis is $A_2 = A - A_1 = 9 - \frac{4}{3} = \frac{23}{3}$.The ratio is $\frac{A_1}{A_2} = \frac{4/3}{23/3} = \frac{4}{23}$.

The ratio in which the $x$-axis divides the area of the region bounded by the curves $y = x^2 - 4x$ and $y = 2x - x^2$ is:

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