The area bounded by the curve y = 4x - x^2 an | vedclass

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(A) The given curve is $y = 4x - x^2$. To find the points where the curve intersects the $X$-axis,we set $y = 0$: $4x - x^2 = 0$ $x(4 - x) = 0$ So,the

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$\frac{32}{3}$

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(A) The given curve is $y = 4x - x^2$. To find the points where the curve intersects the $X$-axis,we set $y = 0$: $4x - x^2 = 0$ $x(4 - x) = 0$ So,the intersection points are $x = 0$ and $x = 4$. The area $A$ is given by the integral: $A = \int_{0}^{4} (4x - x^2) \, dx$ $A = [2x^2 - \frac{x^3}{3}]_{0}^{4}$ $A = (2(4)^2 - \frac{4^3}{3}) - (0)$ $A = (2 	imes 16 - \frac{64}{3})$ $A = 32 - \frac{64}{3}$ $A = \frac{96 - 64}{3} = \frac{32}{3}$ square units.

The area bounded by the curve $y = 4x - x^2$ and the $X$-axis in square units is:

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