The area bounded by the curves y = x(x - 3)^2 | vedclass

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(D) To find the area bounded by the curves $y = x(x - 3)^2$ and $y = x$,we first find their points of intersection by setting $x(x - 3)^2 = x$.$x((x -

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$8$

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(D) To find the area bounded by the curves $y = x(x - 3)^2$ and $y = x$,we first find their points of intersection by setting $x(x - 3)^2 = x$.$x((x - 3)^2 - 1) = 0$$x(x^2 - 6x + 9 - 1) = 0$$x(x^2 - 6x + 8) = 0$$x(x - 2)(x - 4) = 0$The points of intersection are $x = 0, 2, 4$.The area $A$ is given by the integral:$A = \int_{0}^{2} |x(x - 3)^2 - x| dx + \int_{2}^{4} |x - x(x - 3)^2| dx$For $x \in [0, 2]$,$x(x - 3)^2 \ge x$,so the integral is $\int_{0}^{2} (x^3 - 6x^2 + 8x) dx$.$= [\frac{x^4}{4} - 2x^3 + 4x^2]_{0}^{2} = (\frac{16}{4} - 2(8) + 4(4)) - 0 = 4 - 16 + 16 = 4$.For $x \in [2, 4]$,$x \ge x(x - 3)^2$,so the integral is $\int_{2}^{4} (x - (x^3 - 6x^2 + 9x)) dx = \int_{2}^{4} (-x^3 + 6x^2 - 8x) dx$.$= [-\frac{x^4}{4} + 2x^3 - 4x^2]_{2}^{4} = (-64 + 128 - 64) - (-4 + 16 - 16) = 0 - (-4) = 4$.Total Area $= 4 + 4 = 8 	ext{ sq. units}$.

The area bounded by the curves $y = x(x - 3)^2$ and $y = x$ is (in sq. units) :

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