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Question Number 162651 by Mathematification last updated on 31/Dec/21

	Answered by tounghoungko last updated on 31/Dec/21
	$y=−x(x+a)^2 =−x^3 −2ax^2 −a^2 x y′=−3x^2 −4ax−a^2 = 0 3x^2 +4ax+a^2 =0 (3x+a)(x+a)=0 { ((x=−(1/3)a)),((x=−a)) :} y′′=−6x−4a >0 (for minimum) x<−(2/3)a so min at x=−(1/3)a =(2/3); a=−2 ∴ y= −x(x−2)^2 intersect at x−axis { ((x=0)),((x=2)) :} The yellow area is A =−∫_0 ^( 2) −x(x−2)^2 dx A= ∫_0 ^( 2) (x^3 −4x^2 +4x)dx A= (1/4).16−(4/3).8+2.2^2 = (4/3)$
	$y = - x {(x + a)}^{2} = - x^{3} - 2 {a x}^{2} - a^{2} x$ $y^{'} = - 3 x^{2} - 4 a x - a^{2} = 0$ $3 x^{2} + 4 a x + a^{2} = 0$ $(3 x + a) (x + a) = 0 {\begin{cases} x = - \frac{1}{3} a \\ x = - a \end{cases}$ $y^{″} = - 6 x - 4 a > 0 (f o r m i n i m u m)$ $x < - \frac{2}{3} a s o m i n a t x = - \frac{1}{3} a = \frac{2}{3}; a = - 2$ $∴ y = - x {(x - 2)}^{2}$ $i n t e r s e c t a t x - a x i s {\begin{cases} x = 0 \\ x = 2 \end{cases}$ $T h e y e l l o w a r e a i s$ $A = - \int_{0}^{2} - x {(x - 2)}^{2} d x$ $A = \int_{0}^{2} (x^{3} - 4 x^{2} + 4 x) d x$ $A = \frac{1}{4} . 16 - \frac{4}{3} . 8 + 2 . 2^{2} = \frac{4}{3}$
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