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Question-162651




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)
$$\:{y}=−{x}\left({x}+{a}\right)^{\mathrm{2}} \:=−{x}^{\mathrm{3}} −\mathrm{2}{ax}^{\mathrm{2}} −{a}^{\mathrm{2}} {x} \\ $$$$\:{y}'=−\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}{ax}−{a}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\:\:\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{ax}+{a}^{\mathrm{2}} =\mathrm{0}\: \\ $$$$\:\left(\mathrm{3}{x}+{a}\right)\left({x}+{a}\right)=\mathrm{0}\:\begin{cases}{{x}=−\frac{\mathrm{1}}{\mathrm{3}}{a}}\\{{x}=−{a}}\end{cases} \\ $$$$\:\:{y}''=−\mathrm{6}{x}−\mathrm{4}{a}\:>\mathrm{0}\:\left({for}\:{minimum}\right) \\ $$$$\:\:{x}<−\frac{\mathrm{2}}{\mathrm{3}}{a}\:{so}\:{min}\:{at}\:{x}=−\frac{\mathrm{1}}{\mathrm{3}}{a}\:=\frac{\mathrm{2}}{\mathrm{3}};\:{a}=−\mathrm{2} \\ $$$$\:\:\therefore\:{y}=\:−{x}\left({x}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\:{intersect}\:{at}\:{x}−{axis}\:\begin{cases}{{x}=\mathrm{0}}\\{{x}=\mathrm{2}}\end{cases} \\ $$$$\:{The}\:{yellow}\:{area}\:{is}\: \\ $$$$\:{A}\:=−\int_{\mathrm{0}} ^{\:\mathrm{2}} \:−{x}\left({x}−\mathrm{2}\right)^{\mathrm{2}} \:{dx}\: \\ $$$$\:{A}=\:\int_{\mathrm{0}} ^{\:\mathrm{2}} \left({x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}\right){dx} \\ $$$$\:{A}=\:\frac{\mathrm{1}}{\mathrm{4}}.\mathrm{16}−\frac{\mathrm{4}}{\mathrm{3}}.\mathrm{8}+\mathrm{2}.\mathrm{2}^{\mathrm{2}} \:=\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$

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