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Question Number 154142 by EDWIN88 last updated on 14/Sep/21
Answered by mr W last updated on 14/Sep/21
y=a(x−4)(x+2)28=a(−4)(+2)2⇒a=−12y=−12(x−4)(x+2)2dydx=−12[(x+2)2+2(x−4)(x+2)]=0(x+2)(x−2)=0⇒x=−2(rejected)⇒x=2⇒xP=2,yP=−12(2−4)(2+2)2=16eqn.ofOP:y=8xeqn.ofQR:y=8−x88−x8=8x⇒xR=6465A=∫04ydx−12×8×6465A=−12∫04(x−4)(x+2)2dx−25665A=−12∫04(x3−12x−16)dx−25665A=−12[444−6×42−16×4]−25665A=48−25665=286465≈44.06
Commented by iloveisrael last updated on 15/Sep/21
yes
Answered by iloveisrael last updated on 15/Sep/21
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