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f-0-3-x-gt-0-f-0-3-f-3-8-3-0-f-x-2-f-x-1-dx-4-3-f-2-




Question Number 205826 by tri26112004 last updated on 31/Mar/24
f_([0;3]) (x)>0  f(0)=3  f(3)=8  ∫^3 _0 (([f′(x)]^2 )/(f(x)+1))dx = (4/3)  f(2)=¿
f[0;3](x)>0f(0)=3f(3)=803[f(x)]2f(x)+1dx=43f(2)=¿

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