※ 引述《johnnyzsefb (AJ)》之铭言： : x : 1.suppose f:R→R:f(x)=∫(t^2)(3+t^4)^0.5 dt +2 : 　　　　　　　　　　　1 : (iii)determine the range of f. : (v) show that 4/3 < f(0) < 2 : 以上我不知如何下手 (iii) The range of f is R . (v) 因为当 0 < t < 1 时 , 0 < (t^2)(√(3 + t^4)) < (t^2)(√(3+1)) = (2)(t^2) 1 1 1 所以 ∫ 0 dt < ∫ (t^2)(√(3 + t^4)) dt < ∫ (2)(t^2) dt 0 0 0 1 2 => 0 < ∫ (t^2)(√(3 + t^4)) dt < --- 0 3 1 -2 => 0 > -∫ (t^2)(√(3 + t^4)) dt > --- 0 3 0 -2 => 0 > ∫ (t^2)(√(3 + t^4)) dt > --- 1 3 0 2 4 => 2 > ∫ (t^2)(√(3 + t^4)) dt + 2 > 2 - --- = --- 1 3 3 4 0 => --- < ∫ (t^2)(√(3 + t^4)) dt + 2 < 2 3 1 4 => --- < f(0) < 2 3 : 3.(ii) 1 : is the improper integral ∫ln(x)/(1-x^2) dx convergent? : 0 : 这题原想说用比较测试法但是找不到合适的~ : 先说谢谢啦~ --

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