第1个回答 2018-12-16
F'(x) = 3x/(x^2-x+1), 得驻点 x = 0,
F(0) = 0,
F(1) = ∫<0, 1> 3tdt/(t^2-t+1)
= (3/2)∫<0, 1> (2t-1+1)dt/(t^2-t+1)
= (3/2) {∫<0, 1> d(t^2-t+1)/(t^2-t+1) + ∫<0, 1> d(t-1/2)/[(t-1/2)^2+3/4]}
= (3/2)[ln(t^2-t+1)+(2/√3)arctan((2t-1)/√3)]<0, 1>
= (3/2)(2/√3)π/3 = π/√3,
最小值 F(0) = 0, 最大值 F(1) = π/√3本回答被网友采纳