Question

I(X)=∫0-X (2t-1)/(t^2-t+1)dt在[0,2]上的最大值和最小值

Answer 1

ƒ(x) = ∫(0→x) (2t - 1)/(t² - t + 1) dt,where x∈[0,2]

ƒ'(x) = (2x - 1)/(x² - x + 1)

ƒ''(x) = (- 2x² + 2x + 1)/(x² - x + 1)²

ƒ'(x) = 0

==> 2x - 1 = 0

==> x = 1/2

ƒ''(1/2) = 8/3 > 0,取得极小值

极小值ƒ(1/2) = ln(3/4) ≈ - 0.287682

在端点,ƒ(0) = 0 > ƒ(1/2),ƒ(2) = ln(3) ≈ 1.09861 > ƒ(1/2)

于是,最小值是ln(3/4),最大值是ln(3)