ƒ(x) = ∫(0→x) (t - 1)(t - 2)² dt
ƒ'(x) = (x - 1)(x - 2)²
ƒ''(x) = (x - 1) • 2(x - 2) + (x - 2)²
= (x - 2)[2(x - 1) + (x - 2)]
= (x - 2)(2x - 2 + x - 2)
= (x - 2)(3x - 4)
令 ƒ'(x) = 0
則 x - 1 = 0 或 x - 2 = 0
即 x = 1 或 x = 2
ƒ''(1) = (1 - 2)(3 - 4) = (- 1)(- 1) > 0,取得極小值
當(dāng) x < 1 時(shí),ƒ'(x) < 0,遞減
ƒ''(2) = (2 - 2)(3 • 2 - 4) = 0,不確定,于是用一階導(dǎo)數(shù)測(cè)試
當(dāng) 1 < x < 2 時(shí),ƒ'(x) > 0,遞增
當(dāng) x > 2 時(shí),ƒ'(x) > 0,遞增
所以 x = 2 處不是極點(diǎn)
ƒ(1) = ∫(0→1) (t - 1)(t - 2)² dt = - 17/12
所以極小值是- 17/12
遞減區(qū)間為(- ∞,1],遞增區(qū)間為[1,+ ∞)