求函数y=∫上限x下限0,(t-1)(t-2)^2*dt的单调区间及极值

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
则 x - 1 = 0 或 x - 2 = 0
即 x = 1 或 x = 2
ƒ''(1) = (1 - 2)(3 - 4) = (- 1)(- 1) > 0,取得极小值
当 x < 1 时,ƒ'(x) < 0,递减
ƒ''(2) = (2 - 2)(3 • 2 - 4) = 0,不确定,于是用一阶导数测试
当 1 < x < 2 时,ƒ'(x) > 0,递增
当 x > 2 时,ƒ'(x) > 0,递增
所以 x = 2 处不是极点
ƒ(1) = ∫(0→1) (t - 1)(t - 2)² dt = - 17/12
所以极小值是- 17/12
递减区间为(- ∞,1],递增区间为[1,+ ∞)