Question

函数f(x)=∫(x∧2-t)e∧-t∧2dt积分上下限为1～x∧2的单调递增区间

Answer 1

f(x)
=∫(1->x^2) (x^2-t)e^(-t^2) dt
=x^2.∫(1->x^2) e^(-t^2) dt -∫(1->x^2) t.e^(-t^2) dt
f'(x)
=2x∫(1->x^2) e^(-t^2) dt +2x^3.e^(-x^4) -2x^3.e^(-x^4)
=2x∫(1->x^2) e^(-t^2) dt
f'(x) = 0
x=0 or 1 or -1

f''(x) =2∫(1->x^2) e^(-t^2) dt +4x^2. e^(-x^4)
f''(0) = -2∫(0->1) e^(-t^2) dt <0 (max)
f''(1) = 4. e^(-1) >0 (min)
f''(-1) = 4. e^(-1) >0 (min)

单调递增区间
增加: [-1, 0] U [-1, ∞)
减小: (- ∞, -1] U [0 , 1]

追问

由f'(x)=0是如何得到x为0或±1呢

追答

f'(x) = 0
2x∫(1->x^2) e^(-t^2) dt =0
=>x=0 or ∫(1->x^2) e^(-t^2) dt
=> x=0 or x^2=1
=> x=0 or x=1 or -1