Question

求F(x)=∫0到x te^(-t^2) dt 的极值。

如题。答案是：当x=0时有极小值F(0)=0。求过程。谢谢！

Answer 1

F(x) = ∫(0->x) t*e^(-t²) dt
= ∫(0->x) e^(-t²) d(t²/2)
= (1/2) * -e^(-t²)：(0->x)
= (-1/2)[e^(-x²) - e^0]
= 1/2 - (1/2)e^(-x²)

F'(x) = x*e^(-x²)
F''(x) = (1-2x²)*e^(-x²)
令F'(x) = 0
x*e^(-x²) = 0
x = 0 或 [e^-(x²) = 0 (无解)]
F''(0) = (1-0)*1 = 1 > 0，所以有极小值

极小值为F(0)
= 1/2 - (1/2)e^0
= 1/2 - 1/2
= 0