Question

这道微积分题怎么做？

如图 求过程 谢谢了

Answer 1

I = -∫<下0, 上1>x^2dx∫<下x, 上1>e^(-t^2)dt
D: 0 ≤ x ≤1, x ≤ t ≤ 1. 交换积分次序 0 ≤ t ≤ 1， 0 ≤ x ≤t
I = -∫<0, 1>e^(-t^2)dt∫<0, t>x^2dx = ∫<0, 1>e^(-t^2)dt[x^3/3]<0, t>
= -(1/3)∫<0, 1>t^3e^(-t^2)dt = -(1/6)∫<0, 1>t^2e^(-t^2)d(t^2) 令 u = t^2
= -(1/6)∫<0, 1>ue^(-u)du = (1/6)∫<0, 1>ude^(-u)
= (1/6)[ue^(-u)]<0, 1> - (1/6)∫<0, 1>e^(-u)du
= (1/6)e^(-1) + (1/6)[e^(-u)]<0, 1> = (1/6)[2/e - 1]

Answer 2

还需要做下去吗？

Answer 3

n-->∞时
原式-->√[n(1+n)/2]-√[n(n-1)/2]
=n/{√[n(1+n)/2]-√[n(n-1)/2]}
=√2/[√(1+1/n)+√(1-1/n)]
-->√2/2.