Question

请问这道微积分题怎么做呢？

答案详细一点

Answer 1

let

u= t/x
du = (1/x) dt
t=0, u=0
t=1, u=1/x
∫(0->1) e^[ -(t/x)^2] dt
=∫(0->1/x) e^(-u^2) (x du)
=x.∫(0->1/x) e^(-u^2) du
=x.∫(0->1/x) e^(-t^2) dt
d/dx { ∫(0->1) e^[ -(t/x)^2] dt }
=d/dx { x.∫(0->1/x) e^(-t^2) dt }
= ∫(0->1/x) e^(-t^2) dt + x. [ e^[ -(1/x)^2] . ( -1/x^2)
= ∫(0->1/x) e^(-t^2) dt - (1/x). [ e^[ -(1/x)^2]

Answer 2