Question

求问这道定积分怎么做

Answer 1

-x^2 +2x =-(x-1)^2 + 1
let
u=x-1
du =dx
x=1, u=0
x=+∞, u=+∞
∫(1->+∞) x^2. e^(-x^2+2x) dx
=∫(0->+∞) (u+1)^2. e^(-u^2+1) du
=e. ∫(0->+∞) (u+1)^2. e^(-u^2) du
=e. ∫(0->+∞) (u^2+2u +1) .e^(-u^2) du
=e.[ ∫(0->+∞) u^2.e^(-u^2) du +∫(0->+∞) (2u+1).e^(-u^2) du ]
=e.[ ∫(0->+∞) u^2.e^(-u^2) du -[e^(-u^2)]|(0->+∞) +√π/2]
=e.[ ∫(0->+∞) u^2.e^(-u^2) du +1 +√π/2]
=e.[ 1 +√π/2 -(1/2)∫(0->+∞) u de^(-u^2) ]
=e.[ 1 +√π/2 -(1/2)[u.e^(-u^2)]|(0->+∞) + (1/2)∫(0->+∞) e^(-u^2) du ]
=e.[ 1 +√π/2 -0 + (1/2)(√π/2) ]
=e.[ 1 + (3/4)√π]