Question

设f(x)=∫(0,x)e^(-t^2+2t)dt,求∫(0,1)f(x)(x-1)^2 dx.

Answer 1

f '(x)=e^(-x^2+2x)
则 ∫[0,1]f(x)(x-1)^2 dx
=1/3∫[0,1]f(x) d((x-1)^3)
=1/3[0,1]f(x)(x-1)^3-1/3∫[0,1] (x-1)^3*f '(x) dx 前一部分上下限代入后均为0
=-1/3∫[0,1] (x-1)^3*e^(-x^2+2x) dx
=-1/3∫[0,1] (x-1)^3*e^(1-(x-1)^2) dx
=-1/6∫[0,1] (x-1)^2*e^(1-(x-1)^2) d(x-1)^2
令(x-1)^2=u
=-1/6∫[0,1] u*e^(1-u) du
=-e/6∫[0,1] u*e^(-u) du
=e/6∫[0,1] u de^(-u)
=e/6[0,1] ue^(-u)-e/6∫[0,1] e^(-u)du
=e/6*e^(-1)+e/6*[0,1] e^(-u)
=1/6+1/6-e/6
=(2-e)/6