f(x)=∫[1-->x^2] e^(-t^2)dt,求∫[0-->1} xf(x)dx
首先由前一式求导得：f '(x)=2xe^(-x^4)
且用x=1代入后得：f(1)=0
∫[0-->1] xf(x)dx
=(1/2)∫[0-->1] f(x)d(x^2)
=(1/2)x^2f(x)-(1/2)∫[0-->1] (x^2)f '(x)dx 前一项将上下限1,0代入相减
=0-(1/2)∫[0-->1] 2(x^3)e^(-x^4)dx
=-∫[0-->1] x^3*e^(-x^4)dx
=-1/4∫[0-->1] e^(-x^4)d(x^4)
=1/4e^(-x^4) [0-->1]
=1/4e^(-1)-1/4
=1/4(1/e-1)