设f(x+1)=xe^-x,求∫f(x)dx上限2下限0
问题描述：
设f(x+1)=xe^-x,求∫f(x)dx上限2下限0
答
f(x+1)=xe^-x
令x=x-1代入得
f(x)=(x-1)e^[-(x-1)]=e(x-1)e^(-x)
∫[0,2]f(x)dx
=∫[0,2]e(x-1)e^(-x)dx
=e∫[0,2]xe^(-x)dx-e∫[0,2]e^(-x)dx (第一项用分步积分法）
=-e∫[0,2]xde^(-x)+ee^(-x)[0,2]
=[-exe^(-x)-ee^(-x)][0,2]+ee^(-x)[0,2]
=-2/e
答
f(x(=(x-1)e^[-(x-1)]
原式=-∫(x-1)e^[-(x-1)]d[-(x-1)]
=-∫(x-1)de^[-(x-1)]
=-(x-1)e^[-(x-1)]+∫e^[-(x-1)]d(x-1)
=-(x-1)e^[-(x-1)]-∫e^[-(x-1)]d[-(x-1)]
=-(x-1)e^[-(x-1)]-e^[-(x-1)] (0~2)
=-xe^[-(x-1)] (0~2)
=-2/e