设f(x+1)=xe^-x,求∫f(x)dx上限2下限0_数学解答

设f(x+1)=xe^-x,求∫f(x)dx上限2下限0

人气：470 ℃　时间：2020-03-27 15:27:24

解答

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