设f(x+1)=xe^-x,求∫f(x)dx上限2下限0
人气:298 ℃ 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
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