因為xe^x為f（x）的一個原函數(shù)
所以f(x)=（xe^x）’=e^x+xe^x=(1+x)e^x
∫【0,1】xf '(x)dx
=∫【0,1】x（1+x）e^xdx
=∫【0,1】xe^xdx+∫【0,1】x²e^xdx
=xe^x【0,1】-∫【0,1】e^xdx+x²e^x【0,1】-2∫【0,1】xe^xdx
=e-e^x【0,1】+e-2∫【0,1】xe^xdx
=e-e+1+e-2∫【0,1】xe^xdx
=1+e-2×1
e-1恩。不過能再求一下后面的那個定積分么？算了三四次都沒一個對的。。。因為xe^x為f（x）的一個原函數(shù)所以f(x)=（xe^x）’=e^x+xe^x=(1+x)e^x∫【0,1】xf '(x)dx=∫【0,1】x（1+x）e^xdx=∫【0,1】xe^xdx+∫【0,1】x²e^xdx=xe^x【0,1】-∫【0,1】e^xdx+x²e^x【0,1】-2∫【0,1】xe^xdx=e-e^x【0,1】+e-2∫【0,1】xe^xdx=e-e+1+e-2∫【0,1】xe^xdx=1+e-2×1=e-1