Question

求定积分:上限是(ln2)下限是(0)xe^-xdx

Answer 1

∫(0,ln2) xe^(-x)dx
=∫(0,ln2)(-x)e^(-x)d(-x)
=∫(0,ln2)(-x)d(e^(-x))
=(-x)e^(-x)|(0,ln2)-∫(0,ln2)e^(-x)d(-x)
=-(1/2)*ln2-e^(-x)|(0,ln2)
=-(1/2)*ln2-(1/2-1)
=-(1/2)*ln2+(1/2)
=(1-ln2)/2
用到分部积分法
有不懂欢迎追问

Answer 2

利用分部积分法
∫﹙0→ln2﹚【xe^﹙-x﹚】dx
＝－∫﹙0→ln2﹚【xe^﹙-x﹚】d﹙－x﹚
＝－∫﹙0→ln2﹚x d[e^﹙-x﹚]
＝－xe^﹙-x﹚｜﹙0→ln2﹚＋∫﹙0→ln2﹚【e^﹙-x﹚】dx
＝－xe^﹙-x﹚｜﹙0→ln2﹚－e^﹙-x﹚｜﹙0→ln2﹚
＝[1－ln2]/2

Answer 3

∵∫xe^-xdx
=- ∫xd(e^-x)
=-xe^-x+∫e^-xdx
=-xe^-x-∫e^-xd-x
=-xe^-x-e^-x+C
=(-x-1)e^-x+C
∴ ∫(ln2,0)xe^-xdx
=(-ln2-1)e^-ln2+C+(-0-1)e^-0-C
=(-ln2-1)/2-1
=-ln2/2-3/2