问题
解答题
求定积分∫10(xex2+x2e2)dx.
答案
(xex2+x2e2)dx=∫ 10
xex2dx+∫ 10
x2e2dx.∫ 10
其中
xex2dx=∫ 10 1 2
ex2dx2=∫ 10
ex21 2
=1 0
(e-1)1 2
∫01(x2e2)dx=e2∫01x2dx=e2×
=[
]x3 3 10 e2 3
∴∫10(xex2+x2e2)dx=-∫01(xex2+x2e2)dx=-[
(e-1)+1 2
]=-e2 3
-e2 3
e+1 2 1 2