问题 解答题

求定积分∫10(xex2+x2e2)dx.

答案

10
(xex2+x2e2)dx=
10
xex2dx+
10
x2e2dx.

其中

10
xex2dx=
1
2
10
ex2dx2=
1
2
ex2
.
1
0
=
1
2
(e-1)

01(x2e2)dx=e201x2dx=e2×

[
x3
3
]
10
=
e2
3

∴∫10(xex2+x2e2)dx=-∫01(xex2+x2e2)dx=-[

1
2
(e-1)+
e2
3
]=-
e2
3
-
1
2
e+
1
2

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