Question

求定積分∫₁⁰（xex2+x²e²）dx．

Answer 1

∫

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=e²∫₀¹x²dx=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