∫∫（根号x＋y）dxdy
1 4x
=∫ dx∫（根号x＋y）dy
0 x
1 |4x
=∫ (2/3)(x+y)^(3/2)| dx
0 |x
1
=∫ (2/3)(x+4x)^(3/2)-(2/3)(x+x)^(3/2)dx
0
1
=(2/3)(5^(3/2)-2^(3/2))∫ x^(3/2)dx
0
|1
=(2/3)(5根号5-2根号2) (2/5)x^(5/2)|
|0
=(4/15)(5根号5-2根号2)(1-0)
=(20根号5-8根号5)/15