2 3
∫ ∫ 3x^2 + 3xy + 2 dx dy =
1 0
Integrating first with respect to x
3
[ (x^3 + (3/2)x^2y + 2x ] = 27 + (27/2)y + 6 = 33 + (27/2)y
0
Then integrate with respect to y
2
∫ 33 + (27/2)y dy =
1
2
[ 33y + (27 /4)y^2 ] = [ 66 +27 ] - [ 33 + 27/4] = 213 / 4
1