∫ ∫ (8xy+9x2·y2)dxdy, где D: x=1, y=–x3, y=∛x
∫ ∫ (8xy+9x2·y2)dxdy= ∫1 0 (∫ ∛x–x3((8xy+9x2·y2)dy)dx=
= ∫1 0( (8xy2/2)+(9x2y3/3))| ∛x–x3dx=
=∫1 0( (4x·(x2/3)–4·x·x6+(3x2·x)+3x2·x9)dx=
= ∫1 0( (4x5/3–4·x7+3x3+3x11)dx=
=(4·x8/3/(8/3) –4·x8/8 +3x4/4 +3x12/12)|10=
=32/3 –(1/2) +(3/4)+(1/4)= (32/3)+(1/2)=65/3
О т в е т 65/3.