∫ ∫ (x–y)dxdy;
D
где D:
x=0; y=0; x+y=2.
считаем внутренний интеграл
= ∫20 (xy – (y2/2))|y=2–xy=0 dx=
= ∫20 (x·(2–x) – ((2–x)2/2)– 0) dx=
= ∫20 (2x–x2 – (4–4x+x2)/2) dx=
=∫20 (2x–x2 – (4–4x+x2)/2) dx=
=(1/2)∫20(8x – 3x2–4)dx=
=(1/2)·((8x2/2) –3·(x3/3) –4x)|20=
=(1/2)·(4·4–8–8)=0