Задача 27276 ...
Условие
∫∫(3x2–y2) dx dy
D: y=0, y=x, x+y=2;
математика
1147
Решение
★
= ∫^(1) _(0)dx ∫^(x) _(0)(3x^2-y^2)dy+∫^(2) _(1)dx ∫^(2-x) _(0)(3x^2-y^2)dy=
= ∫^(1) _(0)(3x^2y-(y^3/3))|^(x) _(0)dx+∫^(2) _(1)(3x^2y-(y^3/3))|^(2-x) _(0)=
= ∫^(1) _(0)(3x^2*x-(x^3/3))dx+∫^(2) _(1)(3x^2*(2-x)-((2-x)^3/3))dx=
=(8/3)*(x^4/4)|^(1)_(0)+(1/3) ∫^(2)_(1)(-8x^3+12x^2+12x-8)dx=
=(8/12)+(1/3)*(-2x^4+4x^3+6x^2-8x)|^(2)_(1)=
=(2/3)+(1/3)*(-2*2^4+4*2^3+6*2^2-8*+2*1^4-4*1^3 -6*1^2+8*1)=
=(2/3)+(1/3)*0=(2/3)