∫∫x^2dxdy,积分区域为y=2x-x^2和y=x-2围成的区域

Question

∫∫x^2dxdy,积分区域为y=2x-x^2和y=x-2围成的区域

Answer 1

y=2x-x^2
y=x-2
联立成方程组,
解得
x=2 或 x=-1
y=0 或 y=-3
∴y=2x-x^2和y=x-2的交点坐标是(-1,-3)(2,0)
∫∫x^2dxdy
=∫(-1->2)x^2dx∫(x-2 ->2x-x^2)dy
=∫(-1->2)x^2dx(2x-x^2-x+2)
=∫(-1->2)x^2dx(-x^2+x+2)
=∫(-1->2)(-x^4+x^3+2x)dx
=(-x^5/5+x^4/4+x^2)|(-1->2)
=-1/5(2^5-(-1)^5)+1/4(2^4-(1)^1)+(2^2-(-1)^2)
=-1/5(32+1)+1/4(16-1)+3
=-33/5+15/4+3
=-57/20+3
=3/20