求二重积分∫∫D(x+1)dδ,D:由曲线y=x^2,y=x 围成的区域

2024-11-07 16:51:26
推荐回答(2个)
回答(1):

解:积分区域D为:0《x《1,x²《y《x
∫∫D(x+1)dδ
= ∫(0,1) (x+1)dx ∫(x²,x)dy
= ∫(0,1) (x+1)(x-x²)dx
= ∫(0,1) (x - x³)dx
= [x²/2-x^4/4] |(0,1)
=1/2-1/4=1/4

回答(2):

∫∫D (x + 1) dσ
= ∫(0,1) ∫(x²,x) (x + 1) dydx
= ∫(0,1) (x + 1) [y] |(x²,x) dx
= ∫(0,1) (x + 1)(x - x²) dx
= ∫(0,1) (1 + x) * x(1 - x) dx
= ∫(0,1) (x - x³) dx
= [x²/2 - x⁴/4] |(0,1)
= 1/2 - 1/4
= 1/4