求积分∫∫1⼀(x^4+y^2)dxdy,其中D=(x,y)x≥1,y≥x^2

2025-02-25 12:30:48
推荐回答(1个)
回答1:

(1)原式=∫dx∫(y/x)dy
=∫((x-x/2)/x)dx
=∫(1/2)dx
=(1/2)(2-1)
=1/2;
(2)原式=∫dx∫xdy
=∫x√(4-x²)dx
=(-1/2)∫√(4-x²)d(4-x²)
=(-1/2)(2/3)(0-2)
=2/3;
(3)原式=∫dy∫(y/x)dx
=∫y(2lny)dy
=2∫y*lnydy
=(y²*lny)│-∫ydy (应用分部积分法)
=4ln2-3/2.