Question

∫∫|xy-1|dxdy

∫∫ |xy-1| dxdy D= 0≤x≤2 0≤y≤2 绝对值打开后 怎么写分段函数?
答案是：3/2+2ln2

Answer 1

简单计算一下，答案如图所示

Answer 2

for xy ≤ 1
|xy-1| = -xy+1
for xy> 1
|xy-1| = xy-1
xy ≤ 1
=> x≤ 1/y for 0≤y≤2
I =∫∫ |xy-1| dxdy
= ∫∫ -xy+1 dxdy ( 0≤x≤y ) + ∫∫ xy-1 dxdy (y