∫[下限0上限1]∫[下限0上限1](x+y)dxdy

题目

∫[下限0上限1]∫[下限0上限1](x+y)dxdy

答案

=∫(0,1)dx∫(0,1)(x+y)dy
=∫(0,1)(xy+y^2/2)(0,1)dx
=∫(0,1)(x+1/2)dx
=∫(0,1)(x+1/2)d(x+1/2)
=(x+1/2)^2/2 (0,1)
=(1+1/2)^2/2-(0+1/2)^2/2
=1