计算∫∫D(x+1)dxdy其中D是由直线x=0,y=0,x+y=1所围成的有界区域

2025-06-26 23:37:50
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回答1:

∫∫D(x+1)dxdy
=∫[0,1]∫[0,1-x] (x+1)dydx
=∫[0,1] y[0,1-x] (x+1)dx
=∫[0,1] (1-x^2)dx
=(x-x^3/3)[0,1]
=2/3