计算二次积分∫01dy∫y1y2e-x2dx．
【正确答案】：由于{(x，y)∣0≤y≤1，y≤x≤1}={(x，y)∣0≤x≤1，0≤y≤x} 所以 ∫01dy∫y1y2e-x2dx = ∫01dx∫0xy2e-x2dy= 1/3∫01x3e-x2dx =-(1/6)∫01x2de-x2=-(1/6) (x2e-x2∣01-∫01e-x2dx2) =-(1/6)(e-1+e-x2∣01)=1/6-1/3e
计算二次积分∫01dy∫y1y2e-x2dx．
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