一道积分题∫[0,1]dx∫[x,1]e^(-y^2)dy
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∫[0,1]dx∫[x,1]e^(-y^2)dy
=∫[0,1]dy∫[y,0]e^(-y^2)dx
=∫[0,1]ye^(-y^2)dy
=(1/2)∫[0,1]e^(-y^2)d(y^2)
=(1/2)[-e^(-y^2)] | [0->1]
=(1/2)[1-e^(-1)]
∫[0,1]dx∫[x,1]e^(-y^2)dy
=∫[0,1]dy∫[y,0]e^(-y^2)dx
=∫[0,1]ye^(-y^2)dy
=(1/2)∫[0,1]e^(-y^2)d(y^2)
=(1/2)[-e^(-y^2)] | [0->1]
=(1/2)[1-e^(-1)]
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