Question

一道积分题∫[0,1]dx∫[x,1]e^(-y^2)dy

Answer 1

交换积分顺序
∫[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)]