设f(x)= ∫0-x e^(-y+2y)dy 求∫0-1 [(1-x)^2]f(x)dx

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设f(x)= ∫0-x e^(-y+2y)dy 求∫0-1 [(1-x)^2]f(x)dx
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设f(x)= ∫0-x e^(-y+2y)dy 求∫0-1 [(1-x)^2]f(x)dx
设f(x)= ∫0-x e^(-y+2y)dy 求∫0-1 [(1-x)^2]f(x)dx

设f(x)= ∫0-x e^(-y+2y)dy 求∫0-1 [(1-x)^2]f(x)dx
是f(x)= ∫0-x e^(-y^2+2y)dy吧
交换积分次序即可
∫[0,1] [(1-x)^2]f(x)dx
=∫[0,1] [(1-x)^2]∫[0,x] e^(-y^2+2y)dydx
=∫[0,1]e^(-y^2+2y)∫[y,1] (1-x)^2dxdy
然后自己先算算吧