f(y-x) = sin(y-x), x ≤ y ≤ 1+x
f(y-x) = 0, 其它
I = ∫<0, 1>sinxdx ∫<x, 1+x>sin(y-x)dy
= ∫<0, 1>sinxdx [-cos(y-x)]<x, 1+x>
= (1-cos1) ∫<0, 1>sinxdx = (1-cos1) [-cosx]<0, 1> = (1-cos1)^2
f(y-x) = 0, 其它
I = ∫<0, 1>sinxdx ∫<x, 1+x>sin(y-x)dy
= ∫<0, 1>sinxdx [-cos(y-x)]<x, 1+x>
= (1-cos1) ∫<0, 1>sinxdx = (1-cos1) [-cosx]<0, 1> = (1-cos1)^2
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第1个回答 2018-08-18
根号部分非负，y的取值明显不能小于负a呀，所以是上面那部分本回答被网友采纳
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