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I think ∫[x+y] = ∫[y] as limit is 0-1 which is just [y] and hence 0 as y is between 0 and 1.

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Fix $y=a$, then $[x+y]=0 \;\:\forall x\in[0,1-a)$ and it is $[x+y]=1 \;\: \forall x\in[1-a,1]$, then: $$\int_0^1[x+y]dx=\int_0^{1-a}0dx+\int_{1-a}^1 1dx=1-(1-a)=a=y$$