Case 1:$\alpha$>=0
Let α=1.23, α+1=2.23,
=$\int_{1.23}^{2} x dx$ +$\int_{2}^{2.23} x dx$
=$\int_{1.23}^{2} 1$ + $\int_{2}^{2.23} 2 dx$ ($ \because$ floor between $1.23, 2 = 1$, similarly floor of 2 and 2.23 is 2. )
= 0.77 + 2*(0.23)
=0.77+0.46=1.23 ($\alpha$)
Case 2:$\alpha$<0
Let α=-0.2, α+1=0.8,
= $\int_{-0.2}^{0} x dx$+ $\int_{0}^{0.8} x dx$
= $\int_{-0.2}^{0} -1 dx$ + $\int_{0}^{0.8} 0 dx$
= -1(0-(-0.2) + 2*(0)
=0.2 ($\alpha$)
In both case, $\int_{\alpha}^{\alpha +1} x dx$ = $\alpha$
Answer : A.