# ISI2014-DCG-31

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For real $\alpha$, the value of $\int_{\alpha}^{\alpha+1} [x]dx$, where $[x]$ denotes the largest integer less than or equal to $x$, is

1. $\alpha$
2. $[\alpha]$
3. $1$
4. $\dfrac{[\alpha] + [\alpha +1]}{2}$
in Calculus
retagged
0
B?
2

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, $\int_{\alpha}^{\alpha +1} x dx$ = $\alpha$ Hope, it helps.

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