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**Kmath** First,

$\displaystyle \int_{0}^1\sqrt{[x]}dx= \int_{0}^a\sqrt{[x]}dx+\int_{1}^2\sqrt{[x]}dx+\cdots +\int_{a-1}^a\sqrt{[x]}dx $

second, note that in the first interval (integral)$\displaystyle [x]=\sqrt{0}$

in the second interval (integral) $\displaystyle [x]=\sqrt{1}$

and so on, unitil

the ath integral $\displaystyle [x]=\sqrt{a-1}$, so

$\displaystyle \int_{0}^1\sqrt{[x]}dx= \int_{0}^a0dx+\int_{1}^21dx+\cdots +\int_{a-1}^a\sqrt{a-1}dx $

$\displaystyle = \int_{1}^21dx+\cdots +\int_{a-1}^a\sqrt{a-1}dx =1+2+\cdots+\sqrt{a-1}$ as desired.