2004 a STEP thread...

\begin{array}{rl}[br]\displaystyle\int_0^a \sqrt{[x]} \, dx & = \displaystyle\int_0^1 \sqrt{0} \, dx + \int_1^2 \sqrt{1} \, dx + \cdots + \int_{a-1}^a \sqrt{a-1} \, dx \\ \br \\[br]& = \displaystyle\sum_{r=0}^{a-1} \sqrt{r} \end{array}

\begin{array}{rl}[br]\displaystyle\int_0^a 2^{[x]} \, dx & = \displaystyle\int_0^1 2^0 \, dx + \int_1^2 2^1 \, dx + \cdots \int_{a-1}^a 2^{a-1} \, dx \\ \br \\[br]& = \displaystyle\sum_{r=0}^{a-1} 2^r = \frac{1 - 2^{a}}{1-2} = 2^a - 1 \end{array}

\begin{array}{rl}[br]\displaystyle\int_0^a 2^{[x]} \, dx & = \displaystyle\int_0^{[a]} 2^{[x]} \, dx + \int_{[a]}^{a-[a]} 2^{[x]} \, dx \\ \br \\[br]& \displaystyle = 2^{[a]} - 1 + (a-[a])2^{[a]} = \boxed{(a-[a]+1)2^{[a]} -1} \end{array}