Upside down box integral

$\begin{aligned} I & = \int_0^{10} \lceil x \rceil \lfloor x \rfloor \ dx = \sum_{k=0}^9 \int_k^{k+1} \lceil x \rceil \lfloor x \rfloor \ dx = \sum_{k=1}^9 (k+1)k = \sum_{k=1}^9 (k^2+k) = \frac {9(10)(19)}6 + \frac {9(10)}2 = \boxed{330} \end{aligned}$

The above integral is just a series of rectangles (each of width 1), which the sum of areas can be computed per:

$\Sigma_{k=1}^{10} k^2-k = \frac{(10)(11)(21)}{6} - \frac{(10)(11)}{2} = \boxed{330}.$