Question

The integral ${\int }_0^{16}\left[x\right]dx,$ where [ .] denotes the greatest function equals ...............

Answer 1

${\int }_0^{16}\left[x\right]dx={\int }_0^{16}(x-\left\{x\right\})dx$

$={\int }_0^{16}xdx-{\int }_0^{16}\left\{x\right\}dx={\left[\right(\frac{x^2}{2}\left)\right]}_0^{16}-16{\int }_0^1\left\{x\right\}dx$

$=\left(\frac{{16}^2}{2}\right)-16\left(\right(\frac{1}{2})\times 1\times 1)$

$=128-8=120$