Question

If $\left[x\right]$ denotes the greater integer less than or equal to $x$,then the value of the integral ${\int }_0^2{x}^2\left[x\right]dx.$

Answer 1

$\because \left[x\right]\le x{\int }_0^2{x}^2\left[x\right]dx\left[x\right]=00\le x<1\left[x\right]=11\le x<2{\int }_0^1{x}^2\times 0dx+{\int }_1^2{x}^{2}dx0+{\left[\frac{x^3}{3}\right]}_1^2\Rightarrow \frac{8}{3}-\frac{1}{3}=\frac{7}{3}$