Question

${\int }_0^{\sqrt{2}}\left[x^2\right]dx$ is equal to (where $[.]$ denotes greatest integer function).

Answer 1

${\int }_0^{\sqrt{2}}\left[x^2\right]={\int }_0^1\left[x^2\right]+{\int }_1^{\sqrt{2}}\left[x^2\right]$

$={\int }_0^{1}odx+{\int }_1^{\sqrt{2}}1dx$

$=0+\sqrt{2}-1$

$=\sqrt{2}-1$