Question

$\underset{0}{\overset{\sqrt{2}}{\int }}\left[x^2\right]dx.$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\underset{0}{\overset{\sqrt{2}}{\int }}\left[x^2\right]dx \\ =\underset{0}{\overset{1}{\int }}\left[x^2\right]dx+\underset{1}{\overset{\sqrt{2}}{\int }}\left[x^2\right]dx \\ =\underset{0}{\overset{1}{\int }}\left(0\right)dx+\underset{1}{\overset{\sqrt{2}}{\int }}\left(1\right)dx\left(\because \left[x^2\right]=\left\{\begin{array}{l}00<x<1\\ 11<x<\sqrt{2}\end{array}\right.\right) \\ =0+{\left[x\right]}_1^{\sqrt{2}} \\ =\sqrt{2}-1 \end{gathered}$$