Question

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

• A. $\sqrt{2}-1$
• B. $2(\sqrt{2}-1)$
• C. $\sqrt{2}$
• D. $noneofthese$

Answer 1

Answer: (A)

$\sqrt{2}-1$

$f\left(x\right)=\{\begin{array}{c}00\le x<1\\ 1\le x<\sqrt{2}\end{array}$

${\int }_0^{\sqrt{2}}\left[x^2\right]dx={\int }_0^{1}0.dx+{\int }_1^{\sqrt{2}}1.dx$

$=0+[x{]}_1^{\sqrt{2}}$

$=\sqrt{2}-1$