Question

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

• A. $\sqrt{2}-1$
• B. $1-\sqrt{2}$
• C. $2(\sqrt{2}-1)$
• D. $\sqrt{3}-1$

Answer 1

Answer: (A)

$\sqrt{2}-1$

We have to find ${\int }_0^{\sqrt{2}}\left[x^2\right]dx$, (where $[.]$ represnts greatest integer function)

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

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

$=\sqrt{2}-1$

$\therefore {\int }_0^{\sqrt{2}}\left[x^2\right]dx=\sqrt{2}-1$