Question

$\stackrel{\sqrt{2}}{\underset{0}{\int }}\left[x^2\right]dx$ is equal to (where [.] denotes greatest integer function).

• A. $\sqrt{2}-1$
• B. $2(\sqrt{2}-1)$
• C. $\sqrt{2}$
• D. None of these

Answer 1

The correct option is A $\sqrt{2}-1$

We have $\left[x^2\right]=\{\begin{array}{c}0\text{for}0\le x<1\\ 1\text{for}1\le x<\sqrt{2}\end{array}$.

Now,

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

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

$=\stackrel{1}{\underset{0}{\int }}0.dx$$+\stackrel{\sqrt{2}}{\underset{1}{\int }}1.dx$ [ Using definition of $\left[x^2\right]$]

$=\sqrt{2}-1$.