Question

The value of ${\int }_0^{\sqrt{2}}\left[x^2\right]·dx$, where $\left[x\right]$ is the greatest integer function, is

• A. $2-\sqrt{2}$
• B. $2+\sqrt{2}$
• C. $\sqrt{2}-1$
• D. $\sqrt{2}-2$

Answer 1

The correct option is C

$\sqrt{2}-1$

Explanation for correct option

Given: ${\int }_0^{\sqrt{2}}\left[x^2\right]·dx$

Let $I={\int }_0^{\sqrt{2}}\left[x^2\right]·dx$

${\left[x\right]}^2=\left\{\begin{array}{l}0,0\le x<1\\ 1,1\le x<\sqrt{2}\end{array}\right.$

$$\begin{gathered} \Rightarrow I={\int }_0^{\sqrt{2}}\left[x^2\right]·dx \\ \Rightarrow I={\int }_0^1\left[x^2\right]·dx+{\int }_1^{\sqrt{2}}\left[x^2\right]·dx \\ \Rightarrow I={\int }_0^{1}0·dx+{\int }_1^{\sqrt{2}}1·dx \\ \Rightarrow I=0+{\left|x\right|}_1^{\sqrt{2}} \\ \Rightarrow I=\sqrt{2}-1 \end{gathered}$$

Hence, option (C) is correct.