Question

${\int }_0^2\left[x^2\right]dx=$

• A. $2-\sqrt{2}$
• B. $2+\sqrt{2}$
• C. $\sqrt{2}-1$
• D. $-\sqrt{2}-\sqrt{3}$
• E. None of these

Answer 1

The correct option is E

None of these

Explanation for correct option:

Analysing the given integration:

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

Here we have $\left[x^2\right]$ which is a greatest integer function, therefore,

$$\begin{gathered} \left[x^2\right]=0x\in \left[0,1\right] \\ \left[x^2\right]=1x\in \left[1,\sqrt{2}\right] \\ \left[x^2\right]=2x\in \left[\sqrt{2},\sqrt{3}\right] \\ \left[x^2\right]=3x\in \left[\sqrt{3},\sqrt{4}\right] \end{gathered}$$

Hence we can write the given integral as follows,

$\begin{array}{rcl}I& =& {\int }_0^2\left[x^2\right]dx\\ & =& {\int }_0^{1}0dx+{\int }_1^{\sqrt{2}}1dx+{\int }_{\sqrt{2}}^{\sqrt{3}}2dx+{\int }_{\sqrt{3}}^{\sqrt{4}}3dx\\ & =& 0+\left(\sqrt{2}-1\right)+2\left(\sqrt{3}-\sqrt{2}\right)+3\left(\sqrt{4}-\sqrt{3}\right)\\ & =& \sqrt{2}-1+2\sqrt{3}-2\sqrt{2}+6-3\sqrt{3}\\ & =& 5-\sqrt{2}-\sqrt{3}\end{array}$

Hence, the value of given integral is not equal to any value given in options.

Hence, option (E) is the correct answer.