Question

The value of ${\int }_0^2[x^2-1]dx$ dx, where [x] denotes the greatest integer function is given by:

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

Answer 1

The correct option is D -1

${\int }_0^2[x^2-1]dx\left[\because \left[x+1\right]=\left[x\right]+1\right]\Rightarrow {\int }_0^1\left[x^2\right]dx+{\int }_1^{\sqrt{2}}\left[x^2\right]dx+{\int }_{\sqrt{2}}^{\sqrt{3}}\left[x^2\right]dx+{\int }_{\sqrt{3}}^2\left[x^2\right]dx-{\int }_0^{2}dx\Rightarrow 0+{\int }_1^{\sqrt{2}}dx+{\int }_{\sqrt{2}}^{\sqrt{3}}dx+{\int }_{\sqrt{3}}^{2}dx-{\int }_0^{2}dx\Rightarrow {\left[x\right]}_1^{\sqrt{2}}+{\left[x\right]}_{\sqrt{2}}^{\sqrt{3}}+{\left[x\right]}_{\sqrt{3}}^2-{\left[x\right]}_0^2\Rightarrow \sqrt{2}-1+\sqrt{3}-\sqrt{2}+2-\sqrt{3}-2\Rightarrow -1$