Question

${\int }_0^2\left[x^2\right]dx\mid$ is equal to

• (A) $2-\sqrt{2}$
• (B) $2+\sqrt{2}$
• (C) $\sqrt{2}-1$
• (D) $-\sqrt{2}-\sqrt{3}+5$

Answer 1

The Correct Option is D

Explanation:
$\mathrm{I}={\int }_0^2\left[{\mathrm{x}}^2\right]\mathrm{d}\mathrm{x}$$={\int }_0^1\left[{\mathrm{x}}^2\right]\mathrm{d}\mathrm{x}+{\int }_1^{\sqrt{2}}\left[{\mathrm{x}}^2\right]\mathrm{d}\mathrm{x}+{\int }_{\sqrt{2}}^{\sqrt{3}}\left[{\mathrm{x}}^2\right]\mathrm{d}\mathrm{x}+{\int }_{\sqrt{3}}^2\left[{\mathrm{x}}^2\right]\mathrm{d}\mathrm{x}$$={\int }_0^{1}0\mathrm{d}\mathrm{x}+{\int }_1^{\sqrt{2}}1\mathrm{d}\mathrm{x}+{\int }_{\sqrt{2}}^{\sqrt{3}}2\mathrm{d}\mathrm{x}+{\int }_{\sqrt{3}}^{2}3\mathrm{d}\mathrm{x}$$=[\mathrm{x}{]}_1^{\sqrt{2}}+[2\mathrm{x}{]}_{\sqrt{2}}^{\sqrt{3}}+[3\mathrm{x}{]}_{\sqrt{3}}^2$$=\sqrt{2}-1+2\sqrt{3}-2\sqrt{2}+6-3\sqrt{3}$$=5-\sqrt{3}-\sqrt{2}$