Question

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

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

Answer 1

Answer: (D)

$-\sqrt{2}-\sqrt{3}+5$

${\int }_0^2\left[x^2\right]dx$
$={\int }_0^{1}0dx+{\int }_1^{\sqrt{2}}1dx+{\int }_{\sqrt{2}}^{\sqrt{3}}2dx+{\int }_{\sqrt{3}}^{2}3dx$
$=0+(\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})+3(2-\sqrt{3})$
$=\sqrt{2}-1+2\sqrt{3}-2\sqrt{2}+6-3\sqrt{3}$
$=-\sqrt{2}-\sqrt{3}+5$