Question

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

• A. $-4+\sqrt{3}+\sqrt{2}$
  
• B. $4+\sqrt{3}-\sqrt{2}$
  
• C. $4+2\sqrt{3}+\sqrt{2}$
  
• D. None of these

Answer 1

The correct option is A

$-4+\sqrt{3}+\sqrt{2}$

${\int }_0^2\left(\right[x{]}^2-\left[x^2\right])dx={\int }_0^1([x{]}^2-[x^2\left]\right)dx+{\int }_1^2\left(\right[x{]}^2-\left[x^2\right])dx={\int }_0^1(0-0)dx+{\int }_1^{2}1.dx-{\int }_1^2[x^2]dx$
[ $\because$ If x lies between 0 and 1, then $x^2$ also lies between 0 and 1]
$=0+1-\left[{\int }_1^{\sqrt{2}}\right[x^2]dx+{\int }_{\sqrt{2}}^{\sqrt{3}}[x^2]dx+{\int }_{\sqrt{3}}^2[x^2\left]dx\right]=1-[{\int }_1^{\sqrt{2}}1.dx+{\int }_{\sqrt{2}}^{\sqrt{3}}2.dx+{\int }_{\sqrt{3}}^{2}3.dx]=1-(\sqrt{2}-1)-2(\sqrt{3}-\sqrt{2})-3(2-\sqrt{3})=-4+\sqrt{3}+\sqrt{2}$