Question

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

Answer 1

Consider the given integral.

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

Since,

$\left[x^2\right]=0forx\in [0,1]$

$\left[x^2\right]=1forx\in [1,\sqrt{2}]$

$\left[x^2\right]=0forx\in [\sqrt{2},\sqrt{3}]$

$\left[x^2\right]=forx\in [\sqrt{3},2]$

Therefore,

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

$I={\int }_0^{1}0dx+{\int }_1^{\sqrt{2}}1dx+{\int }_{\sqrt{2}}^{\sqrt{3}}2dx+{\int }_{\sqrt{3}}^{2}3dx$

$I=0+{x|}_1^{\sqrt{2}}+2{x|}_{\sqrt{2}}^{\sqrt{3}}+3{x|}_{\sqrt{3}}^2$

$I=(\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})+3(2-\sqrt{3})$

$I=\sqrt{2}-1+2\sqrt{3}-2\sqrt{2}+6-3\sqrt{3}$

$I=5-(\sqrt{2}+\sqrt{3})$

Hence, the value of integral is $5-(\sqrt{2}+\sqrt{3})$.