Question

The value of the integral, $\underset{1}{\overset{3}{\int }}[x^2-2x-2]dx,$ where $\left[x\right]$ denotes the greatest integer less than or equal to $x,$ is

• A. $-4$
• B. $-5$
• C. $-\sqrt{2}-\sqrt{3}-1$
• D. $-\sqrt{2}-\sqrt{3}+1$

Answer 1

The correct option is C $-\sqrt{2}-\sqrt{3}-1$

$I=\underset{1}{\overset{3}{\int }}[x^2-2x+1-3]dx$
$\Rightarrow I=\underset{1}{\overset{3}{\int }}-3dx+\underset{1}{\overset{3}{\int }}\left[\right(x-1{)}^2]dx$
Put $x-1=t\Rightarrow dx=dt$
$I=(-6)+\underset{0}{\overset{2}{\int }}\left[t^2\right]dt$
$\Rightarrow I=-6+\underset{0}{\overset{1}{\int }}0dt+\underset{1}{\overset{\sqrt{2}}{\int }}1dt+\underset{\sqrt{2}}{\overset{\sqrt{3}}{\int }}2dt+\underset{\sqrt{3}}{\overset{2}{\int }}3dt$
$\Rightarrow I=-6+(\sqrt{2}-1)+2\sqrt{3}-2\sqrt{2}+6-3\sqrt{3}$
$\Rightarrow I=-1-\sqrt{2}-\sqrt{3}$