# The value of the integral,

Question:

The value of the integral, $\int_{1}^{3}\left[x^{2}-2 x-2\right] \mathrm{d} x$, where $[\mathrm{x}]$ denotes the greatest integer less than or equal to $x$, is:

1. (1) $-4$

2. (2) $-5$

3. (3) $-\sqrt{2}-\sqrt{3}-1$

4. (4) $-\sqrt{2}-\sqrt{3}+1$

Correct Option: , 3

Solution:

$\mathrm{I}=\int_{1}^{3}-3 d x+\int_{1}^{3}\left[(x-1)^{2}\right] d x$

Put $x-1=t ; d x=d t$'

$I=(-6)+\int_{0}^{2}\left[t^{2}\right] d t$

$I=-6+\int_{0}^{1} 0 d t+\int_{1}^{\sqrt{2}} 1 d t+\int_{\sqrt{2}}^{\sqrt{3}} 2 d t+\int_{\sqrt{3}}^{2} 3 d t$

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

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