The value of the integral,

Question:

The value of the integral, $\int_{1}^{3}\left[x^{2}-2 x-2\right] d x$

where $[x]$ denotes the greatest integer less than or equal to $\mathrm{X}$, is :

 

  1. $-\sqrt{2}-\sqrt{3}+1$

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

  3. $-5$

  4. $-4$


Correct Option: , 2

Solution:

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

$=\int_{1}^{2}\left[x^{2}\right]-3 \int_{1}^{3} d x$

$=\int_{1}^{3} 0 \mathrm{dx}+\int_{1}^{\sqrt{2}} 1 \cdot \mathrm{dx}+\int_{\sqrt{2}}^{\sqrt{3}} 2 \cdot \mathrm{d} \mathrm{x}+\int_{\sqrt{3}}^{2} 3 \cdot \mathrm{dx}-6$

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

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

 

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