The value of the integral int_{1}^{3} [x^{2}- | vedclass

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(B) Let $I = \int_{1}^{3} [x^{2}-2x-2] dx$. We can rewrite the expression inside the bracket as $x^{2}-2x-2 = (x-1)^{2}-3$. Since $3$ is an integer,we

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

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(B) Let $I = \int_{1}^{3} [x^{2}-2x-2] dx$. We can rewrite the expression inside the bracket as $x^{2}-2x-2 = (x-1)^{2}-3$. Since $3$ is an integer,we can use the property $[f(x)+n] = [f(x)]+n$ for $n \in \mathbb{Z}$. So,$I = \int_{1}^{3} [(x-1)^{2}] dx - \int_{1}^{3} 3 dx$. Let $t = x-1$,then $dt = dx$. When $x=1, t=0$ and when $x=3, t=2$. $I = \int_{0}^{2} [t^{2}] dt - [3x]_{1}^{3} = \int_{0}^{2} [t^{2}] dt - 6$. Now,we split the integral $\int_{0}^{2} [t^{2}] dt$ based on the values of $[t^{2}]$: For $0 \le t For $1 \le t For $\sqrt{2} \le t For $\sqrt{3} \le t $I = \int_{0}^{1} 0 dt + \int_{1}^{\sqrt{2}} 1 dt + \int_{\sqrt{2}}^{\sqrt{3}} 2 dt + \int_{\sqrt{3}}^{2} 3 dt - 6$. $I = 0 + (\sqrt{2}-1) + 2(\sqrt{3}-\sqrt{2}) + 3(2-\sqrt{3}) - 6$. $I = \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3} - 6$. $I = -\sqrt{2} - \sqrt{3} - 1$.

The value of the integral $\int_{1}^{3} [x^{2}-2x-2] dx$,where $[x]$ denotes the greatest integer function,is:

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