A) \[2-\sqrt{2}\]
B) \[2+\sqrt{2}\]
C) \[\sqrt{2}-1\]
D) \[\sqrt{2}-2\]
Correct Answer: C
Solution :
\[I=\int_{0}^{\sqrt{2}}{[{{x}^{2}}]\,\,dx}\]\[=\int_{\,0}^{\,1}{[{{x}^{2}}]\,dx+}\int_{\,1}^{\,\sqrt{2}}{[{{x}^{2}}]\,\,dx}\] \[=\int_{\,0}^{\,1}{\,0\,dx+}\int_{\,1}^{\,\sqrt{2}}{\,dx}\]\[=[x]_{1}^{\sqrt{2}}=\sqrt{2}-1\].
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