A) \[2+\sqrt{2}\]
B) \[2-\sqrt{2}\]
C) \[-2+\sqrt{2}\]
D) \[-2-\sqrt{2}\]
Correct Answer: B
Solution :
\[\int_{0}^{1.5}{[{{x}^{2}}]dx=\int_{0}^{1}{[{{x}^{2}}]dx+\int_{1}^{\sqrt{2}}{[{{x}^{2}}]dx+\int_{\sqrt{2}}^{1.5}{[{{x}^{2}}]dx}}}}\] \[=0+\int_{1}^{\sqrt{2}}{1dx+\int_{\sqrt{2}}^{1.5}{2dx=\sqrt{2}-1+3-2\sqrt{2}=2-\sqrt{2}}}\].
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