A) \[4+2\sqrt{2}\]
B) \[2+\sqrt{2}\]
C) \[2-\sqrt{2}\]
D) none of these
Correct Answer: C
Solution :
Let \[I=\int_{0}^{1.5}{[{{x}^{2}}]}\,dx\] \[=\int_{0}^{1}{[{{x}^{2}}]dx+}\int_{1}^{\sqrt{2}}{[{{x}^{2}}]dx+\int_{\sqrt{2}}^{1.5}{[{{x}^{2}}]dx}}\] \[=\int_{0}^{1}{0\,dx+}\int_{1}^{\sqrt{2}}{1\,dx+\int_{\sqrt{2}}^{1.5}{2dx}}\] \[=0+[x]_{1}^{\sqrt{2}}+2[x]_{\sqrt{2}}^{1.5}\] \[=(\sqrt{2}-1)+2(1.5-\sqrt{2})\] \[=2-\sqrt{2}\]
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