A) \[3\sqrt{2}\]
B) \[\frac{3}{\sqrt{2}}\]
C) \[\frac{3}{\sqrt{2}}-1\]
D) infinite
Correct Answer: B
Solution :
\[y=\left[ \frac{3}{{{x}^{2}}}+1 \right]=\left\{ \begin{matrix} 0, & x\in \left( \sqrt{2},\,\infty \right) \\ \begin{align} & 1, \\ & 2, \\ & 3, \\ \end{align} & \begin{align} & x\in \left( \frac{1}{\sqrt{2}},\,\sqrt{2} \right) \\ & x\in \left( 0,\,\frac{1}{\sqrt{2}} \right) \\ & x=0 \\ \end{align} \\ \end{matrix} \right.\] \[\int\limits_{0}^{1/\sqrt{2}}{2dx+\int\limits_{1/\sqrt{2}}^{\sqrt{2}}{1dx=\sqrt{2}+\sqrt{2}-\frac{1}{\sqrt{2}}=\frac{3}{\sqrt{2}}}}\]
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