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Question Number 64176 by aliesam last updated on 15/Jul/19

Answered by MJS last updated on 15/Jul/19

$$\mathrm{the}\mathrm{minimum}\mathrm{of}f\left(x\right)=\sqrt{1+x^2}\mathrm{is}f\left(0\right)=1\Rightarrow \mathrm{the}$$$$\mathrm{area}\mathrm{between}\mathrm{the}x-\mathrm{axis}\mathrm{and}f\left(x\right)\mathrm{in}[-1;1]$$$$\mathrm{is}\mathrm{greater}\mathrm{than}2\Rightarrow 2<{\underset{-1}{\int }}^1\sqrt{1+x^2}dx$$$$\mathrm{the}\mathrm{maximum}\mathrm{of}f\left(x\right)\mathrm{in}[-1;1]\mathrm{is}f(-1)=f\left(1\right)=\sqrt{2}$$$$\Rightarrow \mathrm{the}\mathrm{area}\mathrm{is}\mathrm{less}\mathrm{than}2\sqrt{2}\Rightarrow \underset{-1}{\stackrel{1}{\int }}\sqrt{1+x^2}dx<2\sqrt{2}$$$$1<2<{\underset{-1}{\int }}^1\sqrt{1+x^2}dx<2\sqrt{2}<4$$

Commented by aliesam last updated on 15/Jul/19

$$godblessyou$$

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