Question Number 64176 by aliesam last updated on 15/Jul/19
Answered by MJS last updated on 15/Jul/19
![the minimum of f(x)=(√(1+x^2 )) is f(0)=1 ⇒ the area between the x−axis and f(x) in [−1; 1] is greater than 2 ⇒ 2<∫^1 _(−1) (√(1+x^2 ))dx the maximum of f(x) in [−1; 1] is f(−1)=f(1)=(√2) ⇒ the area is less than 2(√2) ⇒ ∫_(−1) ^1 (√(1+x^2 ))dx<2(√2) 1<2<∫^1 _(−1) (√(1+x^2 ))dx<2(√2)<4](https://www.tinkutara.com/question/Q64182.png)
$$\mathrm{the}\:\mathrm{minimum}\:\mathrm{of}\:{f}\left({x}\right)=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\mathrm{is}\:{f}\left(\mathrm{0}\right)=\mathrm{1}\:\Rightarrow\:\mathrm{the} \\ $$$$\mathrm{area}\:\mathrm{between}\:\mathrm{the}\:{x}−\mathrm{axis}\:\mathrm{and}\:{f}\left({x}\right)\:\mathrm{in}\:\left[−\mathrm{1};\:\mathrm{1}\right] \\ $$$$\mathrm{is}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{2}\:\Rightarrow\:\mathrm{2}<\underset{−\mathrm{1}} {\int}^{\mathrm{1}} \sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$\mathrm{the}\:\mathrm{maximum}\:\mathrm{of}\:{f}\left({x}\right)\:\mathrm{in}\:\left[−\mathrm{1};\:\mathrm{1}\right]\:\mathrm{is}\:{f}\left(−\mathrm{1}\right)={f}\left(\mathrm{1}\right)=\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{area}\:\mathrm{is}\:\mathrm{less}\:\mathrm{than}\:\mathrm{2}\sqrt{\mathrm{2}}\:\Rightarrow\:\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx}<\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mathrm{1}<\mathrm{2}<\underset{−\mathrm{1}} {\int}^{\mathrm{1}} \sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx}<\mathrm{2}\sqrt{\mathrm{2}}<\mathrm{4} \\ $$
Commented by aliesam last updated on 15/Jul/19
$${god}\:{bless}\:{you} \\ $$