Question Number 199468 by Calculusboy last updated on 04/Nov/23
Answered by witcher3 last updated on 04/Nov/23
![(x/(2021π))⇔ ∫_0 ^1 (1/e^t^(2021) )(1+t^(2022) )^(1/(2022)) dt=I ∫_0 ^1 e^(−t^(2021) ) (1+t^(2022) )^(1/(2022)) dt e^x ≥1+x⇒e^(−x) ≤(1/(1+x)),∀x≥0 prof x→e^x is convexe graphe of x→e^x is up to tangent in/zero e^x ≥e^0 (x−0)+e^0 =x+1 (1+t)^a ≤1+at,∀a∈[0,1[ t>0 t→(1+t)^a concave ⇒(1+t)^a ≤at+1 {tangent in zero} e^(−t^(2021) ) ≤(1/(1+t^(2021) )) (1+t^(2022) )^(1/(2022)) ≤1+(t^(2022) /(2022))<1+(t^(2021) /(2022))<1+t^(2021) ,∀t∈[0,1] I≤∫_0 ^1 ((1+(t^(2022) /(2022)))/(1+t^(2021) ))dt<∫_0 ^1 ((1+t^(2021) )/(1+t^(2021) ))dt=1 ∫_0 ^(2021π) (1/e^(((x/(2021π)))^(2021) ) )(((1+((x/(2021π)))^(2022) ))^(1/(2022)) dx<1](https://www.tinkutara.com/question/Q199533.png)
$$\frac{\mathrm{x}}{\mathrm{2021}\pi}\Leftrightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{e}^{\mathrm{t}^{\mathrm{2021}} } }\left(\mathrm{1}+\mathrm{t}^{\mathrm{2022}} \right)^{\frac{\mathrm{1}}{\mathrm{2022}}} \mathrm{dt}=\mathrm{I} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{−\mathrm{t}^{\mathrm{2021}} } \left(\mathrm{1}+\mathrm{t}^{\mathrm{2022}} \right)^{\frac{\mathrm{1}}{\mathrm{2022}}} \mathrm{dt} \\ $$$$\mathrm{e}^{\mathrm{x}} \geqslant\mathrm{1}+\mathrm{x}\Rightarrow\mathrm{e}^{−\mathrm{x}} \leqslant\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}},\forall\mathrm{x}\geqslant\mathrm{0} \\ $$$$\mathrm{prof}\:\mathrm{x}\rightarrow\mathrm{e}^{\mathrm{x}} \:\mathrm{is}\:\mathrm{convexe}\:\:\mathrm{graphe}\:\mathrm{of}\:\mathrm{x}\rightarrow\mathrm{e}^{\mathrm{x}} \:\mathrm{is}\:\mathrm{up}\:\mathrm{to}\:\mathrm{tangent}\:\mathrm{in}/\mathrm{zero} \\ $$$$\mathrm{e}^{\mathrm{x}} \geqslant\mathrm{e}^{\mathrm{0}} \left(\mathrm{x}−\mathrm{0}\right)+\mathrm{e}^{\mathrm{0}} =\mathrm{x}+\mathrm{1} \\ $$$$\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{a}} \leqslant\mathrm{1}+\mathrm{at},\forall\mathrm{a}\in\left[\mathrm{0},\mathrm{1}\left[\:\mathrm{t}>\mathrm{0}\right.\right. \\ $$$$\mathrm{t}\rightarrow\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{a}} \:\:\mathrm{concave}\:\Rightarrow\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{a}} \leqslant\mathrm{at}+\mathrm{1}\:\:\left\{\mathrm{tangent}\:\mathrm{in}\:\mathrm{zero}\right\} \\ $$$$\mathrm{e}^{−\mathrm{t}^{\mathrm{2021}} } \leqslant\frac{\mathrm{1}}{\mathrm{1}+\mathrm{t}^{\mathrm{2021}} } \\ $$$$\left(\mathrm{1}+\mathrm{t}^{\mathrm{2022}} \right)^{\frac{\mathrm{1}}{\mathrm{2022}}} \leqslant\mathrm{1}+\frac{\mathrm{t}^{\mathrm{2022}} }{\mathrm{2022}}<\mathrm{1}+\frac{\mathrm{t}^{\mathrm{2021}} }{\mathrm{2022}}<\mathrm{1}+\mathrm{t}^{\mathrm{2021}} ,\forall\mathrm{t}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\mathrm{I}\leqslant\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}+\frac{\mathrm{t}^{\mathrm{2022}} }{\mathrm{2022}}}{\mathrm{1}+\mathrm{t}^{\mathrm{2021}} }\mathrm{dt}<\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}+\mathrm{t}^{\mathrm{2021}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2021}} }\mathrm{dt}=\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2021}\pi} \frac{\mathrm{1}}{\mathrm{e}^{\left(\frac{\mathrm{x}}{\mathrm{2021}\pi}\right)^{\mathrm{2021}} } }\sqrt[{\mathrm{2022}}]{\left(\mathrm{1}+\left(\frac{\mathrm{x}}{\mathrm{2021}\pi}\right)^{\mathrm{2022}} \right.}\mathrm{dx}<\mathrm{1} \\ $$$$\: \\ $$$$ \\ $$$$ \\ $$