Question Number 204878 by mnjuly1970 last updated on 29/Feb/24
$$ \\ $$$$\:\:{prove}\:{that}: \\ $$$$ \\ $$$$\:\:\:\:\sqrt[{\mathrm{4}}]{{e}}\:<\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {e}^{\:{t}^{\mathrm{2}} } {dt}<\:\frac{\mathrm{1}\:+\:{e}}{\mathrm{2}} \\ $$
Answered by witcher3 last updated on 29/Feb/24
![let f(x)=e^x^2 f′(x)=2xe^x^2 ;f′′(x)=(2+4x^2 )e^x^2 ≥0 f is convex Let C_f ={(x,f(x));x∈R^2 } Γ: tangent to Cf at(1/2) Γ=f′((1/2))(x−(1/2))+f((1/2))=e^(1/4) (x−(1/2))+e^(1/4) =e^(1/4) x+(e^(1/4) /2);Since f is convex⇒f(x)≥xe^(1/4) +(e^(1/4) /2) ⇒∫_0 ^1 e^x^2 dx≥∫_0 ^1 (xe^(1/4) +(e^(1/4) /2))dx=e^(1/4) =(e)^(1/4) f(x)=e^x^2 applie lagrange theorem in [0,t] ⇒∃_c ∈[0,t] suche f(t)=f(0)+tf′(c) f′′(x)≥0⇒f′ is increase ⇒f′(c)≤f′(t) ⇒f(t)=1+t(2ce^c^2 )≤1+2t^2 e^t^2 ⇒∫_0 ^1 f(t)dt≤∫_0 ^1 (1+2t^2 e^t^2 )dt=∫_0 ^1 1dt+∫_0 ^1 t.(2te^t^2 )dt u=t,v′=2te^t^2 IBP⇒ ∫_0 ^1 e^t^2 dt≤1+[te^t^2 ]_0 ^1 −∫_0 ^1 e^t^2 dt⇒2∫_0 ^1 e^t^2 dt≤1+e ⇒∫_0 ^1 e^t^2 dt≤((1+e)/2) ⇔(e)^(1/4) ≤∫_0 ^1 e^t^2 dt≤((1+e)/2)](https://www.tinkutara.com/question/Q204880.png)
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{e}^{\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{2xe}^{\mathrm{x}^{\mathrm{2}} } ;\mathrm{f}''\left(\mathrm{x}\right)=\left(\mathrm{2}+\mathrm{4x}^{\mathrm{2}} \right)\mathrm{e}^{\mathrm{x}^{\mathrm{2}} } \geqslant\mathrm{0} \\ $$$$\mathrm{f}\:\mathrm{is}\:\mathrm{convex}\:\mathrm{Let}\:\mathrm{C}_{\mathrm{f}} =\left\{\left(\mathrm{x},\mathrm{f}\left(\mathrm{x}\right)\right);\mathrm{x}\in\mathbb{R}^{\mathrm{2}} \right\} \\ $$$$\Gamma:\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{Cf}\:\mathrm{at}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Gamma=\mathrm{f}'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{4}}} \left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$=\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{4}}} \mathrm{x}+\frac{\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{4}}} }{\mathrm{2}};\mathrm{Since}\:\mathrm{f}\:\mathrm{is}\:\mathrm{convex}\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\geqslant\mathrm{xe}^{\frac{\mathrm{1}}{\mathrm{4}}} +\frac{\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{4}}} }{\mathrm{2}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{\mathrm{x}^{\mathrm{2}} } \mathrm{dx}\geqslant\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{xe}^{\frac{\mathrm{1}}{\mathrm{4}}} +\frac{\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{4}}} }{\mathrm{2}}\right)\mathrm{dx}=\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{4}}} =\sqrt[{\mathrm{4}}]{\mathrm{e}} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{e}^{\mathrm{x}^{\mathrm{2}} } \:\mathrm{applie}\:\mathrm{lagrange}\:\mathrm{theorem}\:\mathrm{in}\:\left[\mathrm{0},\mathrm{t}\right] \\ $$$$\Rightarrow\exists_{\mathrm{c}} \in\left[\mathrm{0},\mathrm{t}\right]\:\mathrm{suche}\:\mathrm{f}\left(\mathrm{t}\right)=\mathrm{f}\left(\mathrm{0}\right)+\mathrm{tf}'\left(\mathrm{c}\right) \\ $$$$\mathrm{f}''\left(\mathrm{x}\right)\geqslant\mathrm{0}\Rightarrow\mathrm{f}'\:\mathrm{is}\:\mathrm{increase}\:\Rightarrow\mathrm{f}'\left(\mathrm{c}\right)\leqslant\mathrm{f}'\left(\mathrm{t}\right) \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{t}\right)=\mathrm{1}+\mathrm{t}\left(\mathrm{2ce}^{\mathrm{c}^{\mathrm{2}} } \right)\leqslant\mathrm{1}+\mathrm{2t}^{\mathrm{2}} \mathrm{e}^{\mathrm{t}^{\mathrm{2}} } \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}\leqslant\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\mathrm{2t}^{\mathrm{2}} \mathrm{e}^{\mathrm{t}^{\mathrm{2}} } \right)\mathrm{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{1dt}+\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{t}.\left(\mathrm{2te}^{\mathrm{t}^{\mathrm{2}} } \right)\mathrm{dt} \\ $$$$\mathrm{u}=\mathrm{t},\mathrm{v}'=\mathrm{2te}^{\mathrm{t}^{\mathrm{2}} } \:\mathrm{IBP}\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{\mathrm{t}^{\mathrm{2}} } \mathrm{dt}\leqslant\mathrm{1}+\left[\mathrm{te}^{\mathrm{t}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{\mathrm{t}^{\mathrm{2}} } \mathrm{dt}\Rightarrow\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{\mathrm{t}^{\mathrm{2}} } \mathrm{dt}\leqslant\mathrm{1}+\mathrm{e} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{\mathrm{t}^{\mathrm{2}} } \mathrm{dt}\leqslant\frac{\mathrm{1}+\mathrm{e}}{\mathrm{2}} \\ $$$$\Leftrightarrow\sqrt[{\mathrm{4}}]{\mathrm{e}}\leqslant\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{\mathrm{t}^{\mathrm{2}} } \mathrm{dt}\leqslant\frac{\mathrm{1}+\mathrm{e}}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 29/Feb/24
$${thanks}\:{alot}\:{sir}\: \\ $$
Commented by witcher3 last updated on 29/Feb/24
$$\mathrm{withe}\:\mathrm{pleasur}\:\mathrm{barak}\left[\mathrm{alah}\:\mathrm{fik}\right. \\ $$
Answered by mr W last updated on 29/Feb/24
Commented by mr W last updated on 01/Mar/24
![x>0 y=e^x^2 ⇒y(0)=1, y(1)=e y′=2xe^x^2 >0 ⇒↗ y′′=2(1+2x)e^x^2 >0 ⇒ ⌣ ∫_0 ^1 e^x^2 dx=area under blue curve <area under red line >area under green line red area =((1+e)/2)×1=((1+e)/2) tangent at x=p: y=[1+2p(p−x)]e^p^2 y(0)=(1+2p^2 )e^p^2 y(1)=(1+2p^2 −2p)e^p^2 green area =((y(0)+y(1))/2)×1=(1+2p^2 −p)e^p^2 with p=(1/2): green area=(1+2×((1/2))^2 −(1/2))e^(((1/2))^2 ) =e^(1/4) green area < blue area < red area ⇒e^(1/4) < ∫_0 ^1 e^x^2 dx <((1+e)/2)](https://www.tinkutara.com/question/Q204884.png)
$${x}>\mathrm{0} \\ $$$${y}={e}^{{x}^{\mathrm{2}} } \:\Rightarrow{y}\left(\mathrm{0}\right)=\mathrm{1},\:{y}\left(\mathrm{1}\right)={e} \\ $$$${y}'=\mathrm{2}{xe}^{{x}^{\mathrm{2}} } >\mathrm{0}\:\Rightarrow\nearrow \\ $$$${y}''=\mathrm{2}\left(\mathrm{1}+\mathrm{2}{x}\right){e}^{{x}^{\mathrm{2}} } >\mathrm{0}\:\Rightarrow\:\smile \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{{x}^{\mathrm{2}} } {dx}={area}\:{under}\:{blue}\:{curve} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:<{area}\:{under}\:{red}\:{line}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:>{area}\:{under}\:{green}\:{line}\: \\ $$$${red}\:{area}\:=\frac{\mathrm{1}+{e}}{\mathrm{2}}×\mathrm{1}=\frac{\mathrm{1}+{e}}{\mathrm{2}} \\ $$$${tangent}\:{at}\:{x}={p}: \\ $$$${y}=\left[\mathrm{1}+\mathrm{2}{p}\left({p}−{x}\right)\right]{e}^{{p}^{\mathrm{2}} } \\ $$$${y}\left(\mathrm{0}\right)=\left(\mathrm{1}+\mathrm{2}{p}^{\mathrm{2}} \right){e}^{{p}^{\mathrm{2}} } \\ $$$${y}\left(\mathrm{1}\right)=\left(\mathrm{1}+\mathrm{2}{p}^{\mathrm{2}} −\mathrm{2}{p}\right){e}^{{p}^{\mathrm{2}} } \\ $$$${green}\:{area}\:=\frac{{y}\left(\mathrm{0}\right)+{y}\left(\mathrm{1}\right)}{\mathrm{2}}×\mathrm{1}=\left(\mathrm{1}+\mathrm{2}{p}^{\mathrm{2}} −{p}\right){e}^{{p}^{\mathrm{2}} } \\ $$$${with}\:{p}=\frac{\mathrm{1}}{\mathrm{2}}: \\ $$$${green}\:{area}=\left(\mathrm{1}+\mathrm{2}×\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\right){e}^{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} } ={e}^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$${green}\:{area}\:<\:{blue}\:{area}\:<\:{red}\:{area} \\ $$$$\Rightarrow{e}^{\frac{\mathrm{1}}{\mathrm{4}}} <\:\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{{x}^{\mathrm{2}} } {dx}\:<\frac{\mathrm{1}+{e}}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 01/Mar/24
$${thank}\:{you}\:{so}\:{much}\:{sir}\:{W} \\ $$