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Question Number 204878 by mnjuly1970 last updated on 29/Feb/24

$p r o v e t h a t :$ $\sqrt[4]{e} < \int_{0}^{1} e^{t^{2}} d t < \frac{1 + e}{2}$
	Answered by witcher3 last updated on 29/Feb/24
	$let f(x)=e^x^2 f′(x)=2xe^x^2 ;f′′(x)=(2+4x^2 )e^x^2 ≥0 f is convex Let C_f ={(x,f(x));x∈R^2 } Γ: tangent to Cf at(1/2) Γ=f′((1/2))(x−(1/2))+f((1/2))=e^(1/4) (x−(1/2))+e^(1/4) =e^(1/4) x+(e^(1/4) /2);Since f is convex⇒f(x)≥xe^(1/4) +(e^(1/4) /2) ⇒∫_0 ^1 e^x^2 dx≥∫_0 ^1 (xe^(1/4) +(e^(1/4) /2))dx=e^(1/4) =(e)^(1/4) f(x)=e^x^2 applie lagrange theorem in [0,t] ⇒∃_c ∈[0,t] suche f(t)=f(0)+tf′(c) f′′(x)≥0⇒f′ is increase ⇒f′(c)≤f′(t) ⇒f(t)=1+t(2ce^c^2 )≤1+2t^2 e^t^2 ⇒∫_0 ^1 f(t)dt≤∫_0 ^1 (1+2t^2 e^t^2 )dt=∫_0 ^1 1dt+∫_0 ^1 t.(2te^t^2 )dt u=t,v′=2te^t^2 IBP⇒ ∫_0 ^1 e^t^2 dt≤1+[te^t^2 ]_0 ^1 −∫_0 ^1 e^t^2 dt⇒2∫_0 ^1 e^t^2 dt≤1+e ⇒∫_0 ^1 e^t^2 dt≤((1+e)/2) ⇔(e)^(1/4) ≤∫_0 ^1 e^t^2 dt≤((1+e)/2)$
	$let f (x) = e^{x^{2}}$ $f^{'} (x) = {2 xe}^{x^{2}}; f^{″} (x) = (2 + {4 x}^{2}) e^{x^{2}} ⩾ 0$ $f is convex Let C_{f} = {(x, f (x)); x \in R^{2}}$ $Γ : tangent to Cf at \frac{1}{2}$ $Γ = f^{'} (\frac{1}{2}) (x - \frac{1}{2}) + f (\frac{1}{2}) = e^{\frac{1}{4}} (x - \frac{1}{2}) + e^{\frac{1}{4}}$ $= e^{\frac{1}{4}} x + \frac{e^{\frac{1}{4}}}{2}; Since f is convex \Rightarrow f (x) ⩾ {xe}^{\frac{1}{4}} + \frac{e^{\frac{1}{4}}}{2}$ $\Rightarrow \int_{0}^{1} e^{x^{2}} dx ⩾ \int_{0}^{1} ({xe}^{\frac{1}{4}} + \frac{e^{\frac{1}{4}}}{2}) dx = e^{\frac{1}{4}} = \sqrt[4]{e}$ $f (x) = e^{x^{2}} applie lagrange theorem in [0, t]$ $\Rightarrow \exists_{c} \in [0, t] suche f (t) = f (0) + {tf}^{'} (c)$ $f^{″} (x) ⩾ 0 \Rightarrow f^{'} is increase \Rightarrow f^{'} (c) ⩽ f^{'} (t)$ $\Rightarrow f (t) = 1 + t ({2 ce}^{c^{2}}) ⩽ 1 + {2 t}^{2} e^{t^{2}}$ $\Rightarrow \int_{0}^{1} f (t) dt ⩽ \int_{0}^{1} (1 + {2 t}^{2} e^{t^{2}}) dt = \int_{0}^{1} 1 dt + \int_{0}^{1} t . ({2 te}^{t^{2}}) dt$ $u = t, v^{'} = {2 te}^{t^{2}} IBP \Rightarrow$ $\int_{0}^{1} e^{t^{2}} dt ⩽ 1 + {[{te}^{t^{2}}]}_{0}^{1} - \int_{0}^{1} e^{t^{2}} dt \Rightarrow 2 \int_{0}^{1} e^{t^{2}} dt ⩽ 1 + e$ $\Rightarrow \int_{0}^{1} e^{t^{2}} dt ⩽ \frac{1 + e}{2}$ $\Leftrightarrow \sqrt[4]{e} ⩽ \int_{0}^{1} e^{t^{2}} dt ⩽ \frac{1 + e}{2}$
	Commented by mnjuly1970 last updated on 29/Feb/24

	$t h a n k s a l o t s i r$
	Commented by witcher3 last updated on 29/Feb/24

	$withe pleasur barak [alah fik$
	Answered by mr W last updated on 29/Feb/24

	Commented by mr W last updated on 01/Mar/24

	$x > 0$ $y = e^{x^{2}} \Rightarrow y (0) = 1, y (1) = e$ $y^{'} = 2 {x e}^{x^{2}} > 0 \Rightarrow↗$ $y^{″} = 2 (1 + 2 x) e^{x^{2}} > 0 \Rightarrow ⌣$ $\int_{0}^{1} e^{x^{2}} d x = a r e a u n d e r b l u e c u r v e$ $< a r e a u n d e r r e d l i n e$ $> a r e a u n d e r g r e e n l i n e$ $r e d a r e a = \frac{1 + e}{2} \times 1 = \frac{1 + e}{2}$ $t a n g e n t a t x = p :$ $y = [1 + 2 p (p - x)] e^{p^{2}}$ $y (0) = (1 + 2 p^{2}) e^{p^{2}}$ $y (1) = (1 + 2 p^{2} - 2 p) e^{p^{2}}$ $g r e e n a r e a = \frac{y (0) + y (1)}{2} \times 1 = (1 + 2 p^{2} - p) e^{p^{2}}$ $w i t h p = \frac{1}{2} :$ $g r e e n a r e a = (1 + 2 \times {(\frac{1}{2})}^{2} - \frac{1}{2}) e^{{(\frac{1}{2})}^{2}} = e^{\frac{1}{4}}$ $g r e e n a r e a < b l u e a r e a < r e d a r e a$ $\Rightarrow e^{\frac{1}{4}} < \int_{0}^{1} e^{x^{2}} d x < \frac{1 + e}{2}$
	Commented by mnjuly1970 last updated on 01/Mar/24

	$t h a n k y o u s o m u c h s i r W$
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