prove-that-e-1-4-lt-0-1-e-t-2-dt-lt-1-e-2-

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) = {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