Question Number 177173 by Shrinava last updated on 01/Oct/22 | ||
Answered by Peace last updated on 02/Oct/22 | ||
$${let}\:{f}\left({x}\right)={e}^{−\frac{\mathrm{1}}{{x}+\mathrm{1}}} ,{applie}\:{Taylors}\:{lagrange}\:{formul}\:\Rightarrow\exists{c}\in\left[\mathrm{0},\mathrm{1}\right]\:{such}\:{that} \\ $$$${f}\left({x}\right)={f}\left(\mathrm{0}\right)+{xf}'\left({c}\right)={e}^{−\mathrm{1}} +{x}.\frac{{e}^{−\frac{\mathrm{1}}{\mathrm{1}+{c}}} }{\left(\mathrm{1}+{c}\right)^{\mathrm{2}} } \\ $$$${We}\:{have}\:\forall{c}\in\left[\mathrm{0},\mathrm{1}\right]\:\frac{{e}^{−\frac{\mathrm{1}}{\mathrm{1}+{c}}} }{\left(\mathrm{1}+{c}\right)^{\mathrm{2}} }\leqslant\mathrm{1}\Leftrightarrow{t}^{\mathrm{2}} {e}^{−{t}} \leqslant{e}^{−\mathrm{1}} ,\forall{t}\in\left[\frac{\mathrm{1}}{\mathrm{2}},\mathrm{1}\right] \\ $$$${Wich}\:{is}\:{True}\:\frac{{e}^{{t}} }{{t}^{\mathrm{2}} }={g}\left({t}\right),{g}'\left({t}\right)=\frac{{t}\left({t}−\mathrm{2}\right){e}^{{t}} }{{t}^{\mathrm{4}} }<\mathrm{0},\forall{t}\in\left[\frac{\mathrm{1}}{\mathrm{2}},\mathrm{1}\right] \\ $$$${g}\left({t}\right)\geqslant{g}\left(\mathrm{1}\right)={e}\Rightarrow{et}^{\mathrm{2}} \geqslant{e}^{{t}} \Leftrightarrow{t}^{\mathrm{2}} {e}^{−{t}} \leqslant{e}^{−\mathrm{1}} \Rightarrow \\ $$$$\Rightarrow{f}\left({x}\right)\leqslant{e}^{−\mathrm{1}} +{xe}^{−\mathrm{1}} \\ $$$$\int_{{a}} ^{{b}} {f}\left({x}\right){dx}\leqslant{e}^{−\mathrm{1}} \int_{{a}} ^{{b}} \left(\mathrm{1}+{x}\right){dx}={e}^{−\mathrm{1}} \left({b}−{a}+\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}}\right)=\frac{\left({b}−{a}\right)}{\mathrm{2}{e}}\left(\mathrm{2}+{b}+{a}\right) \\ $$ | ||