Question Number 47182 by maxmathsup by imad last updated on 05/Nov/18 | ||
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}} \sqrt{\mathrm{1}−\sqrt{{x}}}{dx}\: \\ $$ | ||
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Nov/18 | ||
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Nov/18 | ||
$${graph}\:{of}\:{e}^{−{x}} \sqrt{\mathrm{1}−\sqrt{{x}}\:} \\ $$$$\mathrm{0}.\mathrm{50}\:>\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{x}} \sqrt{\mathrm{1}−\sqrt{{x}}}\:\:{dx}>\mathrm{0} \\ $$ | ||
Commented by maxmathsup by imad last updated on 11/Nov/18 | ||
$${let}\:{determine}\:{a}\:{approximat}\:{value}\:{of}\:{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}} \sqrt{\mathrm{1}−\sqrt{{x}}}{dx}\:{we}\:{have} \\ $$$${e}^{{u}} =\mathrm{1}+\frac{{u}}{\mathrm{1}!}\:+\frac{{u}^{\mathrm{2}} }{\mathrm{2}!}\:+....\Rightarrow{e}^{−{x}} =\mathrm{1}−{x}\:+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:−...\Rightarrow\mathrm{0}\leqslant\:{e}^{−{x}} \leqslant\mathrm{1}−{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{0}\leqslant{e}^{−{x}} \sqrt{\mathrm{1}−\sqrt{{x}}}\leqslant\left(\mathrm{1}−{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\sqrt{\mathrm{1}−\sqrt{{x}}}\:\Rightarrow\mathrm{0}\leqslant\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}} \sqrt{\mathrm{1}−\sqrt{{x}}}{dx}\leqslant\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\sqrt{\mathrm{1}−\sqrt{{x}}}{dx} \\ $$$${changement}\:\sqrt{\mathrm{1}−\sqrt{{x}}}={t}\:{give}\:\mathrm{1}−\sqrt{{x}}={t}^{\mathrm{2}} \:\Rightarrow\sqrt{{x}}=\mathrm{1}−{t}^{\mathrm{2}} \:\Rightarrow{x}=\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} ={t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\sqrt{\mathrm{1}−\sqrt{{x}}}{dx}=−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} −\mathrm{1}\:+\frac{\left({t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}\right){t}\left(\mathrm{4}{t}^{\mathrm{3}} −\mathrm{4}{t}\right){dt} \\ $$$$=−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{2}{t}^{\mathrm{4}} \:+\mathrm{4}{t}^{\mathrm{2}} \:+\left(\:{t}^{\mathrm{8}} +\mathrm{4}{t}^{\mathrm{4}} \:+\mathrm{1}\:+\mathrm{2}\left(−\mathrm{2}{t}^{\mathrm{6}} \:+{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} \right)\left({t}^{\mathrm{4}} −{t}^{\mathrm{2}} \right){dt}\right.\right. \\ $$$$=−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{2}{t}^{\mathrm{4}} \:+\mathrm{4}{t}^{\mathrm{2}} \:+{t}^{\mathrm{8}} \:+\mathrm{6}{t}^{\mathrm{4}} \:−\mathrm{4}{t}^{\mathrm{6}} −\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{1}\right)\left({t}^{\mathrm{4}} −{t}^{\mathrm{2}} \right){dt} \\ $$$$=−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left({t}^{\mathrm{8}} −\mathrm{4}{t}^{\mathrm{6}} −\mathrm{2}{t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{4}} −{t}^{\mathrm{2}} \right){dt} \\ $$$$=−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\:{t}^{\mathrm{12}} −{t}^{\mathrm{10}} −\mathrm{4}{t}^{\mathrm{10}} \:+\mathrm{4}{t}^{\mathrm{8}} \:−\mathrm{2}{t}^{\mathrm{8}} \:+\mathrm{2}{t}^{\mathrm{6}} −\mathrm{4}{t}^{\mathrm{6}} \:+\mathrm{4}{t}^{\mathrm{4}} \right){dt} \\ $$$$=−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left({t}^{\mathrm{12}} −\mathrm{5}{t}^{\mathrm{10}} \:+\mathrm{2}{t}^{\mathrm{8}} \:−\mathrm{2}{t}^{\mathrm{6}} \:+\mathrm{4}{t}^{\mathrm{4}} \right){dt} \\ $$$$=−\mathrm{2}\left[\frac{{t}^{\mathrm{13}} }{\mathrm{13}}\:−\frac{\mathrm{5}{t}^{\mathrm{11}} }{\mathrm{11}}\:+\frac{\mathrm{2}{t}^{\mathrm{9}} }{\mathrm{9}}\:−\frac{\mathrm{2}{t}^{\mathrm{7}} }{\mathrm{7}}\:+\frac{\mathrm{4}{t}^{\mathrm{5}} }{\mathrm{5}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=−\mathrm{2}\left(\:\frac{\mathrm{1}}{\mathrm{13}}\:−\frac{\mathrm{5}}{\mathrm{11}}\:+\frac{\mathrm{2}}{\mathrm{9}}\:−\frac{\mathrm{2}}{\mathrm{7}}\:+\frac{\mathrm{4}}{\mathrm{5}}\right)=−\frac{\mathrm{2}}{\mathrm{13}}\:+\frac{\mathrm{10}}{\mathrm{11}}\:−\frac{\mathrm{4}}{\mathrm{9}}\:+\frac{\mathrm{4}}{\mathrm{7}}\:−\frac{\mathrm{8}}{\mathrm{5}}\:=\alpha_{\mathrm{0}} \:\Rightarrow\mathrm{0}<\:{A}\leqslant\alpha_{\mathrm{0}} \\ $$ | ||