Question Number 156455 by SANOGO last updated on 11/Oct/21
Commented by prakash jain last updated on 11/Oct/21
$$\mathrm{0} \\ $$
Commented by SANOGO last updated on 11/Oct/21
$${demonstration}\:{si}\:{possible} \\ $$
Commented by prakash jain last updated on 11/Oct/21
$$\mathrm{put}\:{t}\:=\:\frac{\mathrm{1}}{{u}}\: \\ $$$$\mathrm{calculate}\:\mathrm{2}{I} \\ $$
Commented by prakash jain last updated on 11/Oct/21
$${one}\:{with}\:{t}\:{and}\:{another}\:{one}\:{with} \\ $$$${t}=\frac{\mathrm{1}}{{u}} \\ $$
Answered by Mathspace last updated on 11/Oct/21
$${I}=\int_{\frac{\mathrm{1}}{{x}}} ^{{x}} {f}\left({t}\right){dt}\:\Rightarrow{I}=_{{t}=\frac{\mathrm{1}}{{u}}} \:\int_{{x}} ^{\frac{\mathrm{1}}{{x}}} {f}\left(\frac{\mathrm{1}}{{u}}\right)\frac{−{du}}{{u}^{\mathrm{2}} } \\ $$$$=\int_{\frac{\mathrm{1}}{{x}}} ^{{x}} \frac{−{u}^{\mathrm{2}} {f}\left({u}\right)}{{u}^{\mathrm{2}} }{du}\:=−\int_{\frac{\mathrm{1}}{{x}}} ^{{x}} {f}\left({u}\right){du} \\ $$$$=−\left(\:{F}\left({x}\right)−{F}\left(\frac{\mathrm{1}}{{x}}\right)\right)={F}\left(\frac{\mathrm{1}}{{x}}\right)−{F}\left({x}\right) \\ $$$${F}\:{est}\:{la}\:{primitive}\:{de}\:{f} \\ $$$$\left(\:{je}\:{pense}\:{qu}\:{il}\:{ya}\:{erreur}\:{dans}\right. \\ $$$$\left.{la}\:{Q}!\right) \\ $$
Commented by SANOGO last updated on 11/Oct/21
$${merci}\:{bien} \\ $$
Commented by prakash jain last updated on 12/Oct/21
$${I}=\int_{\mathrm{1}/{x}} ^{{x}} \frac{\mathrm{1}}{{u}^{\mathrm{2}} }{f}\left(\frac{\mathrm{1}}{{u}}\right){du}=\int_{\mathrm{1}/{x}} ^{{x}} {f}\left({u}\right){du} \\ $$$${x}^{\mathrm{2}} {f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{{x}}\right)=\mathrm{0}\:\:\left({x}>\mathrm{0}\right) \\ $$$${f}\left({x}\right)+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{f}\left(\frac{\mathrm{1}}{{x}}\right)=\mathrm{0} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{1}/{x}} ^{{x}} \frac{\mathrm{1}}{{u}^{\mathrm{2}} }{f}\left(\frac{\mathrm{1}}{{u}}\right){du}+\int_{\mathrm{1}/{x}} ^{{x}} {f}\left({u}\right){du} \\ $$
Commented by SANOGO last updated on 12/Oct/21
$${merci}\:{bien} \\ $$