Question Number 138407 by tugu last updated on 13/Apr/21
$${if}\:{the}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}}\underset{\mathrm{1}} {\overset{{x}} {\int}}\left(\mathrm{2}{t}−{F}\:'\left({t}\right)\right){dt}\:\:\Rightarrow\:{what}\:{the}\:{F}\:'\left(\mathrm{1}\right)\:{value}\:{using}\:{the}\:{Leibnitz}\:{formula}. \\ $$
Answered by ajfour last updated on 13/Apr/21
![F ′(x)=−((F(x))/x)+(1/x)[2x−F ′(x)] (x+1)F ′(x)+F(x)=2x xF ′(x)+F(x)=2x−F ′(x) d[xF(x)]=[2x−F ′(x)]dx xF(x)=x^2 −F(x)+c F(x)=((x^2 +c)/(x+1)) xF(x)=(x^2 −1)−F(x)+F(1) F(1)=1−x^2 +x^2 +c F(1)=c+1=((c+1)/2) ⇒ c=−1 F(1)=0 F(x)=((x^2 −1)/(x+1))=x−1 F ′(x)=1 ⇒ F ′(1)=1 ★ check x(x−1)=∫_( 1) ^( x) (2t−1)dt x(x−1)=x^2 −1−(x−1) ⇒ x(x−1)=x(x−1) ✓](Q138430.png)
$${F}\:'\left({x}\right)=−\frac{{F}\left({x}\right)}{{x}}+\frac{\mathrm{1}}{{x}}\left[\mathrm{2}{x}−{F}\:'\left({x}\right)\right] \\ $$$$\left({x}+\mathrm{1}\right){F}\:'\left({x}\right)+{F}\left({x}\right)=\mathrm{2}{x} \\ $$$${xF}\:'\left({x}\right)+{F}\left({x}\right)=\mathrm{2}{x}−{F}\:'\left({x}\right) \\ $$$${d}\left[{xF}\left({x}\right)\right]=\left[\mathrm{2}{x}−{F}\:'\left({x}\right)\right]{dx} \\ $$$${xF}\left({x}\right)={x}^{\mathrm{2}} −{F}\left({x}\right)+{c} \\ $$$${F}\left({x}\right)=\frac{{x}^{\mathrm{2}} +{c}}{{x}+\mathrm{1}} \\ $$$${xF}\left({x}\right)=\left({x}^{\mathrm{2}} −\mathrm{1}\right)−{F}\left({x}\right)+{F}\left(\mathrm{1}\right) \\ $$$${F}\left(\mathrm{1}\right)=\mathrm{1}−{x}^{\mathrm{2}} +{x}^{\mathrm{2}} +{c} \\ $$$${F}\left(\mathrm{1}\right)={c}+\mathrm{1}=\frac{{c}+\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:{c}=−\mathrm{1} \\ $$$${F}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${F}\left({x}\right)=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}+\mathrm{1}}={x}−\mathrm{1} \\ $$$${F}\:'\left({x}\right)=\mathrm{1}\:\:\Rightarrow\:{F}\:'\left(\mathrm{1}\right)=\mathrm{1}\:\:\bigstar \\ $$$${check} \\ $$$${x}\left({x}−\mathrm{1}\right)=\int_{\:\mathrm{1}} ^{\:{x}} \left(\mathrm{2}{t}−\mathrm{1}\right){dt} \\ $$$${x}\left({x}−\mathrm{1}\right)={x}^{\mathrm{2}} −\mathrm{1}−\left({x}−\mathrm{1}\right) \\ $$$$\Rightarrow\:\:{x}\left({x}−\mathrm{1}\right)={x}\left({x}−\mathrm{1}\right)\:\checkmark \\ $$
Commented by ajfour last updated on 13/Apr/21
$${ah}!\:\:{nobody}\:{likes}\:{this}.. \\ $$