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Question Number 138407 by tugu last updated on 13/Apr/21
if the F(x)=(1/x)∫_1 ^x (2t−F ′(t))dt  ⇒ what the F ′(1) value using the Leibnitz formula.
$${if}\:{the}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}}\underset{\mathrm{1}} {\overset{{x}} {\int}}\left(\mathrm{2}{t}−{F}\:'\left({t}\right)\right){dt}\:\:\Rightarrow\:{what}\:{the}\:{F}\:'\left(\mathrm{1}\right)\:{value}\:{using}\:{the}\:{Leibnitz}\:{formula}. \\ $$
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) ✓
$${F}\:'\left({x}\right)=−\frac{{F}\left({x}\right)}{{x}}+\frac{\mathrm{1}}{{x}}\left[\mathrm{2}{x}−{F}\:'\left({x}\right)\right] \\ $$$$\left({x}+\mathrm{1}\right){F}\:'\left({x}\right)+{F}\left({x}\right)=\mathrm{2}{x} \\ $$$${xF}\:'\left({x}\right)+{F}\left({x}\right)=\mathrm{2}{x}−{F}\:'\left({x}\right) \\ $$$${d}\left[{xF}\left({x}\right)\right]=\left[\mathrm{2}{x}−{F}\:'\left({x}\right)\right]{dx} \\ $$$${xF}\left({x}\right)={x}^{\mathrm{2}} −{F}\left({x}\right)+{c} \\ $$$${F}\left({x}\right)=\frac{{x}^{\mathrm{2}} +{c}}{{x}+\mathrm{1}} \\ $$$${xF}\left({x}\right)=\left({x}^{\mathrm{2}} −\mathrm{1}\right)−{F}\left({x}\right)+{F}\left(\mathrm{1}\right) \\ $$$${F}\left(\mathrm{1}\right)=\mathrm{1}−{x}^{\mathrm{2}} +{x}^{\mathrm{2}} +{c} \\ $$$${F}\left(\mathrm{1}\right)={c}+\mathrm{1}=\frac{{c}+\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:{c}=−\mathrm{1} \\ $$$${F}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${F}\left({x}\right)=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}+\mathrm{1}}={x}−\mathrm{1} \\ $$$${F}\:'\left({x}\right)=\mathrm{1}\:\:\Rightarrow\:{F}\:'\left(\mathrm{1}\right)=\mathrm{1}\:\:\bigstar \\ $$$${check} \\ $$$${x}\left({x}−\mathrm{1}\right)=\int_{\:\mathrm{1}} ^{\:{x}} \left(\mathrm{2}{t}−\mathrm{1}\right){dt} \\ $$$${x}\left({x}−\mathrm{1}\right)={x}^{\mathrm{2}} −\mathrm{1}−\left({x}−\mathrm{1}\right) \\ $$$$\Rightarrow\:\:{x}\left({x}−\mathrm{1}\right)={x}\left({x}−\mathrm{1}\right)\:\checkmark \\ $$
Commented by ajfour last updated on 13/Apr/21
ah!  nobody likes this..
$${ah}!\:\:{nobody}\:{likes}\:{this}.. \\ $$

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