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Question Number 156475 by SANOGO last updated on 11/Oct/21

soit:∫_o ^x f(t)=x+∫_0 ^x t^2 f(t)dt determiner f(1)

$${soit}:\int_{{o}} ^{{x}} {f}\left({t}\right)={x}+\int_{\mathrm{0}} ^{{x}} {t}^{\mathrm{2}} {f}\left({t}\right){dt} \\ $$$${determiner}\:{f}\left(\mathrm{1}\right) \\ $$

Commented by SANOGO last updated on 11/Oct/21

ok thank you

$${ok}\:{thank}\:{you} \\ $$

Commented by mr W last updated on 11/Oct/21

f(x)=1+x^2 f(x) f(x)=(1/(1−x^2 )) f(1) doesn′t exist.

$${f}\left({x}\right)=\mathrm{1}+{x}^{\mathrm{2}} {f}\left({x}\right) \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${f}\left(\mathrm{1}\right)\:{doesn}'{t}\:{exist}. \\ $$

Commented by SANOGO last updated on 11/Oct/21

can you detail for me so i understand?

$${can}\:{you}\:{detail}\:{for}\:{me}\: \\ $$$${so}\:{i}\:{understand}? \\ $$

Commented by mr W last updated on 11/Oct/21

(d/dx)[∫_(a(x)) ^(b(x)) f(t)dt]=f[b(x)]((db(x))/dx)−f[a(x)]((da(x))/dx)

$$\frac{{d}}{{dx}}\left[\int_{{a}\left({x}\right)} ^{{b}\left({x}\right)} {f}\left({t}\right){dt}\right]={f}\left[{b}\left({x}\right)\right]\frac{{db}\left({x}\right)}{{dx}}−{f}\left[{a}\left({x}\right)\right]\frac{{da}\left({x}\right)}{{dx}} \\ $$

Answered by Mathspace last updated on 11/Oct/21

by derivation we get f(x)=1+x^2 f(x) ⇒(1−x^2 )f(x)=1 ⇒ f(x)=(1/(1−x^2 )) ⇒f(1) dont exist!

$${by}\:{derivation}\:{we}\:{get} \\ $$$${f}\left({x}\right)=\mathrm{1}+{x}^{\mathrm{2}} {f}\left({x}\right)\:\Rightarrow\left(\mathrm{1}−{x}^{\mathrm{2}} \right){f}\left({x}\right)=\mathrm{1}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\:\Rightarrow{f}\left(\mathrm{1}\right)\:{dont}\:{exist}! \\ $$

Commented by SANOGO last updated on 11/Oct/21

thank you

$${thank}\:{you} \\ $$

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