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If $\int^x_0 f (t) dt =x+ \int^1_x t f (t) dt$, find value of $f(1)$

solution:-

$\int^x_0 f (t) dt =x+ \int^1_x t f (t) dt$

$\int^x_0 f (t) dt =x+ \int^0_x t f (t) dt$ + $\int^1_0 t f (t) dt$

$\int^x_0 f (t) dt =x- \int^x_0 t f (t) dt$ + $\int^1_0 t f (t) dt$

$\int^x_0 f (t) dt + \int^x_0 t f (t) dt$ =$x + $$\int^1_0 t f (t) dt$

I think, I am not in the right track

Help me to find the value of $f(1)$

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    $\begingroup$ Differentiate with respect to $x$, that will be sufficient. $\endgroup$ – Start wearing purple Jun 2 '13 at 13:31
  • $\begingroup$ I think we may assume that $f$ is continuous. $\endgroup$ – Maddy Jun 2 '13 at 13:32
  • $\begingroup$ @O.L. Yeah your method looks good. I am getting ans $1/2$ .Am I right ?? $\endgroup$ – rst Jun 2 '13 at 13:35
  • $\begingroup$ @O.L. Thanks a lot $\endgroup$ – rst Jun 2 '13 at 13:36
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Differentiate both sides with respect to $x$. By the Fundamental Theorem of Calculus, we get $$f(x)=1-xf(x)$$ So $f(x)=\frac{1}{1+x}$

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  • $\begingroup$ Yeah,I am also getting the same $\endgroup$ – rst Jun 2 '13 at 13:37
  • $\begingroup$ how you got $-xf(x)$ can you please explain@daniel $\endgroup$ – David Jun 18 '15 at 9:41
  • $\begingroup$ It comes from the Fundamental Theorem of Calculus $\endgroup$ – preferred_anon Jun 18 '15 at 10:51
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    $\begingroup$ Something is wrong here: see math.stackexchange.com/questions/2648960/… $\endgroup$ – Robert Z Feb 14 '18 at 10:47
  • $\begingroup$ @RobertZ Wow, I didn't expect that! I'll update my answer later to reflect it. $\endgroup$ – preferred_anon Feb 14 '18 at 11:25

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