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

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 + $$\int^1_0 t f (t) dt I think, I am not in the right track Help me to find the value of f(1) • Differentiate with respect to x, that will be sufficient. – Start wearing purple Jun 2 '13 at 13:31 • I think we may assume that f is continuous. – Maddy Jun 2 '13 at 13:32 • @O.L. Yeah your method looks good. I am getting ans 1/2 .Am I right ?? – rst Jun 2 '13 at 13:35 • @O.L. Thanks a lot – rst Jun 2 '13 at 13:36 ## 1 Answer Differentiate both sides with respect to x. By the Fundamental Theorem of Calculus, we get$$f(x)=1-xf(x)$$So$f(x)=\frac{1}{1+x}$• Yeah,I am also getting the same – rst Jun 2 '13 at 13:37 • how you got$-xf(x)\$ can you please explain@daniel – David Jun 18 '15 at 9:41
• It comes from the Fundamental Theorem of Calculus – preferred_anon Jun 18 '15 at 10:51
• Something is wrong here: see math.stackexchange.com/questions/2648960/… – Robert Z Feb 14 '18 at 10:47
• @RobertZ Wow, I didn't expect that! I'll update my answer later to reflect it. – preferred_anon Feb 14 '18 at 11:25