TITLE: If $\int^x_0 f (t) dt =x+ \int^1_x t f (t) dt$, find value of $f(1)$
QUESTION [5 upvotes]: 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 $f(1)$

REPLY [8 votes]: Differentiate both sides with respect to $x$. By the Fundamental Theorem of Calculus, we get
$$f(x)=1-xf(x)$$
So $f(x)=\frac{1}{1+x}$
Overdue Edit: As pointed out by Robert Z in the comments, this question is both incorrect, and asked verbatim elsewhere (with a more complete answer). I have voted to close on this basis.