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

Question

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 $f(1)$

@O.L. Yeah your method looks good. I am getting ans $1/2$ .Am I right ?? — rst, Jun 02 '13 at 13:35

preferred_anon · Accepted Answer · 2020-10-02T22:46:15.213

8

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.

edited Oct 02 '20 at 22:46

answered Jun 02 '13 at 13:36

preferred_anon

17,121

Yeah,I am also getting the same – rst Jun 02 '13 at 13:37
how you got $-xf(x)$ can you please explain@daniel – David Jun 18 '15 at 09:41
It comes from the Fundamental Theorem of Calculus – preferred_anon Jun 18 '15 at 10:51
2

Something is wrong here: see https://math.stackexchange.com/questions/2648960/find-f1-from-equation-int-0xftdt-x-int-x1tftdt – 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

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

1 Answers1

Linked