$\underset{0}{\overset{x}{\int }}f\left(t\right)dt=x+\underset{x}{\overset{1}{\int }}t\cdot f\left(t\right)dt$, then the value of $f\left(1\right)=$

$\underset{0}{\overset{x}{\int }}f\left(t\right)dt=x+\underset{x}{\overset{1}{\int }}t\cdot f\left(t\right)dt$

$\frac{d}{dx}\cdot [\underset{0}{\overset{x}{\int }}f\left(t\right)\cdot dt]=\frac{d}{dx}[x+\underset{x}{\overset{1}{\int }}t\cdot f\left(t\right)\cdot dt]$

Newton – Leibnitz rule :

$\frac{d}{dx}\left[\underset{\varphi \left(x\right)}{\overset{\Psi \left(x\right)}{\int }}f\left(t\right)dt\right]=f\left(\Psi \left(x\right)\right)\cdot \frac{d}{dx}\left(\Psi \left(x\right)\right)-f(\Psi \left(x\right)\cdot \frac{d}{dx}\left(\varphi \left(x\right)\right))$

$\Psi \left(x\right),\varphi \left(x\right)$ are different equations

$(f\left(x\right)\cdot 1-0)=1+(0-x\cdot f\left(x\right)\left(1\right))$

$\Rightarrow f\left(x\right)=1-x\cdot f\left(x\right)$

$\Rightarrow f\left(x\right)+x\cdot f\left(x\right)=1$

$\Rightarrow f\left(x\right)(1+x)=1$

$\Rightarrow f\left(x\right)=\frac{1}{x+1}$

$\Rightarrow f\left(1\right)=\frac{1}{1+1}=\frac{1}{2}$.