Question

If ${\int }_0^{x}f\left(t\right)dt=x+{\int }_x^{1}tf\left(t\right)dt$

then the value of $f\left(1\right)$ is?

• A. $\frac{1}{2}$
• B. $0$
• C. $1$
• D. $-\frac{1}{2}$

Answer 1

The correct option is A $\frac{1}{2}$

${\int }_0^{x}f\left(t\right)dt=x+{\int }_x^{1}tf\left(t\right)dt$
Differentiate both sides w.r.t $x$ using Leibnitz rule,
$f\left(x\right)=1-xf\left(x\right)$

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