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 $\frac{1}{2}$

We have
${\int }_0^{x}f\left(t\right)dt=x+{\int }_x^{1}tf\left(t\right)dt$
Differentiating w.r. to x, we get
$f\left(x\right)=1-xf\left(x\right)$
$\Rightarrow f\left(x\right)+xf\left(x\right)=1\Rightarrow f\left(x\right)=\frac{1}{x+1}\Rightarrow f\left(1\right)=\frac{1}{2}$