Question

If $\underset{0}{\overset{x}{\int }}f\left(t\right)dt=x+\underset{x}{\overset{1}{\int }}tf\left(t\right)dt,$ 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 $\frac{1}{2}$

$\underset{0}{\overset{x}{\int }}f\left(t\right)dt=x+\underset{x}{\overset{1}{\int }}tf\left(t\right)dt$
Differentiating both sides using Leibnitz's rule, we have:
$\Rightarrow f\left(x\right)=1+0-xf\left(x\right)$
$\Rightarrow f\left(x\right)=1-xf\left(x\right)$
$\Rightarrow f\left(x\right)=\frac{1}{x+1}\Rightarrow f\left(1\right)=\frac{1}{2}$