Practicing Success
If $\int\limits_0^x f(t) d t=x+\int\limits_x^1 t f(t) d t$, then the value of $f(1)$, is |
1/2 0 1 -1/2 |
1/2 |
We have, $\int\limits_0^x f(t) d t=x+\int\limits_x^1 t f(t) d t$ $\Rightarrow \frac{d}{d x}\left\{\int\limits_0^x f(t) d t\right\}=\frac{d}{d x}\left\{x+\int\limits_x^1 t f(t) d t\right\}$ $\Rightarrow f(x)=1+0-x f(x)$ [Using Leibnitz's rule] $\Rightarrow f(x)=1-x f(x)$ $\Rightarrow f(x)=\frac{1}{x+1} \Rightarrow f(1)=\frac{1}{2}$ |