if f(x)=x+∫01 t(x+t)f(t)dt, then the value of 232f(0) is equal to__.

$f (x) = x + x \int_{0}^{1} t f (t) d t + \int_{0}^{1} t^{2} f (t) d t$

$∴ f (x) = x (1 + A) + B$ where

$A = \int_{0}^{1} t f (t) d t$ and $B = \int_{0}^{1} t^{2} f (t) d t$

Now, $A = \int_{0}^{1} t [t (1 + A) + B] d t = {\frac{t^{3}}{3} (1 + A)|}_{0}^{1} + {\frac{B}{2} t^{2}|}_{0}^{1} = \frac{1 + A}{3} + \frac{B}{2}$

or $4 A - 3 B = 2$ (1)

Again, ${B = \int_{0}^{1} t^{2} [t (1 + A) + B] d t = \frac{t^{4} (1 + A)}{4} + \frac{B t^{3}}{3}]}_{0}^{1}$

= $= \frac{1 + A}{4} + \frac{B}{3}$

or $8 B - 3 A = 3$ (2)

Solving equations (1) and (2), we have $B = \frac{18}{23} = f (0)$ .

if $f (x) = x + \int_{0}^{1} t (x + t) f (t) d t$ , then the value of $\frac{23}{2} f (0)$ is equal to__.