If `f(x)=x+int_0^1t(x+t)f(t)dt`, then the value of `23/2\ f(0)` is equal to _________

Answer

9

Explanation

`f(x)=x+x int_(0)^(1)int_(0)^(1)tf(t)dt+int_(0)^(1)t^(2)f(t)dt`
`:.f(x)=x(1+A)+B`, where
`A=int_(0)^(1)tf(t)dt` an `B=int_(0)^(1)t^(2)f(t)dt`
Now `A=int_(0)^(1)t[t(1+A)+B]dt=(t^(3))/3(1+A)|_(0)^(1)+B/2t^(2)|_(0)^(1)=(1+A)/3+B/2`
or `4A-3B=2`.............1
Again `B=int_(0)^(1)t^(2)[t(1+A)+B]dt=(t^(4)(1+A))/4+(Bt^(3))/3]_(0)^(1)`
`=(1+A)/4+B/3`.............2
or `8B-3A=3`
Solving equations 1 and 2 we have `B-18/23=f(0)`.