Let f : `[1,oo] to[2,oo]` differentiable function such that f (1) = 2 . If `[1,oo] to[2,oo]`

Answer

`(8//3)`

Explanation

Given , f `(1)=(1)/(3)and6 int_(1)^(x)f(t)dt = 3xf (x) -x^(3),AAx le 1` Using Newton - Leibnitz formula. Differentiating both sides
`rArr6(x)*1-0=3f(x)+3xf'(x)-3x^(2)`
`rArr 3xf' (x) -3 f(x) = 3x^(2) rArr f' (x) -(1)/(x)f(x)=x`
`rArr (xf'(x)-f'(x))/(x^(2))=1rArr (d)/(dx){(x)/(x)}=1`
On integrating both sides , we get
`rArr (f(x))/(x')=x+c` " " [`:' f(1)=(1)/(3)]`
`(1)/(3)=1+crArrc=(2)/(3)andf(x)=x^(2)-(2)/(3)x`
`:. f(2)=4-(4)/(3)=(8)/(3)`
NOTE Here , f (1) = 2 , does not satisfy given function.
`:. f(1) = (1)/(3)`
For that f (x) `=x^(2)-(2)/(3)x andf(2)=4-(4)/(3)=(8)/(3)`