$\underset{0}{\overset{\mathrm{x}}{\int }}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}=\mathrm{x}\mathrm{f}\left(\mathrm{x}\right)-\frac{4}{3}{\mathrm{x}}^3+{\mathrm{ax}}^2$

differentiate with respect to x.

            f (x) = x f ' (x) + f (x) – 4x² + 2ax

 $\therefore$      f ' (x) = 4x – 2a

            f (x) = 2x² – 2ax + c

                        c = a² – a – 6

            f (x) = 2x² – 2ax + a² – a – 6

Case-I:

F(0) < 0

 A $\in$ (–2,3)                      ……..(1) 

Case-II:

$\begin{array}{l}\mathrm{D}\ge 0\\ \\ \mathrm{f}\left(0\right)\ge 0\\ \frac{-\mathrm{b}}{2\mathrm{a}}>0\end{array}\}\mathrm{a}\in [3,1+\sqrt{3}].....\left(2\right)$      

sum = – 1 + 0 + 1 + 2 + 3 + 4 = 9 Ans.