A cubic $f\left(x\right)$vanishes at $x = - 2$ and has relative minimum" /"maximum at$x=-1$ and $x=\frac{1}{3}$ such that ${\int }_{-1}^1 f\left(x\right)dx=\frac{14}{3}$ Then $f\left(x\right)$ is

Since $f (x$), which is of degree 3, has relative minimum/maximum at $x = - 1$ and $x=\frac{1}{3}$ Therefore $x=-1$

$x=\frac{1}{3}$ are roots of $f^{\mathrm{\prime }}\left(x\right)=0$ Thus, $x+1$ and $3x-1$ are factors of $f^{\mathrm{\prime }}\left(x\right)$ Consequently, we have

$\begin{array}{l}{f}^{\mathrm{\prime }}\left(x\right)=\lambda (x+1)(3x-1)=\lambda \left(3x^2+2x-1\right).\\ \Rightarrow f\left(x\right)=\lambda \left(x^3+x^2-x\right)+c.\end{array}$

Now,

$f(-2)=0$

$\Rightarrow c=2\lambda$

We have

$\begin{array}{l}{\int }_{-1}^1 f\left(x\right)dx=\frac{14}{3}\\ \Rightarrow {\int }_{-1}^1 \left[\lambda \left(x^3+x^2-x\right)+c\right]dx=\frac{14}{3}\\ \Rightarrow \lambda {\int }_{-1}^1 x^2+{\int }_{-1}^1 cdx=\frac{14}{3}\Rightarrow \frac{2\lambda }{3}+2c=\frac{14}{3}\Rightarrow \lambda +3c=7\dots \left(\mathrm{ii}\right)\end{array}$

Solving (i) and (ii), we get $\lambda =1,c=2$

Hence, $f\left(x\right)=x^3+x^2-x+2$