If $f(x)=x^3+3x+2$ and $g(x)$ is the inverse of it, find the area bounded by $g(x)$ and $x$-axis and ordinates $x=-2$ and $x=6$. Is there any way to solve it without finding inverse of the function because finding I can't find inverse?

Here are the graphs of $f(x) = x^3 + 3 x + 2$ (red) and $g(x)$ (green) and the lines $x = -2$ and $x = 6$.

(Large version)
We see that the area $A$ between $g$ and the $x$-axis for $x \in [-2,6]$ is the same as the area between $f$ and the $y$-axis for $y \in [-2, 6]$.
The complication is the change of sign of $f$ at $x = g(0)$.
We could express $A$ as
\begin{align}
A 
= & 
A_1 - \left(- \int\limits_{g(-2)}^{g(0)} f(x) \,dx \right) + A_2 + 
\int\limits_{g(0)}^{0} f(x) \,dx +
A_3 - \int\limits_0^{g(6)} f(x) \, dx \\
&= A_1 + A_2 + A_3
+ \int\limits_{g(-2)}^0 f(x) \,dx
- \int\limits_0^{g(6)} f(x) \, dx
\end{align}
with $A_1 = (g(0) - g(-2))\,(-(-2))$, $A_2 = (0 - g(0))\,(-(-2))$ and $A_3 = (g(6)-0)\,6$.
Their sum reduces to
$$
A_1 + A_2 + A_3 = -2 \, g(-2) + 6 \, g(6)
$$
So we need to know only two values of $g$ and can avoid the complicated $g(0)$:
\begin{align}
g(-2):& -2 = f(x) = x^3 + 3 x + 2 \iff
x = -1 \\
g(6):& 6 = f(x) = x^3 + 3 x + 2 \iff 
4 = x^3 + 3 x \iff
x = 1 \\
\end{align}
This results in:
\begin{align}
A 
&= 8 + \int\limits_{-1}^0 f(x) \, dx - \int\limits_0^1 f(x) \, dx \\
&= 8 
+ \left[\frac{1}{4} x^4 + \frac{3}{2} x^2 + 2 x \right]_{-1}^0
- \left[\frac{1}{4} x^4 + \frac{3}{2} x^2 + 2 x \right]_0^1 \\
&= 8 - (7/4 - 2) - (7/4 + 2) \\
&= 8 - 14/4 \\
&= 9/2
\end{align}
