Answer
No of intersection points of $f(x)$ on $x$ axis is $3$.
Area between the $x$ axis and $f(x)$ is $52 \hspace{0.1cm}\text{sq units}$.
Work Step by Step
The curve intersects the $x$ axis at the following three points in the interval $[-2,4]$, $x=\pm \sqrt{2},0$.
The area of the definite integral $=\int_{-2}^4 (x^3-2x) dx$
Since there are $3$ intersection points we can split the integral into $4$ separate areas.
Note that the curve is symmetric about $0$ in the interval $[-\sqrt{2},\sqrt{2}]$ and hence the area between $f(x)$ and $x$ axis in this interval is twice the area in the interval $[0,\sqrt{2}]$.
$\int_{-\sqrt{2}}^{\sqrt{2}}(x^3-2x) dx=2\times \int_0^{\sqrt{2}}(x^3-2x) dx=2\hspace{0.1cm}\text{sq. units}$
$-\int_{-2}^{-\sqrt{2}}(x^3-2x) dx=1 \hspace{0.1cm} \text{sq. units}$
$\int_{\sqrt{2}}^4 (x^3-2x) dx=49\hspace{0.1cm}\text{sq. units}$
Therefore, the total area $=2+1+49=52\hspace{0.1cm} \text{sq. units}$
