We start by identifying the points of intersection of the curves by solving the equation \(x^3-4x = x\text{.}\)

\begin{gather*}
x^3-4x = x\\
x^3-5x = 0\\
x(x^2-5) = 0
\end{gather*}
One solution is \(x=0\text{,}\) corresponding to the intersection at \((x,y)=(0,0)\text{.}\) Two other solutions come from \(x^2=5\) at \(x = \pm \sqrt{5}\text{.}\) The total area consists of two regions, the first with \(x \in [-\sqrt{5},0]\) and the second with \(x \in [0,\sqrt{5}]\text{.}\)

On the first interval \([-\sqrt{5},0]\text{,}\) the height of increments is given by

\begin{equation*}
h(x) = (x^3-4x) - x = x^3-5x
\end{equation*}
because the cubic polynomial is the top curve. So the area over the interval \([-\sqrt{5},0]\) is computed by

\begin{equation*}
A_1 = \int_{-\sqrt{5}}^{0} x^3-5x \, dx.
\end{equation*}
In order to use the elementary accumulation formulas, the integral needs to start at \(x=0\text{,}\) so we reverse the order of integration and change the sign.

\begin{align*}
A_1 &= \int_{-\sqrt{5}}^{0} x^3-5x \, dx\\
&= -\int_{0}^{-\sqrt{5}} x^3-5x \, dx\\
&= -\int_{0}^{-\sqrt{5}} x^3 \, dx + 5 \int_{0}^{-\sqrt{5}} x \, dx\\
&= -\left( \frac{1}{4}(-\sqrt{5})^4\right) + 5 \left(\frac{1}{2}(-\sqrt{5})^2\right)\\
&= -\frac{25}{4} + \frac{25}{2}\\
&= \frac{25}{4}
\end{align*}
On the second interval \([0,\sqrt{5}]\text{,}\) the height of increments is given by

\begin{equation*}
h(x) = x - (x^3-4x) = 5x-x^3
\end{equation*}
because now the cubic polynomial is the bottom curve. The corresponding area is

\begin{equation*}
A_2 = \int_{0}^{\sqrt{5}} 5x- x^3 \, dx
\end{equation*}
which has a value

\begin{align*}
A_2 &= \int_{0}^{\sqrt{5}} 5x- x^3 \, dx\\
&= 5\int_{0}^{\sqrt{5}} x \, dx - \int_{0}^{\sqrt{5}} x^3 \, dx\\
&= 5\left( \frac{1}{2}(\sqrt{5})^2\right) - \left(\frac{1}{4}(\sqrt{5})^4\right)\\
&= \frac{25}{2} -\frac{25}{4}\\
&= \frac{25}{4}
\end{align*}
As we should expect from symmetry, the two areas are equal \(A_1=A_2\text{.}\)

The total area of the region is

\begin{equation*}
A = A_1 + A_2 = \frac{25}{4} + \frac{25}{4} = \frac{25}{2}\text{.}
\end{equation*}
If we were to think of the distance between the two curves in terms of the absolute value, the integral could be computed over a single interval,

\begin{equation*}
A = \int_{-\sqrt{5}}^{\sqrt{5}} \left| x - (x^3-4x)\right| \, dx
= \int_{-\sqrt{5}}^{\sqrt{5}} \left| 5x - x^3\right| \, dx\text{.}
\end{equation*}
Using SageMath, the absolute value prevents an exact integral.