Evaluate. Then interpret the result in terms of the area above and/or below the x-axis.
$${\int }_{-\frac{1}{2}}^{\frac{1}{2}}(x^3-3x)dx$$
$${\int }_{-\frac{1}{2}}^{\frac{1}{2}}(x^3-3x)dx=◻(\text{ Type an integer or a simplified fraction.) }$$

Step 1 ：The integral of a function over an interval can be interpreted as the area under the curve of the function over that interval. However, if the function dips below the x-axis, the area below the x-axis is subtracted from the total.

Step 2 ：To solve this problem, we need to find the antiderivative of the function $x^3-3x$, which is $\frac{1}{4}{x}^4-\frac{3}{2}{x}^2$.

Step 3 ：Then we need to evaluate this antiderivative at the limits of integration, $-\frac{1}{2}$ and $\frac{1}{2}$, and subtract the lower limit value from the upper limit value.

Step 4 ：The result of the integral is 0. This means that the area above the x-axis is exactly equal to the area below the x-axis over the interval from -1/2 to 1/2.

Step 5 ：The integral of the function $x^3-3x$ from -1/2 to 1/2 is $\boxed{{\displaystyle 0}}$.