Evaluate. Then interpret the result in terms of the area above and/or below the x-axis. \[ \int_{-\frac{1}{2}}^{\frac{1}{2}}\left(x^{3}-3 x\right) d x \] \[ \int_{-\frac{1}{2}}^{\frac{1}{2}}\left(x^{3}-3 x\right) d x=\square(\text { Type an integer or a simplified fraction.) } \]

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 \(\boxed{0}\).