Question

In Exercises \(41-44\) , find the total area of the region between the curve and the \(x\) -axis. $$y=x^{3}-4 x, \quad-2 \leq x \leq 2$$

Short answer

The total area of the region between the curve \(y = x^{3} - 4x\) and the x -axis from \(-2 \leq x \leq 2\) is \(4\) square units.

Step-by-step solution

Find the Roots of the Equation

The roots of the equation can be found by setting \(y = x^{3} - 4x = 0\), which gives \(x = -2, 0, 2\). This shows where the function intersects the x-axis.

Setup the Intervals

Now, the required area is given by the sum of absolute values of the definite integrals over the intervals from \(-2\) to \(0\) and \(0\) to \(2\). That is, \(Area = \int_{-2}^{0}|x^{3}-4x|dx + \int_{0}^{2}|x^{3}-4x|dx\).

Calculate the Area

Perform the definite integrals calculations: \(Area = [\frac{1}{4}x^{4}-x^{2}]_{-2}^{0} + [\frac{1}{4}x^{4}-x^{2}]_{0}^{2}\). Simplify to get \(Area = 4 units^{2}\).