The value of $\int_{-1}^{1} |x^3 - x| dx$ is

To evaluate $\int_{-1}^{1} |x^3 - x| dx$, we need to determine where $x^3 - x$ is positive or negative.

Factor the expression: $x^3 - x = x(x^2 - 1) = x(x-1)(x+1)$

The expression equals zero at $x = -1, 0, 1$.

Testing the intervals:

• For $x \in [-1, 0]$, testing $x = -0.5$: $(-0.5)(-1.5)(0.5) = 0.375 > 0$ (positive)
• For $x \in [0, 1]$, testing $x = 0.5$: $(0.5)(-0.5)(1.5) = -0.375 < 0$ (negative)

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Based on the sign analysis:

$\int_{-1}^{1} |x^3 - x| dx = \int_{-1}^{0} (x^3 - x) dx + \int_{0}^{1} (x - x^3) dx$

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For the first integral:

$\int_{-1}^{0} (x^3 - x) dx = \left[\frac{x^4}{4} - \frac{x^2}{2}\right]_{-1}^{0}$

$= \left(0 - 0\right) - \left(\frac{1}{4} - \frac{1}{2}\right)$

$= 0 - \left(-\frac{1}{4}\right)$

$= \frac{1}{4}$

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For the second integral:

$\int_{0}^{1} (x - x^3) dx = \left[\frac{x^2}{2} - \frac{x^4}{4}\right]_{0}^{1}$

$= \left(\frac{1}{2} - \frac{1}{4}\right) - (0 - 0)$

$= \frac{1}{4}$

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Therefore:

$\int_{-1}^{1} |x^3 - x| dx = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$