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Evaluate ∫|x3 - x| dx for x ∈ [-1,2]

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\(\int\limits_{-1}^2 |x^3 - x|dx \)

\(= \int \limits_{-1}^0(x^3 - x)dx- \int\limits_0^1 (x^3 - x)dx + \int\limits_1^2 (x^3 - x)dx\)

\(= \left[\frac{x^4}4 - \frac{x^2}2\right]_{-1}^0 - \left[\frac{x^4}4 - \frac{x^2}2\right]_{0}^1+ \left[\frac{x^4}4 - \frac{x^2}2\right]_{1}^2\)

\(= \left[(0 - 0) - \left(\frac 14 - \frac 12\right)\right] - \left[\left(\frac 14 - \frac12 \right)-(0 - 0)\right] + \left[(4 - 2) - \left(\frac 14 - \frac 12\right)\right]\)

\(= \frac 14 + \frac 14 + 2+ \frac 14\)

\(=2+\frac 34\)

\(= \frac {11}4\)

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From the integral

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