The value of ∫_-1^1|x^4-x| dx is equal to:

The correct answer is Option (4) → 1

Since \( x^4-x=x(x^3-1) \): on [−1, 0] both factors are ≤ 0, so \( x^4-x\ge 0 \); on [0, 1], x ≥ 0 and \( x^3-1\le 0 \), so \( x^4-x\le 0 \).

    \( \displaystyle \int_{-1}^{1}|x^4-x|\,dx=\int_{-1}^{0}(x^4-x)\,dx+\int_{0}^{1}(x-x^4)\,dx \)

    \( \displaystyle =\left(\frac{1}{5}+\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{5}\right) \)

    \( \displaystyle =\frac{7}{10}+\frac{3}{10} \)

    \( \displaystyle =1 \)