The value of ∫limits_-1^1|x^3- x| dx is

The correct answer is Option (3) → $\frac{1}{2}$

Given: $\int_{-1}^{1} |x^3 - x| \, dx$

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

So sign of $x^3 - x$ changes at $x = -1$, $x = 0$, and $x = 1$

Test intervals:

On $(-1, 0)$: pick $x = -0.5 \Rightarrow (-)(-)(+) = +$ ⇒ $x^3 - x > 0$

On $(0, 1)$: pick $x = 0.5 \Rightarrow (+)(-)(+) = -$ ⇒ $x^3 - x < 0$

So:

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

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

$= \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_{-1}^{0} + \left[ -\frac{x^4}{4} + \frac{x^2}{2} \right]_0^1$

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

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