The value of \(\int_{-1}^{1}\displaystyle |x^3 - x|\,dx\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{1}{2}\)
\(x^3 - x\) is an odd function, so \(|x^3 - x|\) is an even function:
\( \displaystyle \int_{-1}^{1} |x^3 - x|\,dx = 2\int_{0}^{1} |x^3 - x|\,dx \)
On [0, 1], \(x^3 - x = x(x - 1)(x + 1) \le 0\), so \(|x^3 - x| = x - x^3\):
\( \displaystyle = 2\int_{0}^{1} (x - x^3)\,dx \)
\( \displaystyle = 2\left[\frac{x^2}{2} - \frac{x^4}{4}\right]_{0}^{1} \)
\( \displaystyle = 2\left(\frac{1}{2} - \frac{1}{4}\right) = \frac{1}{2} \)
Why the other options are incorrect
Option (1) 0 — Incorrect. 0 is the value of \(\int_{-1}^{1} (x^3 - x)\,dx\) without the modulus (an odd function); with the modulus the integrand is never negative, so the integral is positive.
Option (2) \(\displaystyle\frac{1}{4}\) — Incorrect. This is the integral over [0, 1] only; the part over [−1, 0] adds another \(\frac{1}{4}\).
Option (4) 1 — Incorrect. The value is \(\frac{1}{2}\), not 1.