The value of ∫_-1^1 |x| dx is

The correct answer is Option (2) → 1

Since \(|x| = -x\) for \(x \lt 0\) and \(|x| = x\) for \(x \ge 0\), the integral is split at 0:

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

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

    \( \displaystyle = \frac{1}{2} + \frac{1}{2} = 1 \)

Shortcut: The graph of \(y = |x|\) on [−1, 1] encloses two right triangles with the x-axis, each with base 1 and height 1:

    \( \displaystyle 2 \times \frac{1}{2} \times 1 \times 1 = 1 \)

Why the other options are incorrect

• Option (1) 0 — Incorrect. 0 is \(\int_{-1}^{1}x\,dx\); without the modulus the two halves cancel.

• Option (3) \(\displaystyle\frac{1}{2}\) — Incorrect. \(\frac{1}{2}\) is the integral over [0, 1] only.

• Option (4) 2 — Incorrect. Each half contributes \(\frac{1}{2}\), so the total is 1.