The value of the integral I=∫_-1^1 |x| dx is:

The correct answer is Option (2) → 1

Since |x| = −x for x < 0 and |x| = x for x ≥ 0, split the integral at 0:

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

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

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

Shortcut: |x| is an even function, so \(I = 2\int_{0}^{1}x\,dx = 2\times\frac{1}{2} = 1\). Geometrically, it is two triangles of area \(\frac{1}{2}\) each.