The value of $\int\limits_{-1}^1|x|dx$ is

1

The correct answer is Option (2) → 1

Given integral:

$\displaystyle \int_{-1}^{1} |x|\,dx$

Since $|x| = -x$ for $x < 0$ and $|x| = x$ for $x \ge 0$,

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

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

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

$= \frac{1}{2} + \frac{1}{2} = 1$

Therefore, $\displaystyle \int_{-1}^{1} |x|\,dx = 1$.