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

The absolute value function $|x|$ is defined as:

$|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$

Since $|x|$ changes its definition at $x = 0$, the integral must be split at this point.

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$\int_{-1}^{1}|x|dx = \int_{-1}^{0}|x|dx + \int_{0}^{1}|x|dx$

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For $x \in [-1, 0]$, we have $x \leq 0$, so $|x| = -x$

For $x \in [0, 1]$, we have $x \geq 0$, so $|x| = x$

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

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$\int_{-1}^{0}(-x)dx = \left[-\frac{x^2}{2}\right]_{-1}^{0}$

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

$= 0 + \frac{1}{2}$

$= \frac{1}{2}$

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

$= \frac{(1)^2}{2} - \frac{(0)^2}{2}$

$= \frac{1}{2} - 0$

$= \frac{1}{2}$

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$\int_{-1}^{1}|x|dx = \frac{1}{2} + \frac{1}{2}$

$= 1$

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