Question #f7ed1

Mar 13, 2017

${\int}_{- 1}^{1} \left\mid x \right\mid \mathrm{dx} = 1$

Explanation:

$f \left(x\right)$ is a piecewise function.

$f \left(x\right) = \left\{\begin{matrix}x \mathmr{if} x \ge 0 \\ - x \mathmr{if} x < 0\end{matrix}\right]$

Hence, separate the integral into 2 parts.

${\int}_{- 1}^{1} f \left(x\right) \mathrm{dx} = {\int}_{- 1}^{0} f \left(x\right) \mathrm{dx} + {\int}_{0}^{1} f \left(x\right) \mathrm{dx}$

$= {\int}_{- 1}^{0} \left(- x\right) \mathrm{dx} + {\int}_{0}^{1} x \mathrm{dx}$

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

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

$= 1$