Question

${\int }_{-2}^2\left(x-\left|x\right|\right)dx=$

• A. $0$
• B. $2$
• C. $4$
• D. $-4$

Answer 1

The correct option is D

$-4$

Explanation for the correct option:

Finding the value for the given integral:

Let $I={\int }_{-2}^2\left(x-\left|x\right|\right)dx$

$={\int }_{-2}^{2}xdx-{\int }_{-2}^2\left|x\right|dx$

$={\int }_{-2}^{2}xdx-\left[{\int }_{-2}^0\left(-x\right)dx+{\int }_0^2\left(x\right)dx\right]$ $\left[\mathrm{since}{\int }_{-\mathrm{a}}^{\mathrm{a}}\left|\mathrm{x}\right|\mathrm{dx}={\int }_{-\mathrm{a}}^0\left(-\mathrm{x}\right)\mathrm{dx}+{\int }_0^{\mathrm{a}}\left(\mathrm{x}\right)\mathrm{dx}\right]$

$={\left[\frac{x^2}{2}\right]}_{-2}^2+{\int }_{-2}^{0}xdx-{\int }_0^{2}xdx$

$$\begin{gathered} =\left[\frac{{\left(2\right)}^2}{2}-\frac{{\left(-2\right)}^2}{2}\right]+{\left[\frac{x^2}{2}\right]}_{-2}^0-{\left[\frac{x^2}{2}\right]}_0^2 \\ =0+\left[\frac{0-{\left(-2\right)}^2}{2}\right]-\left[\frac{{\left(2\right)}^2-0}{2}\right] \\ =0-2-2 \\ =-4 \end{gathered}$$

Thus, option D is the correct answer