${\int }_{-5}^5 |\mathrm{x}+2|\mathrm{dx}$ is equal to

Solution:

Here, the given integrand is in the form of absolute function and we define the absolute function as $\left|\mathrm{x}\right|=\mathrm{x},\mathrm{x}\ge 0$
or $\left|\mathrm{x}\right|=-\mathrm{x},\mathrm{x}<0$ by using it, we convert the given integrand in simple form and then integrate it.
Let, $\mathrm{I}={\int }_{-5}^5 |\mathrm{x}+2|\mathrm{dx}$

It can be seen that $(\mathrm{x}+2)\le 0\text{ on }[-5,-2]\text{ and }(\mathrm{x}+2)\ge 0$ on [-2,5]

$\begin{array}{r}\begin{array}{rl}\therefore \mathrm{I}& ={\int }_{-5}^{-2} -(\mathrm{x}+2)\mathrm{dx}+{\int }_{-2}^5 (\mathrm{x}+2)\mathrm{dx}\\ & \left[\because {\int }_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_{\mathrm{a}}^{\mathrm{c}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+{\int }_{\mathrm{c}}^{\mathrm{b}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\right]\end{array}\\ \Rightarrow \mathrm{I}=-{\left[\frac{{\mathrm{x}}^2}{2}+2\mathrm{x}\right]}_{-5}^{-2}+{\left[\frac{{\mathrm{x}}^2}{2}+2\mathrm{x}\right]}_{-2}^5\end{array}$

$\begin{array}{l}=-\left[\frac{(-2{)}^2}{2}+2(-2)-\frac{(-5{)}^2}{2}-2(-5)\right]+\left[\frac{(5{)}^2}{2}+2\left(5\right)-\frac{(-2{)}^2}{2}-2(-2)\right]\\ \\ =-\left[2-4-\frac{25}{2}+10\right]+\left[\frac{25}{2}+10-2+4\right]\\ =-2+4+\frac{25}{2}-10+\frac{25}{2}+10-2+4=29\end{array}$