Value of $\int\limits_{-4}^4|x+2| dx$ is :

20

$I=\int\limits_{-4}^4|x+2| d x$

$|x+2|= \begin{cases}x+2 & x \geq-2~~~since~f(x) = |x+2|=0 \\ -x-2 & x<-2~~~at~x=-2\end{cases}$

so dividing the limit of -4 to 4 to -4 to -2 and -2 to 4

$I=\int\limits_{-4}^{-2}-x-2 d x+\int\limits_{-2}^4 x+2 d x$

$=\left[-\frac{x^2}{2}-2 x\right]_{-4}^{-2}+\left[\frac{x^2}{2}+2 x\right]_{-2}^4$

$=\left(-\frac{(-2)^2}{2}-2(-2)+\frac{(-4)^2}{2}+2(-4)\right)+\left(\frac{(4)^2}{2}+2(4)-\frac{(-2)^2}{2}-2(-2)\right.$

$=\left(-\frac{4}{2}+4+\frac{16}{2}-8\right)+\left(\frac{16}{2}+\frac{8}{2}-\frac{4}{2}+4\right)$

$=(-2+4+8-8)+(8+8-2+4)$

= 2 + 18

= 20