Solution:
Given eq. is $y=|x+3|$
Corresponding values of $x$ and $y$ are given in the following table.
$$\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline$x$ & $-6$ & $-5$ & $-4$ & $-3$ & $-2$ & $-1$ & 0 \\
\hline$y$ & 3 & 2 & 1 & 0 & 1 & 2 & 3 \\
\hline
\end{tabular}$$
On plotting these points, we get the graph of $y=|x+3|$ as follows

We know that, $(x+3) \leq 0$ for $-6 \leq x \leq-3$ and $(x+3) \geq 0$ for $-3 \leq x \leq 0$
$\begin{aligned}
\therefore \int_{-6}^{0}|(x+3)| d x &=-\int_{-6}^{-3}(x+3) d x+\int_{-3}^{0}(x+3) d x \\
&=-\left[\frac{x^{2}}{2}+3 x\right]_{-6}^{-3}+\left[\frac{x^{2}}{2}+3 x\right]_{-3}^{0} \\
&\left.=-\left[\frac{(-3)^{2}}{2}+3(-3)\right)-\left(\frac{(-6)^{2}}{2}+3(-6)\right)\right]+\left[0-\left(\frac{(-3)^{2}}{2}+3(-3)\right)\right] \\
&=-\left[-\frac{9}{2}\right]-\left[-\frac{9}{2}\right] \\
&=9 sq. units
\end{aligned}$