Sketch the graph `y=|x+3|dot` Evaluate `int_(-6)^0|x+3|dxdot` What does the value of this integral represent on the graph?

Explanation

The graph of `y=∣x+3∣` is as shown in the figure. And the shaded region is the required region.
Required area `= int_-6^0 ∣x+3∣ dx`
=` int_-6^-3 ∣x+3∣ dx + int_-3^0 ∣x+3∣ dx`
= `- int_-6^-3 (x+3) dx + int_-3^0 (x+3) dx`
`= [x^2/(2) +3x]_-6^-3 + [ x^2/(2) +3x]_-3^0`
by solving the limits , we get
= `9/(2)` + `9/(2)` = `9` sq. units