Sketch the graph of `y = |x + 3|`and evaluate`int-6 0|x+3|dx`.

Explanation

`y=|x+3|={(-(x+3)", "xlt-3),(" "x+3", "xge -3):}`
When `x lt -3, y= -x -3`
`{:(x,-4,5,6),(y,1,2,3):}`
When `x ge -3`
`{:(x,-1,-2,-3),(y,2,1,0):}`

Plot these on the graph and join them
` :. ` Required area
`=ar(ABC)+ar(OAD)`
`=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)`
`=[((-(-3)^(2))/(2)-3xx(-3))-(-((-6)^(2))/(2)+3xx(-6))]+[0-(-((-3)^(2))/(2)+3xx(-3))]`
`=[((-9)/(2)+9)-(-18+18)]+[(9)/(2)]`
`=(9)/(2)+(9)/(2)`
`=9` sq. units.