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Question
Sketch the graph of y = |x + 3| and evaluate `int_(-6)^0 |x + 3|dx`
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Solution
y = |x + 3|
At x = -3, y = 0
AQ is the line y = x + 3
When x + 3 < 0,
y = -(x + 3) = -x – 3
The graph of the line AP is AP.
∴ y =|x + 3| is shown in the graph.
`int_(-6)^1 |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`
`= - [x^2/2 + 3x]_(- 6)^(- 3) + [x^2/2 + 3x]_(-3)^0`
`= - [x^2/2 + 3x]_(-6)^(-3) + [x^2/2 + 3x]_(-3)^0`
`= - [(9/2 - 9) - (36/2 - 18) + 0 - (9/2 - 9)]`
`= - [- 9/2 - 0] + 9/2`
`= 9/2 + 9/2`
= 9 square unit
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