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Question
By using the properties of the definite integral, evaluate the integral:
`int_0^4 |x - 1| dx`
Solution
`int_0^4 abs (x - 1) dx`
Define,
`abs(x - 1) = {(-(x-1), if x-1<0, or x < 1),(x-1, if x - 1>=0, or x>=1):}`
`int_0^1 abs (x - 1) dx + int_1^4 abs(x - 1) dx`
`int_0^1 - (x - 1) "dx" + int_1^4 (x - 1) dx`
`= - [x^2/2 - x]_0^1 + [x^2/2 - x]_1^4`
`= [(1/2 - 1) - 0] + (16/2 - 4) - (1/2 - 1)`
`= 1/2 + 4 + 1/2`
`= (1 + 8 + 1)/2`
= 5
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