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Question
Evaluate: `int_(-1)^1 |5x - 3| "d"x`
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Solution
Let I = `int_(-1)^1 |5x - 3| "d"x`
|5x − 3| = − (5x − 3) when (5x − 3) < 0 i.e. x < `3/5`
= 5x – 3 when (5x – 3) > 0 i.e., x > `3/5`
∴ I = `int_(-1)^(3/5) |5x - 3| "d"x + int_(3/5)^1|5x - 3| "d"x`
= `int_(-1)^(3/5) -(5x - 3) "d"x + int_(3/5)^1 (5x - ) "d"x`
= `-5int_(-1)^(3/5)x "d"x + 3int_(-1)^(3/5) "d"x + 5 int_(3/5)^1x "d"x - 3int_(3/5)^1 "d"x`
= `-5/2[x^2/2]_(-1)^(3/5) + 3[x]_(-1)^(3/5) + 5[x^2/2]_(3/5)^1 - 3[x]_(3/5)^1`
= `-5/2[(3/5)^2 - (-1)^2] + 3[3/5 - (-1)] + 5/2[(1)^2 - (3/2)^2] - 3(1 -3/5)`
= `5/2(9/25 - 1) + 3(3/5 + 1) + 5/2(1 - 9/25) - 3(2/5)`
= `-5/2((-16)/25) + 3(8/5) + 5/2(16/25) - 6/5`
= `8/5 + 24/5 + 8/5 - 6/5`
= `34/5`
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