# Evaluate (definite integral) abs(2x-5)/(x+1)dx from 0 to 6

You need to evaluate the definite integral such that:

`int_0^6 |2x-5|/(x+1)dx`

You need to use the absolute value property such that:

`|2x-5| = {(2x-5, 2x-5gt=0),(5-2x,2x-5lt0):}`

Since the function `y=2x-5`  is negative if `x in [0, 2.5]`  and it is positive if `x in [2.5,6], ` you need to split the integral such that:

`int_0^6 |2x-5|/(x+1)dx = int_0^2.5 (5-2x)/(x+1)dx + int_2.5^6 (2x-5)/(x+1)dx`

Using the property of linearity of integral yields:

`int_0^6 |2x-5|/(x+1)dx = int_0^2.5 5/(x+1)dx - int_0^2.5 (2x)/(x+1)dx + int_2.5^6 (2x)/(x+1)dx - int_2.5^6 5/(x+1)dx`

`int_0^6 |2x-5|/(x+1)dx = 5ln(x+1)|_0^(2.5) - 2int_0^2.5 (x+1-1)/(x+1)dx + 2int_2.5^6 (x+1-1)/(x+1)dx - 5ln(x+1)|_(2.5)^6`

`int_0^6 |2x-5|/(x+1)dx = 5(ln 3.5 - ln 1) - 2int_0^2.5 dx - 2int_0^2.5 1/(x+1)dx + 2int_2.5^6 dx - 2int_2.5^6 1/(x+1)dx - 5(ln 7 - ln 3.5)`

`int_0^6 |2x-5|/(x+1)dx = 5(ln 3.5) - 2(2.5 - 0) - 2(ln 3.5 - ln 1) + 2(6 - 2.5) - 2(ln 7 - ln 3.5) - 5 ln 7 + 5ln 3.5`

`int_0^6 |2x-5|/(x+1)dx = 10ln 3.5 - 5 - 2 ln 3.5 + 12 - 5 - 2 ln7 + 2ln 3.5 - 5 ln 7`

`int_0^6 |2x-5|/(x+1)dx = 10ln 3.5 + 2 - 7 ln 7`

Hence, evaluating the given definite integral yields `int_0^6 |2x-5|/(x+1)dx = 10ln 3.5 + 2 - 7 ln 7.`

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