# What is the integral from 0 to 5 of (x^2-5x+6)/abs(x-2)?How would i go about solving this? Thanks

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### 1 Answer

yes exactly, let us use the property of the absolute value function,

`int_0^5(x-2)(x-3)/ |x-2|dx`

we see that |x-2| = -(x-2) when x-2< 0 that is x<2 and

= (x+2) when x-2>0 that is x>2

now the integral can be seperated as,

`int_0^5(x-2)(x-3) / |x-2|dx` = `int_0^(2-)(x-2)(x-3) / -(x-2)dx` + `int_(2+)^5(x-2)(x-3)/(x-2)dx`

= `int_0^(2-)-(x-3)dx` +`int_(2+)^5(x-3)dx`

Integrating and applying the limits,

= -(x^2/2 - 3x) 0 to 2 + (x^2/2 - 3x) 2 to 5

= -(2^2/2 - 3*2) +0 + (5^2/2 - 3*5) - (2^2/2-3*2)

= -(2-6) +(25/2-15) -(2-6) = -(-4) + (12.5 -15) -(-4)

= 4+(-2.5)+4 = 8-2.5 = 6.5