(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

solve the integral [abs(x+1)(3+abs(x))]/(x+1) between -3 and 1

2. Relevant equations

3. The attempt at a solution

when x<-1 then [abs(x+1)(3+abs(x))]/(x+1) = [-(x+1)(3-x)]/(x+1) = -(3-x)

when -1<x<0 then [abs(x+1)(3+abs(x))]/(x+1) = (x+1)(3-x)/(x+1) = 3-x

when x>0 then [abs(x+1)(3+abs(x))]/(x+1) = (x+1)(3+x)/(x+1) = 3+x

so now i have:

(x-3)dx(between -3 and -1)+(3-x)dx(between -1 and 0)+(3+x)dx(between 0 and 1)

= ((1/2)x^2-3x)(between -3 and -1)+(3x-(1/2)x^2)(between -1 and 0)+(3x+(1/2)x^2)(between 0 and 1)

= (1/2)+3-((9/2)+9)+(0)-(-3-(9/2))+(3+(1/2))-0

= 1/2+3-9+3+3+1/2 = 1

i should have got -3... help??

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# Solving integrals with absolute values

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