# Solving integrals with absolute values

1. Aug 12, 2011

### Shannabel

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??

2. Aug 12, 2011

### I like Serena

Hi again Shannabel!

Can you recalculate (3x-(1/2)x^2)(between -1 and 0)?

3. Aug 12, 2011

### Shannabel

thankyou!! :)

4. Aug 12, 2011

### I like Serena

I take it you did? And that you found the proper solution?

Then you're welcome!