Find the intergration as 0 approaches 3

You have to integrate two equations, one for the case where the value in the absolute signs is negative and one for the case where the value is positive...

So you will integrate one on the interval 0 to 1 and the other on the interval from 1 to 3...

⌠-x+1 dx, x=[0,1] + ⌠x-1 dx, x=[1,3]

(-x^2/2+x)[0,1] + (x^2/2-x)[1,3]

(-0.5+1)+(4.5-3-0.5+1)

(0.5)+(2)

2.5 u^2

It’s d for sure i just did also that’s why