What am I doing wrong integrating with 1-|x|?

[zmalone]

I understand \int^{1}_{-1}1-|x|dx = 1 visually just by graphing it and taking the area of the triangle but for the sake of more complicated examples I'm not exactly sure what step I'm messing up when I use the FTOC:

|x|= x when x>0, -x when x<0

\int^{0}_{-1}1-|x|dx + \int^{1}_{0}1-|x|dx


= \int^{0}_{-1}1-(-x)dx + \int^{1}_{0}1-(x)dx


= \frac{x^2}{2}|^{0}_{-1} - \frac{x^2}{2}|^{1}_{0}

= -1/2 - 1/2 = -1 \neq1

Any input is appreciated and hope this makes sense lol (still getting used to the formula drawer).

 

[tiny-tim]

hi zmalone! welcome to pf!

= \int^{0}_{-1}1-(-x)dx + \int^{1}_{0}1-(x)dx


= \frac{x^2}{2}|^{0}_{-1} - \frac{x^2}{2}|^{1}_{0}

you missed out [x]_-1¹, = 2

 

[zmalone]

Wow :( I don't know why I was taking the derivative of 1 instead its anti-derivative but I certainly confused myself by doing so for the past half hour. Thanks for pointing it out!