help with splitting absolute integral
\[\int\limits_{-2}^{5}(|x-3|-4)\] do i split it into\[\int\limits_{-2}^{3}(|x-3|-4)\]+\[\int\limits_{3}^{5}(|x-3|-4)\]
Well you find out where the part in the absolute value is 0, because that's your turning point. It's only after and before this point that the absolute value really matters, so as a simple example, you have: |dw:1391973515138:dw| But just consider each of these as being two normal, not absolute value graphs before and after, the only difference being that one has a slope that's negative of the other. So you just split it at that point and you solve those integrals of normal lines just like you would any other line.
i know how to sgo about splitting absolute value integrals. it's just i have never gotten one in the above form(with an absolute and another value outside), only this form \[\int\limits_{-2}^{5}(|x-3|)\]
nevermind, i've solved it
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