Can you please show me the steps to integrate: ∫(x/2-x) dx ?? i let u = (2-x) and then?? Help is very appreciated! thanks!!
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Originally Posted by Yehia Can you please show me the steps to integrate: ∫(x/2-x) dx ?? i let u = (2-x) and then?? Then differentiate $\displaystyle u=2-x$ to get $\displaystyle du=-dx,$ besides $\displaystyle x=2-u.$ Can you take it?
Originally Posted by Krizalid Then differentiate $\displaystyle u=2-x$ to get $\displaystyle du=-dx,$ besides $\displaystyle x=2-u.$ Can you take it? I do do that! to get: ∫ (u-2)/u du and that simplifies to : ∫ 1 - 2u^-1 du still stumped though :S
Well done, you almost got it. (Oh, the numerador is actually $\displaystyle 2-u.$) Next step is to note that $\displaystyle \int du=u+k$ and $\displaystyle \int\frac{du}u=\ln|u|+m.$
Originally Posted by Krizalid Well done, you almost got it. (Oh, the numerador is actually $\displaystyle 2-u.$) Next step is to note that $\displaystyle \int du=u+k$ and $\displaystyle \int\frac{du}u=\ln|u|+m.$ But the numerator is MINUS x isnt it? and if x = 2 - u, then -x = u - 2 ? if im not wrong? And i dont understand your second bit (which is right btw, according to the book) can you explain? thanks a lot
Ah yes, you're right, I forgot the minus sign when differentiating. The latter are basic formulae which you should know, I can't help you any further about that, so check your notes.
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