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Conditions for the applicability of u-substitution
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[QUOTE="justthisonequestion, post: 6071932, member: 652269"] What are the conditions for applicability of u-substitution, [I]i.e. [/I]when does it not work? Note that I'm not asking when is it a bad idea (that won't get you any closer to evaluating the integral), but are there any conditions that cause u-sub to yield wrong answers? I started running into what I think is a case of u-sub not working when I was thinking about integrals of odd functions, ex: $$\int_{-\infty}^{\infty}x e^{-a x^2}dx$$ $$u=x^2$$ $$du=2xdx$$ $$\int_{-b}^{b}x e^{-a x^2}dx=\frac{1}{2}\int_{b^2}^{b^2} e^{-a u}dx=0$$ Where the last part equals zero because now the bounds are equal... But this begged the question... why can't I just u-sub in such a way that the bounds on the integral are always equal, and all integrals go to 0? - Obviously there has to be some constraint on the applicability of u-sub. Here is an example where it seems u-sub just leads to the wrong answer... $$\int_{-2}^{1}x^4dx=\frac{33}{5}=6.6$$ With $$u=x^2$$ we get: $$\frac{1}{2}\int_{4}^{1}u^{3/2}du=-\frac{31}{5}=-6.2$$ So what gives? What basic mathematical principle is being violated here? I'm sure I learned this at some point... waaay back when. [/QUOTE]
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Conditions for the applicability of u-substitution
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