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Jlastino
Can someone explain to me why is there a "u^2" ? Here's the solution in wolfram alpha http://www.wolframalpha.com/input/?i=integrate+sqrt+%283-2x%29+x%5E2 I understand how it used substitution to get (3-u^2)^2 and the 1/4 but i can't seem to know where the u^2 came from...help T_T
\[u=\sqrt{3-2x}\implies du=-\frac{dx}{\sqrt{3-2x}}=-\frac{dx}u\]therefor\[dx=-udu\]so then \[\sqrt{3-2x}dx=u\cdot(-udu)=-u^2du\]