Question

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

Answer 1

\[u=\sqrt{3-2x}\implies du=-\frac{dx}{\sqrt{3-2x}}=-\frac{dx}u\]therefor\[dx=-udu\]so then \[\sqrt{3-2x}dx=u\cdot(-udu)=-u^2du\]

Answer 2

thanks!

Answer 3

welcome :)