How do I evaluate the integral using substitution (Calculus 2)?

Starting with the problem

Integral from 0 to 1 of x*sqrt(1-x)

You're right to make u= 1-x, thus du=-x

We have x on the outside, so we multiply by -1 to get it right? BUT dont forget that since you cant just multiply by something and expect the same answer, we balance it out by dividing by -1 on the outside. (since it's 1, you COULD say that you multiply by -1 as well, but it's more accurate to say divide. The idea is that you multiply by 1 (2/2 or 3/3 or 4/4 etc))

So at this point we have
-(Integral from 0 to 1 of du*(u)^(1/2))
where u=1-x du=-x

Now we take the integral
u^(1/2)*du
-((2u^(3/2))/3)

plug in your points etc 0 and 1 to get the answer.

The only thing you can ever bring out of an integral would be a constant that's being multiplied by something.