Post

So this week basically all I did was go over the tests and do the problems... most of which I already knew... so there's not much for me to post this week that comes to my head...

I'll just explain something I guess...

I find a lot of people still are having issues with integration...like, in particular...change of variable (I think it's called this anyway).

Basically this happens whenever you set your u equal to something and when you take the du, you don't have enough x's that you need...Not sure how to explain it other wise...

Basically let's say the problem was...

x sqrt(x+1)

To integrate this, you would set your u equal to x+1...so
u=x+1
du=1 dx

Now, usually the du will have an x in it because the original problem had an x...but this one doesn't. So what we do, is solve the "u=x+1" for x...so we get that x=u-1.

Now we can substitute back in.

(u-1) (for x) times sqrt(u).

So this is (u-1)(u^(1/2)). Now you can distribute the u^(1/2) into the u-1 and then integrate powers like normal...

Once you learn this change of variable, a lot of problems become way easier.

Another thing...

Let's say the problem says h(x) = f^-1(x) or the inverse of f(x). It then gives you f(x) and asks you to find h'(x)...the formula for this is

h'(x) = 1/(f'(f^-1(x)))

And if you don't know how to do f^-1(x) you should probably look that up from advanced math...and while speaking of that, make sure to look over your trig identities from there as well.

Other than that, you people are starting to learn more :-P. Keep up the good work.