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Thanks a lot!

Yea that's exactly what I thought, so I thought it'd just become \int_{1}^{5}f(x)dx and I have no idea how to do that

yeah but the 2 is included in the function - it's not outside it

How?

$Given $\int_{0}^{4}f(x)dx = 6$, find $\int_{3}^{7}f[(x-3)+2]dx This was a multiple choice question in my trials I just did, anyone know how to do this? I thought it would be 8, but I guessed 18

shifts to the right*

i) Differentiate, sub 2ap into x to get the gradient of the tangent (should end up being p). sub into point-gradient formula and sub x = 0 and y = to get the coordinates of B and A respectively, then use midpoint formula for these two points to get M ii) using the x and y coordinates of M, find...

Most schools make a separate class for 4U, so they do the 3U and 4U together. However at my school they kept the 4U students in their 3U classes so technically they do learn 2U stuff, but it's not really needed at all for 4U. And I'm not completely sure why it becomes 4units

Actually it does, you don't. And it doesn't get tested.

Scan the question up. not all of us have the text book Or you can type the question up http://community.boredofstudies.org/showthread.php?t=234261 start with [tex*] and end with [/tex*] without the *

^

No you don't.

Those are not the same.

Well since it was an HSC English marker I figured it would only be appropriate.

Ah ok, thanks guys :) So the marks you receive the day before your atar is your scaled marks, correct?

*practising.

I'm a bit confused

Pretty much this. And if you do 4U maths, you don't do the 2U course

No, I don't think so

:eek: