VCE Methods Question Thread!

[paper-back]

When integrating a curve in the 3rd quadrant, under the x axis, how do I set up my definite integral? If its from x= -2 to -1, which value goes on top? Is -1 on top and -2 on the bottom correct? And do I need a negative sign outside the whole definite integral because its below the x axis?


Thanks 

Yes, the 'higher number' (sorry for the bad wording) goes on the upper terminal
Yeah, If you are finding the area under the x-axis then you put a negative around the integral

[Kaleidoscope]

When integrating a curve in the 3rd quadrant, under the x axis, how do I set up my definite integral? If its from x= -2 to -1, which value goes on top? Is -1 on top and -2 on the bottom correct? And do I need a negative sign outside the whole definite integral because its below the x axis?


Thanks 

Pretty sure you either reverse the limits or make the integral negative, not both. So either the integral from -1 to -2, or the negative integral of -2 to -1. Both give the same answer.

[Reus]

What's a quick way to learn approximation ?

[speedy]

What's a quick way to learn approximation ?

It's on the formula sheet, so no need to memorise.
Maybe just watch a video on it: this seems quite good https://www.youtube.com/watch?v=eWFAogYI5Ls

[Reus]

It's on the formula sheet, so no need to memorise.
Maybe just watch a video on it: this seems quite good https://www.youtube.com/watch?v=eWFAogYI5Ls

Thanks heaps!

[rosieee]

Could someone please explain to me VCAA exam 2 2013, the one about dilations (X'P'= 10 and X'C"=30)?
Why do they minus the coordinates?

[swagsxcboi]

in tech free, when we find the derivative of a function, should we factorise it?
I was always told to, but vcaa assessment report doesn't do it.

[IndefatigableLover]

in tech free, when we find the derivative of a function, should we factorise it?
I was always told to, but vcaa assessment report doesn't do it.

Not sure how they take factorising of derivatives but I know for substitution (so Question 2 of Exam 1), they would like it to be fully factorised... take VCAA 2009, Question 2 for example

^But yeah for me I've always been factorising them out anyway and I'll be continuing to do it for tomorrow's exam anyway LOL

[spectroscopy]

hey guys how would i anti diff (3-4x)^-5

[Robert123]

hey guys how would i anti diff (3-4x)^-5

Chain rule
let u=2-4x
du/dx= -4
y=u^-5
dy/du= -5u^-6
dy/dx=dy/du*du/dx
=20(2-4x)^-6

[Kaleidoscope]

Could someone explain how these two are equal? Thanks.

[IndefatigableLover]

hey guys how would i anti diff (3-4x)^-5

I do it a bit differently but let's see if you can see where I'm coming from!

So you know when you anti-diff it, the power will go up (so it'll be to the power of '-4').

Now you need to somehow find the number that will be put in front of (3-4x)^-4 (since the inside will stay the same).. what I do is I differentiate it but I write out what I know and chuck in a variable and solve for it.

So you know when you differentiate y.(3-4x)^-4, it'll give you (3-4x)^-5 (so you'll be solving for 'y' in this case)!

Basically when you differentiate that, you know that it'll be -4 x -4 x y = 1
Solving that will give you y=1/16 which is the number you put in front of your overall answer so in the end your answer will be:



I know it's not conventional but hopefully you see where I'm coming from!

[swagsxcboi]

Chain rule
let u=2-4x
du/dx= -4
y=u^-5
dy/du= -5u^-6
dy/dx=dy/du*du/dx
=20(2-4x)^-6

that's not anti diff

hey guys how would i anti diff (3-4x)^-5

raise the power (3-4x)^-4
multiply by 1 over derivative of inside and new power

final answer is 1/16 times (3-4x)^-4

[swagsxcboi]

[swagsxcboi]

Not sure how they take factorising of derivatives but I know for substitution (so Question 2 of Exam 1), they would like it to be fully factorised... take VCAA 2009, Question 2 for example

^But yeah for me I've always been factorising them out anyway and I'll be continuing to do it for tomorrow's exam anyway LOL

haha thanks, i'm gonna stick with factorising too! :p