onur369's Methods Question Thread :)

[pi]

I'm stuck with an integral question, only problem here is how to deal with the denominator :/
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Remember the ax+b special rules? Try looking those up

(don't have text on me, and I forget )

[onur369]

Ahhh, I see. 1/ax+b, Thanks!

[onur369]

Sketch the graph of y=1-x^2, showing all the intercepts. Find the area between the curve and the x axis from
a) x=0 to x=1 , I did this and I got 2/3, not 100% sure.
What I am stuck on is the next two
b) x=1 to x=2
c) x=0 to x=2
^ I am not exactly sure what to do ? Do I draw a new line ?

[tony3272]

Sketch the graph of y=1-x^2, showing all the intercepts. Find the area between the curve and the x axis from
a) x=0 to x=1 , I did this and I got 2/3, not 100% sure.
What I am stuck on is the next two
b) x=1 to x=2
c) x=0 to x=2
^ I am not exactly sure what to do ? Do I draw a new line ?

Your answer is correct for part a. For part b), you just do the same method as part a, but just set your terminals as 1 and 2 rather than 0 and 1.

[enwiabe]

Sketch the graph of y=1-x^2, showing all the intercepts. Find the area between the curve and the x axis from
a) x=0 to x=1 , I did this and I got 2/3, not 100% sure.
What I am stuck on is the next two
b) x=1 to x=2
c) x=0 to x=2
^ I am not exactly sure what to do ? Do I draw a new line ?

Your answer is correct for part a. For part b), you just do the same method as part a, but just set your terminals as 1 and 2 rather than 0 and 1.

Ah, but it wants the area, not a simple evaluation of the integral. Think about where an integral becomes negative and how you have to split it into parts and take the absolute (positive) of certain quantities. is positive between -1 and 1 and becomes negative everywhere else. I hope that's enough of a hint!

[onur369]

Thanks Enwiabe, posting this at 5am WHAT are you doing at that time ? lol, shows the effort you put in the site.

[onur369]

VCAA 2006 EXAM1 Q11:

[luken93]

You know the equation, left endpoint (0) and right endpoint (3) and that the area = 45
hence integrate f(x) from 0->3 and let it equal 45 - this will allow you to determine a

From there, it's simply a matter of finding x ints to determine n and m.

[tony3272]

VCAA 2006 EXAM1 Q11:
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lets you solve for a




therefore the equation is
therefore
therefore m=6, n=-2 by the null factor law

[onur369]

I didn't want to open a new thread so I thought Id just ask here. For the current and past students: What exam is the hardest you did, EXAM1&2. I need the most challenging ones since I always come out sexually abused from the SACS.

[luken93]

Kilbaha, any year.

[tony3272]

From what i've generally heard, Itute and Kilbaha are usually quite challenging, (i've only done VCAA ones so far so i cant really be 100% sure.)

[xZero]

neap(kinda)/kilbaha/tsfx/2010 vcaa

[onur369]

Im stuck with a VCAA 2010 question.



I can do a but not b :/

[enwiabe]

It's integration by recognition.

You get



You can now rearrange to arrive at



They now want you to find the anti-derivative of . Are you okay to proceed from here? If you need a further hint let me know.