VCE Specialist 3/4 Question Thread!

[Alwin]

Thanks for the help! but how do you complete the square in that case? :S

yeah, what nliu1995 said. I prefer this method, tho it doesn't work all the time:



Now, Ill look at the . How can I get it from a perfect square? The easiest way is to first get



Okay, so we had






as required.

Clearly, if the question had been say , you would have ended up with some ugly looking numerator, not the perfect square in this case.

Hope it makes sense!

EDIT: Didn't read posting rules.. my mistake.

[LOLs99]

Can u use substitution method or partial fraction to solve the anti derivative of (x-1)/(x+5)? I got slightly different answer

[e^1]

You can simply divide that term to get a simplified answer.

Spoiler

You know what to do next

[LOLs99]

Yep. I know how to do that. Just I found it weird both methods work but different ans. Probably different constant ? Method 1 : partial fractions then anti diff. Method 2: let u= x+5 then substitution method

[BubbleWrapMan]

Yeah, it's a different constant, remember that (where ).

[LOLs99]

Oh ok that clarifies my ques. Thanks Timmeh

[JieSun92]

Need help with Essentials Chapter 2 Extended Response Question 5d).

[dim_sim]

Hey, can someone help me out with 2a?
I'm really having trouble understanding how finding the area between a curve and the y-axis works and when to use it, which is really stressing me out! :/

[lzxnl]

Well ideally you'd integrate with respect to x. Unfortunately VCE decided that integrating by parts was too complicated. Sigh.

If you draw the curve of y = arccos x, you'll find that the area bounded by the curve, the lines x=1 and x=sqrt3 /2 PLUS the area bounded by the curves y=pi/6, x=1, y=arccos x and the x axis sum to the area of the rectangle bounded by the x-axis, x=1, x=sqrt3 /2, y=pi/6, which has an area of pi/12*(2- sqrt 3)
Now for the second area. If we just swap x and y around, we are now looking for area bound between the curves x=pi/6, the y axis, y=cos x and y=1. y=1 is above the curve, so we integrate 1-cos x from 0 to pi/6, which yields pi/6-1/2
Subtract this from the area of the rectangle and you're done!

[Alwin]

Well ideally you'd integrate with respect to x. Unfortunately VCE decided that integrating by parts was too complicated. Sigh.

yeah, indeed it was just too hard VCAA can only spoon feed maths kids, oh diff this function first and maybe surprisingly you'll get the right answer for the integration in the next part..

But gee, could anyone possibly help me with this question,
     Edit: Sorry, don't get stressed, this is not on the course. Just reminded of it when I saw integration by parts. Off Topic.

I integrated by parts the first time, did it a second time (for the x tan(x) i got), then even went for a third time but kept on going round and round in circles

Even gave it to my UMEP teacher, he integrated once by parts then gave up. Spesh teacher gave one look at it and said it's been too long since I integrated by parts

i know at least one of you out there knows how to do it

[stolenclay]

Find the exact value of:

Another method would be

This avoids all the manipulation of the the other fraction, and it looks nicer (I think).

[b^3]

But gee, could anyone possibly help me with this question,

Pretty sure that can't be done with elementary methods. And the answer uses the polylogarithm function. Also since this is the spesh thread... could you not ask it in the spesh thread. We don't want people freaking out thinking that this is on the course and getting confused e.t.c Start a new thread in the mathematics board asking it next time around.

[Alwin]

Pretty sure that can't be done with elementary methods. And the answer uses the polylogarithm function. Also since this is the spesh thread... could you not ask it in the spesh thread. We don't want people freaking out thinking that this is on the course and getting confused e.t.c Start a new thread in the mathematics board asking it next time around.

Point taken, it was off topic sorry. I got excited when I saw integration by parts. Also nliu1995 answered it (pm) since it was he that actually gave it to me haha. thnx for your input tho

[dim_sim]

Well ideally you'd integrate with respect to x. Unfortunately VCE decided that integrating by parts was too complicated. Sigh.

If you draw the curve of y = arccos x, you'll find that the area bounded by the curve, the lines x=1 and x=sqrt3 /2 PLUS the area bounded by the curves y=pi/6, x=1, y=arccos x and the x axis sum to the area of the rectangle bounded by the x-axis, x=1, x=sqrt3 /2, y=pi/6, which has an area of pi/12*(2- sqrt 3)
Now for the second area. If we just swap x and y around, we are now looking for area bound between the curves x=pi/6, the y axis, y=cos x and y=1. y=1 is above the curve, so we integrate 1-cos x from 0 to pi/6, which yields pi/6-1/2
Subtract this from the area of the rectangle and you're done!

I had an epiphany while lying in bed last night and managed to figure it out myself haha.
But thanks! Really appreciate the help as always.

[sin0001]

Need help with part b of the attached question!