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Edexcel C3,C4 June 2013 Thread

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Reply 4480
Avatar for Aikta_S
Aikta_S
1
Original post by Zaphod77
Think of it in steps: (1) Let u=a function of x (or whatever the letters are). (2) Differentiate to find dx. (3) Sub in u where appropriate and replace the dx with whatever you found after differentiating u. (4) If there are limits, find out what values of u correspond to the values of x and replace them on the integral. Hope this helped :smile:


Ahh awesome! Thanks! :biggrin: really not looking forward till tomorrow :/


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Reply 4481
Avatar for jojo1616
jojo1616
Need help to find exact value, let's say what's the exact value of sin(600°)

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Reply 4482
Avatar for ytop2
ytop2
1
Original post by nm786


that made me want to kill myself
Original post by Aikta_S
Ahh awesome! Thanks! :biggrin: really not looking forward till tomorrow :/


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No problem, did it make sense? Same, but hopefully we'll nail the exam! I hope for a standard paper, so then the grade boundaries won't be really high and I won't be panicking due to difficulty :tongue:
Reply 4484
Avatar for Story
Story
Original post by Zaphod77
Think of it in steps: (1) Let u=a function of x (or whatever the letters are). (2) Differentiate to find dx. (3) Sub in u where appropriate and replace the dx with whatever you found after differentiating u. (4) If there are limits, find out what values of u correspond to the values of x and replace them on the integral. Hope this helped :smile:


So if you had to integrate 1/ 4+ square root (x+1)

with limits 5 and 2.
Where u = square root (x+1)

I get up to 2u/4+u, then take 2 out of the integral to get u/ 4+u, is this just uln(4+u)?
Reply 4485
Avatar for kuku2013
kuku2013
hey quick question if we have a improper fraction like in june 2007 Q4a) 2(4x^2+1)/(2x+1)(2x-1) do you not have divide first for long division and then solve into partial fractions or do you just solve straight away?
Original post by jojo1616
Need help to find exact value, let's say what's the exact value of sin(600°)

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Doesn't your calculator do it for you?
If I enter sin(600) I get -sqroot3/2
Reply 4487
Avatar for nm786
nm786
2
Original post by ytop2
that made me want to kill myself

they're not as bad as they look tbh, definitely try them they're good for tomorrow's exam.
(edited 10 years ago)
Original post by jojo1616
Need help to find exact value, let's say what's the exact value of sin(600°)

Split the 600º into 360º segments. i.e. 600 = 360 + 240. so sin(600)=sin(360+240) = sin(240)

sin(240)=sin(180+60) = -sin(60) = -√3/2

You can use a calculator but this way is also useful.
Original post by Story
So if you had to integrate 1/ 4+ square root (x+1)

with limits 5 and 2.
Where u = square root (x+1)

I get up to 2u/4+u, then take 2 out of the integral to get u/ 4+u, is this just uln(4+u)?

No. You can only use the ln rule when the top of the fraction is the differential of the bottom, or a scalar multiple of it. u is not a scalar as it is not a number, it is a function. To integrate that you'd need to use long division or the methods mentioned earlier to separate it into stuff you can integrate :smile:
Original post by kuku2013
hey quick question if we have a improper fraction like in june 2007 Q4a) 2(4x^2+1)/(2x+1)(2x-1) do you not have divide first for long division and then solve into partial fractions or do you just solve straight away?

You have to divide first :smile:
Reply 4491
Avatar for Jullith
Jullith
Can someone help me with part c of this please:

C4 4.png

I'm not sure how to find the area under the axis. :P
Can someone just teach me the whole syllabus please

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Reply 4493
Avatar for kuku2013
kuku2013
Original post by Zaphod77
You have to divide first :smile:


I thought so. had to ask cause on exam solutions the guy went straight into it!!!
Hey,

could anyone explain how the derivative of ln10(log10x) = 1/x


Thanks :smile: !
Original post by kuku2013
I thought so. had to ask cause on exam solutions the guy went straight into it!!!


Oh dear! But yeah, if the top has an equal power of x or higher then you always have to divide through :smile:
Original post by kuku2013
I thought so. had to ask cause on exam solutions the guy went straight into it!!!


You can do it either way.
I think I'm missing something very simple but

Curve x = 3t^2, y = 2t^3
Line y=2x-8
Show they meet at (3,-2)

I subbed in the parameters to the line equation, but how dyou solve a cubic? :colondollar:

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Reply 4498
Avatar for Aikta_S
Aikta_S
1
Original post by Zaphod77
No problem, did it make sense? Same, but hopefully we'll nail the exam! I hope for a standard paper, so then the grade boundaries won't be really high and I won't be panicking due to difficulty :tongue:


Yeahh true true, hopefully it won't be as bad as c3 or anything so fingers crossed! Vectors are a pain D:


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Original post by Zaphod77
Oh dear! But yeah, if the top has an equal power of x or higher then you always have to divide through :smile:


You don't need to divide through for equal powers of x.

The only difference here would be that, instead of splitting it into:

A2x+1+B2x−1\dfrac{A}{2x+1} + \dfrac{B}{2x-1}

You'd split it into:

A2x+1+B2x−1+C(2x+1)(2x−1)\dfrac{A}{2x+1} + \dfrac{B}{2x-1} + \dfrac{C}{(2x+1)(2x-1)}

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